#set document(title: "10.4 Polar Form for Complex Numbers", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 10.4#h(0.6em)Polar Form for Complex Numbers === Introduction In Section 10.3 we represented the sum of two complex numbers graphically as a vector addition. Is there a way to visualize the product or quotient of two complex numbers? One way to explore a new idea is to consider a simple case. What happens to the vector representing a complex number when we multiply the number by #math.equation(block: false, alt: "i")[$i$]? #examplebox("Example 1")[][ Represent #math.equation(block: false, alt: "z equals 2 plus 3 i")[$" " z = 2 + 3 i " "$] and #math.equation(block: false, alt: "i z equals 2 i minus 3")[$" " i z = 2 i − 3 " "$] as vectors in the complex plane. #solutionbox[ The vectors are shown at right. We see that multiplication by #math.equation(block: false, alt: "i")[$i$] corresponds to rotating the vector around the origin by #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] in the counterclockwise direction. #figure(figph[complex numbers as vectors], alt: "complex numbers as vectors", caption: none) ] ] + Let #math.equation(block: false, alt: "z equals 1 plus i ,")[$z = 1 + i ,$] and calculate #math.equation(block: false, alt: "i z , i squared z ,")[$i z , " " i^(2) z ,$] and #math.equation(block: false, alt: "i cubed z")[$i^(3) z$]. + Plot #math.equation(block: false, alt: "z , i z , i squared z ,")[$z , " " i z , " " i^(2) z ,$] and #math.equation(block: false, alt: "i cubed z")[$i^(3) z$] as points on the complex plane. + #math.equation(block: false, alt: "i z equals i minus 1 , i squared z equals minus 1 minus i , i cubed z equals minus i plus 1")[$i z = i − 1 , " " i^(2) z = − 1 − i , " " i^(3) z = − i + 1$] + #figure(figph[rotations of complex number], alt: "rotations of complex number", caption: none) The previous example suggests that multiplication by a complex number results in a rotation. Polar coordinates are well suited to processes that involve rotation, because they use angles to specify location. Thus, we will next represent complex numbers in an alternate polar form. === Polar Form The figure below shows the complex number #math.equation(block: false, alt: "z equals 3 plus 3 i")[$" " z = 3 + 3 i$], represented as a vector in the complex plane. The distance #math.equation(block: false, alt: "r")[$r$] from the origin to #math.equation(block: false, alt: "z")[$z$] is #math.equation(block: true, alt: "r equals the square root of 3 squared plus 3 squared equals 3 the square root of 2")[$r = sqrt(3^(2) + 3^(2)) = 3 sqrt(2)$] and the angle from the real axis to the vector is #math.equation(block: false, alt: "θ equals the fraction π over 4")[$θ = display(frac(π, 4))$]. Using right triangle trigonometry, we see that #math.equation(block: true, alt: "a, equals 3 the square root of 2 cos ⁡ open parenthesis the fraction π over 4 close parenthesis equals 3 the square root of 2 times the fraction 1 over the square root of 2 equals 3; b, equals 3 the square root of 2 sin ⁡ open parenthesis the fraction π over 4 close parenthesis equals 3 the square root of 2 times the fraction 1 over the square root of 2 equals 3")[$a & = 3 sqrt(2) cos ( display(frac(π, 4)) ) = 3 sqrt(2) ⋅ display(frac(1, sqrt(2))) = 3 \ b & = 3 sqrt(2) sin ( display(frac(π, 4)) ) = 3 sqrt(2) ⋅ display(frac(1, sqrt(2))) = 3$] #figure(figph[polar form of complex number], alt: "polar form of complex number", caption: none) In general, we can write the real and imaginary parts of #math.equation(block: false, alt: "z")[$z$] in terms of #math.equation(block: false, alt: "r")[$r$] and #math.equation(block: false, alt: "θ")[$θ$] as #math.equation(block: true, alt: "a equals r cos ⁡ open parenthesis θ close parenthesis and b equals r sin ⁡ open parenthesis θ close parenthesis")[$a = r cos ( θ ) " " " " " " " " "and" " " " " " " " " b = r sin ( θ )$] Thus, the complex number #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] can also be written as #math.equation(block: true, alt: "z equals open parenthesis r cos ⁡ open parenthesis θ close parenthesis close parenthesis plus open parenthesis r sin ⁡ open parenthesis θ close parenthesis close parenthesis i or z equals r open parenthesis cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$z = ( r cos ( θ ) ) + ( r sin ( θ ) ) i " " " " " " " " "or" " " " " " " " " z = r ( cos ( θ ) + i sin ( θ ) )$] . #examplebox("Example 2")[][ Find the polar form for #math.equation(block: false, alt: "z equals the square root of 3 plus i")[$" " z = sqrt(3) + i$]. #solutionbox[ We compute #math.equation(block: true, alt: "r equals the square root of a squared plus b squared equals the square root of 3 plus 1 equals 2 and tan ⁡ open parenthesis θ close parenthesis equals the fraction 1 over the square root of 3")[$r = sqrt(a^(2) + b^(2)) = sqrt(3 + 1) = 2 " " " " " " " " "and" " " " " " " " " tan ( θ ) = display(frac(1, sqrt(3)))$] The angle #math.equation(block: false, alt: "θ")[$θ$] is in the first quadrant, so #math.equation(block: true, alt: "θ equals tan to the power minus 1 ⁡ open parenthesis the fraction 1 over the square root of 3 close parenthesis equals the fraction π over 6")[$θ = tan^(− 1) ( display(frac(1, sqrt(3))) ) = display(frac(π, 6))$] Thus,#math.equation(block: false, alt: "z equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 6 close parenthesis close parenthesis")[$" " " " z = 2 ( cos ( display(frac(π, 6)) ) + i sin ( display(frac(π, 6)) ) )$] #figure(figph[polar form of complex number], alt: "polar form of complex number", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Polar Form for a Complex Number] The complex number #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] can be written in the polar form #math.equation(block: true, alt: "z equals r open parenthesis cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$bold(italic(z)) = bold(italic(r)) ( bold(cos) ( bold(italic(θ)) ) + bold(italic(i)) bold(sin) ( bold(italic(θ)) ) )$] where #math.equation(block: false, alt: "r equals the square root of a squared plus b squared")[$" " " " r = sqrt(a^(2) + b^(2)) " " " "$] and #math.equation(block: false, alt: "θ")[$θ$] is defined by #math.equation(block: true, alt: "a equals r cos ⁡ open parenthesis θ close parenthesis , b equals r sin ⁡ open parenthesis θ close parenthesis , 0 less than or equal to θ less than or equal to 2 π")[$a = r cos ( θ ) , " " " " " " " " b = r sin ( θ ) , " " " " " " " " 0 ≤ θ ≤ 2 π$] The angle #math.equation(block: false, alt: "θ")[$θ$] is called the #strong[argument] of the complex number, and #math.equation(block: false, alt: "r")[$r$] is its length, or #strong[modulus]. ] Find the polar form for #math.equation(block: false, alt: "w equals 1 plus i the square root of 3")[$" " w = 1 + i sqrt(3)$]. #math.equation(block: true, alt: "w equals 2 open parenthesis cos ⁡ the fraction π over 3 plus i sin ⁡ the fraction π over 3 close parenthesis")[$w = 2 ( cos display(frac(π, 3)) + i sin display(frac(π, 3)) )$] Of course, we can always recover the Cartesian form of a complex number from its polar form by evaluating the trigonometric functions. We'll check the result of the previous example: #math.equation(block: true, alt: "z, equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 6 close parenthesis close parenthesis; equals 2 open parenthesis the fraction the square root of 3 over 2 plus i times the fraction 1 over 2 close parenthesis equals the square root of 3 plus i")[$z & = 2 ( cos ( display(frac(π, 6)) ) + i sin ( display(frac(π, 6)) ) ) \ & = 2 ( display(frac(sqrt(3), 2)) + i ⋅ display(frac(1, 2)) ) = sqrt(3) + i$] === Products and Quotients in Polar Form The polar form is especially convenient for computing the product or quotient of two complex numbers. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Product in Polar Form] If #math.equation(block: false, alt: "z sub 1 equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$" " z_(1) = r ( cos ( α ) + i sin ( α ) ) " "$] and #math.equation(block: false, alt: "z sub 2 equals R open parenthesis cos ⁡ open parenthesis β close parenthesis plus i sin ⁡ open parenthesis β close parenthesis close parenthesis")[$" " z_(2) = R ( cos ( β ) + i sin ( β ) ) " "$], then #math.equation(block: true, alt: "z sub 1 z sub 2 equals r R open parenthesis open parenthesis cos ⁡ open parenthesis α plus β close parenthesis plus i sin ⁡ open parenthesis α plus β close parenthesis close parenthesis")[$bold(italic(z))_(1) bold(italic(z))_(2) = bold(italic(r)) bold(italic(R)) \( \( bold(cos) \( bold(italic(α)) + bold(italic(β)) \) + bold(italic(i)) bold(sin) \( bold(italic(α)) + bold(italic(β)) \) \)$] ] This formula, which you will prove in the Homework Problems, says that the product of two complex numbers in polar form is the complex number with modulus #math.equation(block: false, alt: "r R")[$r R$] and argument #math.equation(block: false, alt: "α plus β")[$α + β$]. Thus, to find the product of two complex numbers, we multiply their lengths and add their arguments. The product rule this sentence states, drawn in the complex plane. The gray vector is the fixed multiplier w = 2(cos 60° + i sin 60°); the blue vector is z, with its modulus r and argument α on sliders; the red vector is the product zw. Whatever you do with the sliders, two things hold: the red vector is exactly twice as long as the blue one (the moduli multiply, and ‖w‖ = 2), and it sits exactly 60° counterclockwise of the blue one (the arguments add, and w contributes π/3). So multiplying by w is a rotation-and-stretch — the geometric meaning of complex multiplication this section builds toward De Moivre's theorem. At the starting values z = 2(cos π/6 + i sin π/6), the product is the book's own worked example: zw = 4(cos π/2 + i sin π/2) = 4i, straight up the imaginary axis. The dashed circle of radius 2r traces every product you can reach by turning α alone.{"functions":\[{"color":"\#6a6a6a","expression":"2\*r\*cos(t)","expression2":"2\*r\*sin(t)","id":"product-locus","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"t\*1","expression2":"t\*1.7320508","id":"vector-w","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*r\*cos(al)","expression2":"t\*r\*sin(al)","id":"vector-z","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"r\*cos(al) + 0.1\*cos(t)","expression2":"r\*sin(al) + 0.1\*sin(t)","id":"z-dot","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"t\*2\*r\*cos(al + pi/3)","expression2":"t\*2\*r\*sin(al + pi/3)","id":"vector-zw","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"2\*r\*cos(al + pi/3) + 0.1\*cos(t)","expression2":"2\*r\*sin(al + pi/3) + 0.1\*sin(t)","id":"zw-dot","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-r","label":"Modulus r of z","max":2,"min":0.5,"name":"r","step":0.25,"value":2},{"id":"p-al","label":"Argument \\u03b1 of z, in radians (steps of \\u03c0/12)","max":6.2831853072,"min":0,"name":"al","step":0.2617993878,"value":0.5235987756}\],"title":"z\\u00b7w for fixed w = 2(cos \\u03c0/3 + i sin \\u03c0/3)","version":1,"viewport":{"centerX":0,"centerY":0,"scale":48}}#examplebox("Example 3")[][ Find the product of #math.equation(block: false, alt: "z equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 6 close parenthesis close parenthesis")[$" " z = 2 ( cos ( display(frac(π, 6)) ) + i sin ( display(frac(π, 6)) ) ) " "$] and #math.equation(block: false, alt: "w equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 3 close parenthesis close parenthesis")[$" " w = 2 ( cos ( display(frac(π, 3)) ) + i sin ( display(frac(π, 3)) ) )$]. #solutionbox[ We multiply the moduli, #math.equation(block: false, alt: "2 times 2 equals 4")[$2 ⋅ 2 = 4$], and add the arguments, #math.equation(block: false, alt: "the fraction π over 6 plus the fraction π over 3 equals the fraction π over 2")[$display(frac(π, 6)) + display(frac(π, 3)) = display(frac(π, 2))$]. The polar form of the product is thus #math.equation(block: true, alt: "z w equals 4 open parenthesis cos ⁡ open parenthesis the fraction π over 2 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 2 close parenthesis close parenthesis equals 4 i")[$z w = 4 ( cos ( display(frac(π, 2)) ) + i sin ( display(frac(π, 2)) ) ) = 4 i$] Note that #math.equation(block: false, alt: "z")[$z$] and #math.equation(block: false, alt: "w")[$w$] are the numbers from the previous Example and Exercise, namely #math.equation(block: false, alt: "z equals the square root of 3 plus i")[$" " z = sqrt(3) + i " "$] and #math.equation(block: false, alt: "w equals 1 plus i the square root of 3")[$" " w = 1 + i sqrt(3)$]. You can compute the product #math.equation(block: false, alt: "z w")[$z w$] in Cartesian form to check that you get the same result, #math.equation(block: false, alt: "4 i")[$4 i$]. The figure at right shows the graphs of #math.equation(block: false, alt: "z")[$z$] and #math.equation(block: false, alt: "w")[$w$], and their product, #math.equation(block: false, alt: "z w")[$z w$]. You can see that multiplying #math.equation(block: false, alt: "z")[$z$] by #math.equation(block: false, alt: "w")[$w$] rotates the graph of #math.equation(block: false, alt: "z")[$z$] by the argument of #math.equation(block: false, alt: "w")[$w$], namely #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] or #math.equation(block: false, alt: "60 degrees")[$60^(∘)$]. #figure(figph[product of complex numbers], alt: "product of complex numbers", caption: none) ] ] + Find the polar forms of #math.equation(block: false, alt: "z equals 3 plus 3 i")[$" " z = 3 + 3 i " "$] and #math.equation(block: false, alt: "w equals minus 2 minus 2 i")[$" " w = − 2 − 2 i " "$]. + Compute the product #math.equation(block: false, alt: "z w")[$z w$]. + #math.equation(block: false, alt: "z equals 3 the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction π over 4 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 4 close parenthesis close parenthesis")[$z = 3 sqrt(2) ( cos ( display(frac(π, 4)) ) + i sin ( display(frac(π, 4)) ) )$], #math.equation(block: false, alt: "w equals 2 the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 4 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 4 close parenthesis close parenthesis")[$" " w = 2 sqrt(2) ( cos ( display(frac(5 π, 4)) ) + i sin ( display(frac(5 π, 4)) ) )$] + #math.equation(block: false, alt: "minus 12 i")[$− 12 i$] The quotient of two complex numbers in polar form is computed in a similar fashion. We divide the moduli and subtract the arguments. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quotient in Polar Form] If #math.equation(block: false, alt: "z sub 1 equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$" " z_(1) = r ( cos ( α ) + i sin ( α ) ) " "$] and #math.equation(block: false, alt: "z sub 2 equals R open parenthesis open parenthesis cos ⁡ β close parenthesis plus i sin ⁡ open parenthesis β close parenthesis close parenthesis")[$" " z_(2) = R ( ( cos β ) + i sin ( β ) )$], then #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals the fraction r over R open parenthesis cos ⁡ open parenthesis α minus β close parenthesis plus i sin ⁡ open parenthesis α minus β close parenthesis close parenthesis")[$display(frac(bold(italic(z))_(1), bold(italic(z))_(2))) = display(frac(bold(italic(r)), bold(italic(R)))) ( bold(cos) ( bold(italic(α)) − bold(italic(β)) ) + bold(italic(i)) bold(sin) ( bold(italic(α)) − bold(italic(β)) ) )$] ] #examplebox("Example 4")[][ Find the quotient of #math.equation(block: false, alt: "z equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 6 close parenthesis close parenthesis")[$" " z = 2 ( cos ( display(frac(π, 6)) ) + i sin ( display(frac(π, 6)) ) ) " "$] and #math.equation(block: false, alt: "w equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 3 close parenthesis close parenthesis")[$" " w = 2 ( cos ( display(frac(π, 3)) ) + i sin ( display(frac(π, 3)) ) )$]. #solutionbox[ We divide the moduli, #math.equation(block: false, alt: "the fraction 2 over 2 equals 1")[$display(frac(2, 2)) = 1$], and subtract the arguments, #math.equation(block: false, alt: "the fraction π over 6 minus the fraction π over 3 equals the fraction minus π over 6")[$display(frac(π, 6)) − display(frac(π, 3)) = display(frac(− π, 6))$]. The polar form of the quotient is thus #math.equation(block: true, alt: "the fraction z over w equals 1 open parenthesis cos ⁡ open parenthesis the fraction minus π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction minus π over 6 close parenthesis close parenthesis equals the fraction the square root of 3 over 2 minus the fraction 1 over 2 i")[$display(frac(z, w)) = 1 ( cos ( display(frac(− π, 6)) ) + i sin ( display(frac(− π, 6)) ) ) = display(frac(sqrt(3), 2)) − display(frac(1, 2)) i$] The figure at right shows the graphs of #math.equation(block: false, alt: "z , w")[$z , " " w$], and the quotient #math.equation(block: false, alt: "the fraction z over w")[$display(frac(z, w))$]. #figure(figph[quotient of complex numbers], alt: "quotient of complex numbers", caption: none) ] ] Compute the quotient #math.equation(block: false, alt: "the fraction z over w")[$" " display(frac(z, w))$] for #math.equation(block: false, alt: "z equals 3 plus 3 i")[$" " z = 3 + 3 i " "$] and #math.equation(block: false, alt: "w equals minus 2 minus 2 i")[$" " w = − 2 − 2 i$]. #math.equation(block: true, alt: "the fraction minus 3 over 2")[$display(frac(− 3, 2))$] === Powers and Roots of Complex Numbers Because raising to a power is just repeated multiplication, we can also use the polar form to simplify powers of a complex number. For example, if #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z = r ( cos ( α ) + i sin ( α ) )$], then #math.equation(block: true, alt: "z squared equals z times z equals r times r open parenthesis cos ⁡ open parenthesis α plus α close parenthesis plus i sin ⁡ open parenthesis α plus α close parenthesis close parenthesis")[$z^(2) = z ⋅ z = r ⋅ r ( cos ( α + α ) + i sin ( α + α ) )$] We compute #math.equation(block: false, alt: "z squared")[$z^(2)$] by squaring the modulus, #math.equation(block: false, alt: "r")[$r$], and doubling the argument, #math.equation(block: false, alt: "α")[$α$], so the polar form is #math.equation(block: true, alt: "z squared equals r squared open parenthesis cos ⁡ open parenthesis 2 α close parenthesis plus i sin ⁡ open parenthesis 2 α close parenthesis close parenthesis")[$z^(2) = r^(2) ( cos ( 2 α ) + i sin ( 2 α ) )$] An analogous result holds for all positive integers #math.equation(block: false, alt: "n")[$n$], and is known as De Moivre's Theorem. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[De Moivre's Theorem] If #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$" " z = r ( cos ( α ) + i sin ( α ) ) " "$] is a complex number in polar form, and #math.equation(block: false, alt: "n")[$n$] is a positive integer, then #math.equation(block: true, alt: "z to the power n equals r to the power n open parenthesis cos ⁡ open parenthesis n α close parenthesis plus i sin ⁡ open parenthesis n α close parenthesis close parenthesis")[$bold(italic(z))^(bold(italic(n))) = bold(italic(r))^(bold(italic(n))) ( bold(cos) ( bold(italic(n)) bold(italic(α)) ) + bold(italic(i)) bold(sin) ( bold(italic(n)) bold(italic(α)) ) )$] ] #examplebox("Example 5")[][ Raise #math.equation(block: false, alt: "z equals the square root of 3 plus i")[$z = sqrt(3) + i$] to the fourth power. #solutionbox[ We use the polar form, #math.equation(block: false, alt: "z equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 6 close parenthesis close parenthesis")[$z = 2 ( cos ( display(frac(π, 6)) ) + i sin ( display(frac(π, 6)) ) )$], and apply De Moivre's theorem. Then #math.equation(block: true, alt: "z to the power 4, equals 2 to the power 4 open parenthesis cos ⁡ open parenthesis 4 times the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis 4 times the fraction π over 6 close parenthesis close parenthesis; equals 16 open parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close parenthesis")[$z^(4) & = 2^(4) ( cos ( 4 ⋅ display(frac(π, 6)) ) + i sin ( 4 ⋅ display(frac(π, 6)) ) ) \ & = 16 ( cos ( display(frac(2 π, 3)) ) + i sin ( display(frac(2 π, 3)) ) )$] The Cartesian form of #math.equation(block: false, alt: "z to the power 4")[$z^(4)$] is thus #math.equation(block: false, alt: "16 open parenthesis the fraction minus 1 over 2 plus i the fraction the square root of 3 over 2 close parenthesis")[$16 ( display(frac(− 1, 2)) + i display(frac(sqrt(3), 2)) )$], or #math.equation(block: false, alt: "minus 8 plus 8 i the square root of 3")[$− 8 + 8 i sqrt(3)$]. You can check that you get this same result if you compute #math.equation(block: false, alt: "open parenthesis the square root of 3 plus i close parenthesis to the power 4")[$attach(( sqrt(3) + i ), t: 4)$] by expanding the power. ] ] Compute #math.equation(block: false, alt: "w cubed")[$w^(3)$], for #math.equation(block: false, alt: "w equals minus the square root of 2 plus the square root of 2 i")[$" " w = − sqrt(2) + sqrt(2) i$]. #math.equation(block: true, alt: "4 the square root of 2 plus 4 the square root of 2 i")[$4 sqrt(2) + 4 sqrt(2) i$] DeMoivre's Theorem also works for rational values of #math.equation(block: false, alt: "n")[$n$], so we can compute roots of complex numbers. For example, by applying the theorem with #math.equation(block: false, alt: "n equals the fraction 1 over 2")[$n = display(frac(1, 2))$], we see that one of the square roots of #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z = r ( cos ( α ) + i sin ( α ) )$] is #math.equation(block: true, alt: "z to the power 1 / 2 equals r to the power 1 / 2 open parenthesis cos ⁡ open parenthesis the fraction α over 2 close parenthesis plus i sin ⁡ open parenthesis the fraction α over 2 close parenthesis close parenthesis")[$z^(1 / 2) = r^(1 / 2) ( cos ( display(frac(α, 2)) ) + i sin ( display(frac(α, 2)) ) )$] Now, every number,whether real or complex, has two square roots. To find the other root, remember that we can add a multiple of #math.equation(block: false, alt: "2 π")[$2 π$] to the argument of #math.equation(block: false, alt: "z")[$z$], that is, we can also write the polar form of #math.equation(block: false, alt: "z")[$z$] as #math.equation(block: true, alt: "z equals r open parenthesis cos ⁡ open parenthesis α plus 2 π close parenthesis plus i sin ⁡ open parenthesis α plus 2 π close parenthesis close parenthesis")[$z = r ( cos ( α + 2 π ) + i sin ( α + 2 π ) )$] The second square root of #math.equation(block: false, alt: "z")[$z$] is thus #math.equation(block: true, alt: "r to the power 1 / 2 open parenthesis cos ⁡ open parenthesis the fraction α over 2 plus π close parenthesis plus i sin ⁡ open parenthesis the fraction α over 2 plus π close parenthesis close parenthesis")[$r^(1 / 2) ( cos ( display(frac(α, 2)) + π ) + i sin ( display(frac(α, 2)) + π ) )$] As an example consider #math.equation(block: false, alt: "w equals minus 8 plus 8 i the square root of 3 ,")[$" " w = − 8 + 8 i sqrt(3) ,$] whose polar form is #math.equation(block: true, alt: "16 open parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close parenthesis")[$" " 16 ( cos ( display(frac(2 π, 3)) ) + i sin ( display(frac(2 π, 3)) ) ) " "$] or by adding #math.equation(block: false, alt: "2 π")[$2 π$] to the argument, #math.equation(block: true, alt: "16 open parenthesis cos ⁡ open parenthesis the fraction 8 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 8 π over 3 close parenthesis close parenthesis")[$16 ( cos ( display(frac(8 π, 3)) ) + i sin ( display(frac(8 π, 3)) ) )$] The two square roots of #math.equation(block: false, alt: "w")[$w$] are #math.equation(block: true, alt: "z sub 1, equals 16 to the power 1 / 2 open parenthesis cos ⁡ open parenthesis the fraction π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 3 close parenthesis close parenthesis equals 4 open parenthesis the fraction 1 over 2 plus i the fraction the square root of 3 over 2 close parenthesis equals 2 plus 2 i the square root of 3; z sub 2, equals 16 to the power 1 / 2 open parenthesis cos ⁡ open parenthesis the fraction 4 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 4 π over 3 close parenthesis close parenthesis; equals 4 open parenthesis the fraction minus 1 over 2 plus i the fraction minus the square root of 3 over 2 close parenthesis equals minus 2 minus 2 i the square root of 3")[$z_(1) & = 16^(1 / 2) ( cos ( display(frac(π, 3)) ) + i sin ( display(frac(π, 3)) ) ) = 4 ( display(frac(1, 2)) + i display(frac(sqrt(3), 2)) ) = 2 + 2 i sqrt(3) \ z_(2) & = 16^(1 / 2) ( cos ( display(frac(4 π, 3)) ) + i sin ( display(frac(4 π, 3)) ) ) \ & = 4 ( display(frac(− 1, 2)) + i display(frac(− sqrt(3), 2)) ) = − 2 − 2 i sqrt(3)$] You can verify that both of these numbers are square roots of #math.equation(block: false, alt: "w equals minus 8 plus 8 i the square root of 3")[$w = − 8 + 8 i sqrt(3)$], for instance, #math.equation(block: true, alt: "open parenthesis 2 plus 2 i the square root of 3 close parenthesis squared, equals 4 plus 2 open parenthesis 2 close parenthesis open parenthesis 2 i the square root of 3 close parenthesis plus 12 i squared; equals minus 8 plus 8 i the square root of 3")[$attach(( 2 + 2 i sqrt(3) ), t: 2) & = 4 + 2 ( 2 ) ( 2 i sqrt(3) ) + 12 i^(2) \ & = − 8 + 8 i sqrt(3)$] The graphs of #math.equation(block: false, alt: "w")[$w$] and its two square roots are shown at right. #figure(figph[square roots of complex number], alt: "square roots of complex number", caption: none) It is not hard to show that every number has three complex cube roots, four complex fourth roots, and so on. #examplebox("Example 6")[][ Find three complex cube roots of #math.equation(block: false, alt: "8")[$8$]. #solutionbox[ One polar form of #math.equation(block: false, alt: "z equals 8")[$z = 8$] is #math.equation(block: true, alt: "z equals 8 open parenthesis cos ⁡ 0 plus i sin ⁡ 0 close parenthesis")[$z = 8 ( cos 0 + i sin 0 )$] so one of its cube roots is #math.equation(block: true, alt: "z sub 1 equals 8 to the power 1 / 3 open parenthesis cos ⁡ open parenthesis the fraction 0 over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 0 over 3 close parenthesis close parenthesis equals 2 open parenthesis cos ⁡ 0 plus i sin ⁡ 0 close parenthesis equals 2")[$z_(1) = 8^(1 / 3) ( cos ( display(frac(0, 3)) ) + i sin ( display(frac(0, 3)) ) ) = 2 ( cos 0 + i sin 0 ) = 2$] But by adding multiples of #math.equation(block: false, alt: "2 π")[$2 π$] to the argument, we can also write the polar form of as #math.equation(block: true, alt: "z equals 8 open parenthesis cos ⁡ 2 π plus i sin ⁡ 2 π close parenthesis and z equals 8 open parenthesis cos ⁡ 4 π plus i sin ⁡ 4 π close parenthesis")[$z = 8 ( cos 2 π + i sin 2 π ) " " " " " " " " "and" " " " " " " " " z = 8 ( cos 4 π + i sin 4 π )$] Applying De Moivre's theorem to each of these expressions gives us two more cube roots #math.equation(block: true, alt: "z sub 2, equals 2 open parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close parenthesis equals minus 1 plus i the square root of 3; z sub 3, equals 2 open parenthesis cos ⁡ open parenthesis the fraction 4 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 4 π over 3 close parenthesis close parenthesis equals minus 1 minus i the square root of 3")[$z_(2) & = 2 ( cos ( display(frac(2 π, 3)) ) + i sin ( display(frac(2 π, 3)) ) ) = − 1 + i sqrt(3) \ z_(3) & = 2 ( cos ( display(frac(4 π, 3)) ) + i sin ( display(frac(4 π, 3)) ) ) = − 1 − i sqrt(3)$] The three roots are shown at right. Note that they are evenly spaced around a circle of radius 2. You can check that adding further muliples of #math.equation(block: false, alt: "2 π")[$2 π$] to the argument does not generate any new cube roots. #figure(figph[square roots of complex number], alt: "square roots of complex number", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Roots of a Complex Number] A complex number #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ α plus i sin ⁡ α close parenthesis")[$z = r ( cos α + i sin α )$] in polar form has #math.equation(block: false, alt: "n")[$n$] complex #math.equation(block: false, alt: "n")[$n$]th roots, given by #math.equation(block: true, alt: "z sub k equals r to the power 1 / n open parenthesis cos ⁡ open parenthesis the fraction α plus 2 π k over n close parenthesis plus i sin ⁡ open parenthesis the fraction α plus 2 π k over n close parenthesis close parenthesis")[$bold(italic(z))_(bold(italic(k))) = bold(italic(r))^(1 / bold(italic(n))) ( bold(cos) ( display(frac(bold(italic(α)) + 2 bold(italic(π)) bold(italic(k)), bold(italic(n)))) ) + bold(italic(i)) bold(sin) ( display(frac(bold(italic(α)) + 2 bold(italic(π)) bold(italic(k)), bold(italic(n)))) ) )$] for #math.equation(block: false, alt: "k equals 0 , 1 , 2 , ⋯ , n minus 1")[$k = 0 , " " 1 , " " 2 , ⋯ , " " n − 1$]. ] + Write #math.equation(block: false, alt: "i")[$i$] in polar form. + Find four complex fourth roots of #math.equation(block: false, alt: "i")[$i$]. + #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 2 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 2 close parenthesis")[$cos ( display(frac(π, 2)) ) + i sin ( display(frac(π, 2)) )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 8 close parenthesis")[$cos ( display(frac(π, 8)) ) + i sin ( display(frac(π, 8)) )$], #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 5 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 8 close parenthesis")[$" " cos ( display(frac(5 π, 8)) ) + i sin ( display(frac(5 π, 8)) )$], #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 9 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 9 π over 8 close parenthesis")[$" " cos ( display(frac(9 π, 8)) ) + i sin ( display(frac(9 π, 8)) )$], #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 13 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 13 π over 8 close parenthesis")[$" " cos ( display(frac(13 π, 8)) ) + i sin ( display(frac(13 π, 8)) )$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write with fractional exponents. + #math.equation(block: false, alt: "the cube root of a")[$root(3, a)$] + #math.equation(block: false, alt: "the fraction 1 over the 4th root of n")[$display(frac(1, root(4, n)))$] + #math.equation(block: false, alt: "the 4th root of t cubed")[$root(4, t^(3))$] + #math.equation(block: false, alt: "the cube root of z to the power 5")[$root(3, z^(5))$] + #math.equation(block: false, alt: "the fraction 1 over the square root of x plus 2")[$display(frac(1, sqrt(x + 2)))$] + #math.equation(block: false, alt: "the cube root of open parenthesis w minus 1 close parenthesis squared")[$root(3, ( w − 1 )^(2))$] + #math.equation(block: false, alt: "z cubed the 5th root of w to the power 4")[$z^(3) root(5, w^(4))$] + #math.equation(block: false, alt: "the fraction a squared over the 4th root of b cubed")[$display(frac(a^(2), root(4, b^(3))))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "a to the power 1 / 3")[$a^(1 / 3)$] + #math.equation(block: false, alt: "n to the power minus 1 / 4")[$n^(− 1 / 4)$] + #math.equation(block: false, alt: "t to the power 3 / 4")[$t^(3 / 4)$] + #math.equation(block: false, alt: "z to the power 5 / 3")[$z^(5 / 3)$] + #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis to the power minus 1 / 2")[$( x + 2 )^(− 1 / 2)$] + #math.equation(block: false, alt: "open parenthesis w minus 1 close parenthesis to the power 2 / 3")[$( w − 1 )^(2 / 3)$] + #math.equation(block: false, alt: "z cubed w to the power 4 / 5")[$z^(3) w^(4 / 5)$] + #math.equation(block: false, alt: "a squared b to the power minus 3 / 4")[$a^(2) b^(− 3 / 4)$] ] === Section 10.4 Summary ==== Vocabulary - Argument - Modulus ==== Concepts + Multiplying a complex number by #math.equation(block: false, alt: "i")[$i$] rotates its graph by #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] around the origin. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Polar Form for a Complex Number] The complex number #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] can be written in the polar form #math.equation(block: true, alt: "z equals r open parenthesis cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$z = r ( cos ( θ ) + i sin ( θ ) )$]where #math.equation(block: true, alt: "r equals the square root of a squared plus b squared")[$r = sqrt(a^(2) + b^(2))$]and #math.equation(block: false, alt: "θ")[$θ$] is defined by #math.equation(block: true, alt: "a equals r cos ⁡ open parenthesis θ close parenthesis , b equals r sin ⁡ open parenthesis θ close parenthesis , 0 less than or equal to θ less than or equal to 2 π")[$a = r cos ( θ ) , " " " " " " " " b = r sin ( θ ) , " " " " " " " " 0 ≤ θ ≤ 2 π$]The angle #math.equation(block: false, alt: "θ")[$θ$] is called the #strong[argument] of the complex number, and #math.equation(block: false, alt: "r")[$r$] is its length, or #strong[modulus]. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Product and Quotient in Polar Form] If #math.equation(block: false, alt: "z sub 1 equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z_(1) = r ( cos ( α ) + i sin ( α ) )$] and #math.equation(block: false, alt: "z sub 2 equals R open parenthesis cos ⁡ open parenthesis β close parenthesis plus i sin ⁡ open parenthesis β close parenthesis close parenthesis")[$z_(2) = R ( cos ( β ) + i sin ( β ) )$], then #math.equation(block: true, alt: "z sub 1 z sub 2 equals r R open parenthesis cos ⁡ open parenthesis α plus β close parenthesis plus i sin ⁡ open parenthesis α plus β close parenthesis close parenthesis")[$z_(1) z_(2) = r R ( cos ( α + β ) + i sin ( α + β ) )$]and #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals the fraction r over R open parenthesis cos ⁡ open parenthesis α minus β close parenthesis plus i sin ⁡ open parenthesis α minus β close parenthesis close parenthesis")[$display(frac(z_(1), z_(2))) = display(frac(r, R)) ( cos ( α − β ) + i sin ( α − β ) )$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[De Moivre's Theorem] If #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z = r ( cos ( α ) + i sin ( α ) )$] is a complex number in polar form, and #math.equation(block: false, alt: "n")[$n$] is a positive integer, then #math.equation(block: true, alt: "z to the power n equals r to the power n open parenthesis cos ⁡ open parenthesis n α close parenthesis plus i sin ⁡ open parenthesis n α close parenthesis close parenthesis")[$z^(n) = r^(n) ( cos ( n α ) + i sin ( n α ) )$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Roots of a Complex Number] A complex number #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z = r ( cos ( α ) + i sin ( α ) )$] in polar form has #math.equation(block: false, alt: "n")[$n$] complex #math.equation(block: false, alt: "n")[$n$]th roots, given by #math.equation(block: true, alt: "z sub k equals r to the power 1 / n open parenthesis cos ⁡ open parenthesis the fraction α plus 2 π k over n close parenthesis plus i sin ⁡ open parenthesis the fraction α plus 2 π k over n close parenthesis close parenthesis")[$z_(k) = r^(1 / n) ( cos ( display(frac(α + 2 π k, n)) ) + i sin ( display(frac(α + 2 π k, n)) ) )$]for #math.equation(block: false, alt: "k equals 0 , 1 , 2 , ⋯ , n minus 1")[$k = 0 , " " 1 , " " 2 , ⋯ , " " n − 1$]. ] ==== Study Questions + What happens to the vector representing a complex number when we multiply the number by #math.equation(block: false, alt: "minus i")[$− i$]? + If #math.equation(block: false, alt: "z equals p plus q i")[$z = p + q i$], what do these expressions represent? #math.equation(block: false, alt: "p squared plus q squared , p minus q i , the fraction q over p")[$p^(2) + q^(2) , " " p − q i , " " display(frac(q, p))$] + If #math.equation(block: false, alt: "z sub 1")[$z_(1)$] and #math.equation(block: false, alt: "z sub 2")[$z_(2)$] lie on the unit circle, explain why #math.equation(block: false, alt: "z sub 1 z sub 2")[$z_(1) z_(2)$] and #math.equation(block: false, alt: "the fraction z sub 1 over z sub 2")[$display(frac(z_(1), z_(2)))$] also lie on the unit circle. + Explain how you could use DeMoivre's theorem to compute #math.equation(block: false, alt: "z to the power 2 / 3")[$z^(2 / 3)$]. ==== Skills + Convert from polar form to standard form \#5–12 + Write a complex number in polar form \#13-22 + Find the product or quotient of two complex numbers in polar form \#25–32 + Find a power of a complex number \#33–42 + Find the complex roots of a number \#43–48, 51–52, 55–60 === Homework 10-4 For Problems 1–4, simplify and plot each complex number as a point on the complex plane. #math.equation(block: false, alt: "1 , i , i squared , i cubed")[$1 , " " i , " " i^(2) , " " i^(3)$] and #math.equation(block: false, alt: "i to the power 4")[$i^(4)$] #math.equation(block: true, alt: "1")[$1$], #math.equation(block: true, alt: "i")[$" " i$], #math.equation(block: true, alt: "minus 1")[$" " − 1$],#math.equation(block: true, alt: "minus i")[$" " − i$], #math.equation(block: true, alt: "1")[$" " 1$] #figure(figph[powers of i], alt: "powers of i", caption: none) #math.equation(block: false, alt: "minus 1 , minus i , minus i squared , minus i cubed")[$− 1 , " " − i , " " − i^(2) , " " − i^(3)$] and #math.equation(block: false, alt: "minus i to the power 4")[$− i^(4)$] #math.equation(block: false, alt: "1 plus 2 i")[$1 + 2 i$] and #math.equation(block: false, alt: "i open parenthesis 1 plus 2 i close parenthesis")[$i ( 1 + 2 i )$] #math.equation(block: true, alt: "1 plus 2 i")[$1 + 2 i$], #math.equation(block: true, alt: "minus 2 plus i")[$− 2 + i$] #figure(figph[complex numbers], alt: "complex numbers", caption: none) #math.equation(block: false, alt: "3 minus 4 i")[$3 − 4 i$] and #math.equation(block: false, alt: "i open parenthesis 3 minus 4 i close parenthesis")[$i ( 3 − 4 i )$] For Problems 5–8, write the complex numbers in standard form. Give exact values for your answers. #math.equation(block: true, alt: "6 open parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close parenthesis")[$6 ( cos ( display(frac(2 π, 3)) ) + i sin ( display(frac(2 π, 3)) ) )$] #math.equation(block: true, alt: "minus 3 plus 3 i the square root of 3")[$− 3 + 3 i sqrt(3)$] #math.equation(block: true, alt: "4 open parenthesis cos ⁡ open parenthesis the fraction 7 π over 4 close parenthesis plus i sin ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close parenthesis")[$4 ( cos ( display(frac(7 π, 4)) ) + i sin ( display(frac(7 π, 4)) ) )$] #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 4 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 4 close parenthesis close parenthesis")[$sqrt(2) ( cos ( display(frac(3 π, 4)) ) + i sin ( display(frac(3 π, 4)) ) )$] #math.equation(block: true, alt: "minus 1 plus i")[$− 1 + i$] #math.equation(block: true, alt: "the fraction 3 over 2 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 6 close parenthesis close parenthesis")[$display(frac(3, 2)) ( cos ( display(frac(5 π, 6)) ) + i sin ( display(frac(5 π, 6)) ) )$] For Problems 9–12, write the complex numbers in standard form. Round your answers to hundredths. #math.equation(block: true, alt: "5 open parenthesis cos ⁡ open parenthesis 5.2 close parenthesis plus i sin ⁡ open parenthesis 5.2 close parenthesis close parenthesis")[$5 ( cos ( 5.2 ) + i sin ( 5.2 ) )$] #math.equation(block: true, alt: "2.34 minus 4.21 i")[$2.34 − 4.21 i$] #math.equation(block: true, alt: "3 open parenthesis cos ⁡ open parenthesis 3.5 close parenthesis plus i sin ⁡ open parenthesis 3.5 close parenthesis close parenthesis")[$3 ( cos ( 3.5 ) + i sin ( 3.5 ) )$] #math.equation(block: true, alt: "12 open parenthesis cos ⁡ open parenthesis 115 degrees close parenthesis plus i sin ⁡ open parenthesis 115 degrees close parenthesis close parenthesis")[$12 ( cos ( 115^(∘) ) + i sin ( 115^(∘) ) )$] #math.equation(block: true, alt: "minus 5.07 plus 10.88 i")[$− 5.07 + 10.88 i$] ) #math.equation(block: true, alt: "20 open parenthesis cos ⁡ open parenthesis 250 degrees close parenthesis plus i sin ⁡ open parenthesis 250 degrees close parenthesis close parenthesis")[$20 ( cos ( 250^(∘) ) + i sin ( 250^(∘) ) )$] For Problems 13–16, write the complex numbers in polar form. Give exact values for your answers. #math.equation(block: false, alt: "3 i")[$3 i$] and #math.equation(block: false, alt: "minus 3 i")[$− 3 i$] #math.equation(block: true, alt: "3 open parenthesis cos ⁡ open parenthesis the fraction π over 2 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 2 close parenthesis close parenthesis")[$3 ( cos ( display(frac(π, 2)) ) + i sin ( display(frac(π, 2)) ) )$], #math.equation(block: true, alt: "3 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 2 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 2 close parenthesis close parenthesis")[$3 ( cos ( display(frac(3 π, 2)) ) + i sin ( display(frac(3 π, 2)) ) )$] #math.equation(block: false, alt: "2 plus 2 i")[$2 + 2 i$] and #math.equation(block: false, alt: "2 minus 2 i")[$2 − 2 i$] #math.equation(block: false, alt: "minus 3 minus the square root of 3 i")[$− 3 − sqrt(3) i$] and #math.equation(block: false, alt: "3 minus the square root of 3 i")[$3 − sqrt(3) i$] #math.equation(block: true, alt: "2 the square root of 3 open parenthesis cos ⁡ open parenthesis the fraction 7 π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction 7 π over 6 close parenthesis close parenthesis")[$2 sqrt(3) ( cos ( display(frac(7 π, 6)) ) + i sin ( display(frac(7 π, 6)) ) )$], #math.equation(block: true, alt: "2 the square root of 3 open parenthesis cos ⁡ open parenthesis the fraction 11 π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction 11 π over 6 close parenthesis close parenthesis")[$2 sqrt(3) ( cos ( display(frac(11 π, 6)) ) + i sin ( display(frac(11 π, 6)) ) )$] #math.equation(block: false, alt: "2 the square root of 3 plus 2 i")[$2 sqrt(3) + 2 i$] and #math.equation(block: false, alt: "minus 2 the square root of 3 plus 2 i")[$− 2 sqrt(3) + 2 i$] For Problems 17–22, write the complex numbers in polar form. Round your answers to hundredths. #math.equation(block: false, alt: "minus 4 plus 2 i")[$− 4 + 2 i$] and #math.equation(block: false, alt: "4 minus 2 i")[$4 − 2 i$] #math.equation(block: true, alt: "4.47 open parenthesis cos ⁡ open parenthesis 2.68 close parenthesis plus i sin ⁡ open parenthesis 2.68 close parenthesis close parenthesis ,")[$4.47 ( cos ( 2.68 ) + i sin ( 2.68 ) ) , " "$] #math.equation(block: true, alt: "4.47 open parenthesis cos ⁡ open parenthesis 5.82 close parenthesis plus i sin ⁡ open parenthesis 5.82 close parenthesis close parenthesis")[$4.47 ( cos ( 5.82 ) + i sin ( 5.82 ) )$] #math.equation(block: false, alt: "minus 3 minus 8 i")[$− 3 − 8 i$] and #math.equation(block: false, alt: "3 plus 8 i")[$3 + 8 i$] #math.equation(block: false, alt: "9 minus 5 i")[$9 − 5 i$] and #math.equation(block: false, alt: "9 plus 5 i")[$9 + 5 i$] #math.equation(block: true, alt: "8.60 open parenthesis cos ⁡ open parenthesis 5.78 close parenthesis plus i sin ⁡ open parenthesis 5.78 close parenthesis close parenthesis ,")[$8.60 ( cos ( 5.78 ) + i sin ( 5.78 ) ) , " "$] #math.equation(block: true, alt: "8.60 open parenthesis cos ⁡ open parenthesis 0.51 close parenthesis plus i sin ⁡ open parenthesis 0.51 close parenthesis close parenthesis")[$8.60 ( cos ( 0.51 ) + i sin ( 0.51 ) )$] #math.equation(block: false, alt: "2 plus 6 i")[$2 + 6 i$] and #math.equation(block: false, alt: "2 minus 6 i")[$2 − 6 i$] #math.equation(block: false, alt: "3 plus 4 i , 3 minus 4 i , minus 3 plus 4 i ,")[$3 + 4 i , " " 3 − 4 i , " " − 3 + 4 i ,$] and #math.equation(block: false, alt: "minus 3 minus 4 i")[$− 3 − 4 i$] #math.equation(block: true, alt: "5 open parenthesis cos ⁡ open parenthesis 0.93 close parenthesis plus i sin ⁡ open parenthesis 0.93 close parenthesis close parenthesis ,")[$5 ( cos ( 0.93 ) + i sin ( 0.93 ) ) , " "$] #math.equation(block: true, alt: "5 open parenthesis cos ⁡ open parenthesis 5.36 close parenthesis plus i sin ⁡ open parenthesis 5.36 close parenthesis close parenthesis ,")[$5 ( cos ( 5.36 ) + i sin ( 5.36 ) ) , " "$] #math.equation(block: true, alt: "5 open parenthesis cos ⁡ open parenthesis 2.21 close parenthesis plus i sin ⁡ open parenthesis 2.21 close parenthesis close parenthesis ,")[$5 ( cos ( 2.21 ) + i sin ( 2.21 ) ) , " "$] #math.equation(block: true, alt: "5 open parenthesis cos ⁡ open parenthesis 4.07 close parenthesis plus i sin ⁡ open parenthesis 4.07 close parenthesis close parenthesis")[$5 ( cos ( 4.07 ) + i sin ( 4.07 ) )$] #math.equation(block: false, alt: "1 plus 3 i , 1 minus 3 i , minus 1 plus 3 i ,")[$1 + 3 i , " " 1 − 3 i , " " − 1 + 3 i ,$] and #math.equation(block: false, alt: "minus 1 minus 3 i")[$− 1 − 3 i$] What can you conclude about the polar forms of #math.equation(block: false, alt: "z")[$z$] and #math.equation(block: false, alt: "z bar")[$overline(z)$]? If #math.equation(block: false, alt: "z equals r open parenthesis cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis close parenthesis ,")[$z = r ( cos ( θ ) + i sin ( θ ) ) ,$] then #math.equation(block: false, alt: "z bar equals r open parenthesis cos ⁡ open parenthesis 2 π minus θ close parenthesis plus i sin ⁡ open parenthesis 2 π minus θ close parenthesis")[$overline(z) = r \( cos \( 2 π − θ \) + i sin \( 2 π − θ \)$] What can you conclude about the polar forms of #math.equation(block: false, alt: "z")[$z$] and #math.equation(block: false, alt: "minus z")[$− z$]? For Problems 25–28, find the product #math.equation(block: false, alt: "z sub 1 z sub 2")[$z_(1) z_(2)$] and the quotient #math.equation(block: false, alt: "the fraction z sub 1 over z sub 2")[$display(frac(z_(1), z_(2)))$]. #math.equation(block: true, alt: "z sub 1 equals 4 open parenthesis cos ⁡ open parenthesis the fraction 4 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 4 π over 3 close parenthesis close parenthesis")[$z_(1) = 4 ( cos ( display(frac(4 π, 3)) ) + i sin ( display(frac(4 π, 3)) ) )$] #math.equation(block: true, alt: "z sub 2 equals the fraction 1 over 2 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 6 close parenthesis close parenthesis")[$z_(2) = display(frac(1, 2)) ( cos ( display(frac(5 π, 6)) ) + i sin ( display(frac(5 π, 6)) ) )$] #math.equation(block: true, alt: "z sub 1 z sub 2 equals 2 open parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 6 close parenthesis close parenthesis equals the square root of 3 plus i")[$z_(1) z_(2) = 2 ( cos ( display(frac(π, 6)) ) + i sin ( display(frac(π, 6)) ) ) = sqrt(3) + i$];#math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals 8 open parenthesis cos ⁡ open parenthesis the fraction π over 2 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 2 close parenthesis close parenthesis equals 8 i")[$" " " " display(frac(z_(1), z_(2))) = 8 ( cos ( display(frac(π, 2)) ) + i sin ( display(frac(π, 2)) ) ) = 8 i$] #math.equation(block: true, alt: "z sub 1 equals 6 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 8 close parenthesis close parenthesis")[$z_(1) = 6 ( cos ( display(frac(5 π, 8)) ) + i sin ( display(frac(5 π, 8)) ) )$] #math.equation(block: true, alt: "z sub 2 equals the fraction 3 over 2 open parenthesis cos ⁡ open parenthesis the fraction π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 8 close parenthesis close parenthesis")[$z_(2) = display(frac(3, 2)) ( cos ( display(frac(π, 8)) ) + i sin ( display(frac(π, 8)) ) )$] #math.equation(block: true, alt: "z sub 1 equals 3 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 5 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 5 close parenthesis close parenthesis")[$z_(1) = 3 ( cos ( display(frac(3 π, 5)) ) + i sin ( display(frac(3 π, 5)) ) )$] #math.equation(block: true, alt: "z sub 2 equals 2 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 10 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 10 close parenthesis close parenthesis")[$z_(2) = 2 ( cos ( display(frac(3 π, 10)) ) + i sin ( display(frac(3 π, 10)) ) )$] #math.equation(block: true, alt: "z sub 1 z sub 2 equals 6 open parenthesis cos ⁡ open parenthesis the fraction 9 π over 10 close parenthesis plus i sin ⁡ open parenthesis the fraction 9 π over 10 close parenthesis close parenthesis")[$z_(1) z_(2) = 6 ( cos ( display(frac(9 π, 10)) ) + i sin ( display(frac(9 π, 10)) ) )$]; #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals the fraction 3 over 2 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 10 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 10 close parenthesis close parenthesis")[$" " " " display(frac(z_(1), z_(2))) = display(frac(3, 2)) ( cos ( display(frac(3 π, 10)) ) + i sin ( display(frac(3 π, 10)) ) )$] #math.equation(block: true, alt: "z sub 1 equals 4 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 12 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 12 close parenthesis close parenthesis")[$z_(1) = 4 ( cos ( display(frac(5 π, 12)) ) + i sin ( display(frac(5 π, 12)) ) )$] #math.equation(block: true, alt: "z sub 2 equals 6 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 4 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 4 close parenthesis close parenthesis")[$z_(2) = 6 ( cos ( display(frac(3 π, 4)) ) + i sin ( display(frac(3 π, 4)) ) )$] For Problems 29–32, convert the complex numbers to polar form, then find the product #math.equation(block: false, alt: "z sub 1 z sub 2")[$z_(1) z_(2)$] and the quotient #math.equation(block: false, alt: "the fraction z sub 1 over z sub 2")[$display(frac(z_(1), z_(2)))$]. #math.equation(block: true, alt: "z sub 1 equals 2 i , z sub 2 equals 4 i")[$z_(1) = 2 i , " " z_(2) = 4 i$] #math.equation(block: true, alt: "z sub 1 z sub 2 equals 8")[$z_(1) z_(2) = 8$]; #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals the fraction 1 over 2")[$" " display(frac(z_(1), z_(2))) = display(frac(1, 2))$] #math.equation(block: true, alt: "z sub 1 equals minus 2 i , z sub 2 equals 3 i")[$z_(1) = − 2 i , " " z_(2) = 3 i$] #math.equation(block: true, alt: "z sub 1 equals 2 the square root of 3 minus 2 i")[$z_(1) = 2 sqrt(3) − 2 i$], #math.equation(block: true, alt: "z sub 2 equals minus 1 plus i")[$" " z_(2) = − 1 + i$] #math.equation(block: true, alt: "z sub 1 z sub 2 equals 4 the square root of 2 open parenthesis cos ⁡ the fraction 7 π over 12 plus i sin ⁡ the fraction 7 π over 12 close parenthesis")[$z_(1) z_(2) = 4 sqrt(2) ( cos display(frac(7 π, 12)) + i sin display(frac(7 π, 12)) )$]; #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals 2 the square root of 2 open parenthesis cos ⁡ the fraction 13 π over 12 plus i sin ⁡ the fraction 13 π over 12 close parenthesis")[$" " " " display(frac(z_(1), z_(2))) = 2 sqrt(2) ( cos display(frac(13 π, 12)) + i sin display(frac(13 π, 12)) )$] #math.equation(block: true, alt: "z sub 1 equals the square root of 3 plus i")[$z_(1) = sqrt(3) + i$], #math.equation(block: true, alt: "z sub 2 equals minus 1 plus i the square root of 3")[$" " z_(2) = − 1 + i sqrt(3)$] For Problems 33–38, find the power. #math.equation(block: true, alt: "open parenthesis 2 plus 2 i close parenthesis to the power 5")[$( 2 + 2 i )^(5)$] #math.equation(block: true, alt: "minus 128 minus 128 i")[$− 128 − 128 i$] #math.equation(block: true, alt: "open parenthesis the square root of 2 minus the square root of 2 i close parenthesis to the power 6")[$( sqrt(2) − sqrt(2) i )^(6)$] #math.equation(block: true, alt: "open parenthesis minus 1 plus the square root of 3 i close parenthesis to the power 8")[$( − 1 + sqrt(3) i )^(8)$] #math.equation(block: true, alt: "minus 128 minus 128 the square root of 3 i")[$− 128 − 128 sqrt(3) i$] #math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 plus the fraction the square root of 3 over 2 i close parenthesis to the power 12")[$attach(( display(frac(1, 2)) + display(frac(sqrt(3), 2)) i ), t: 12)$] #math.equation(block: true, alt: "open parenthesis the square root of 3 minus i close parenthesis to the power 10")[$( sqrt(3) − i )^(10)$] #math.equation(block: true, alt: "512 plus 512 the square root of 3 i")[$512 + 512 sqrt(3) i$] #math.equation(block: true, alt: "open parenthesis 1 minus i close parenthesis to the power 20")[$( 1 − i )^(20)$] For Problems 39–42, use De Moivre's theorem to find the reciprocal. #math.equation(block: true, alt: "2 minus 2 i")[$2 − 2 i$] #math.equation(block: true, alt: "the fraction 1 over 4 plus the fraction 1 over 4 i")[$display(frac(1, 4)) + display(frac(1, 4)) i$] #math.equation(block: true, alt: "3 plus the square root of 3 i")[$3 + sqrt(3) i$] #math.equation(block: true, alt: "minus the square root of 2 plus the square root of 6 i")[$− sqrt(2) + sqrt(6) i$] #math.equation(block: true, alt: "the fraction minus the square root of 2 over 8 minus the fraction the square root of 6 over 8 i")[$display(frac(− sqrt(2), 8)) − display(frac(sqrt(6), 8)) i$] #math.equation(block: true, alt: "minus 1 minus i")[$− 1 − i$] For Problems 43–48, + Find the roots and plot them in the complex plane. + Write the roots in standard form. The square roots of #math.equation(block: false, alt: "9 i")[$9 i$] + #math.equation(block: false, alt: "3 open parenthesis cos ⁡ the fraction π over 4 plus i sin ⁡ the fraction π over 4 close parenthesis")[$3 ( cos display(frac(π, 4)) + i sin display(frac(π, 4)) )$], #math.equation(block: false, alt: "3 open parenthesis cos ⁡ the fraction 3 π over 4 plus i sin ⁡ the fraction 3 π over 4 close parenthesis")[$" " 3 ( cos display(frac(3 π, 4)) + i sin display(frac(3 π, 4)) )$] + #math.equation(block: false, alt: "the fraction 3 over the square root of 2 plus the fraction 3 over the square root of 2 i")[$display(frac(3, sqrt(2))) + display(frac(3, sqrt(2))) i$], #math.equation(block: false, alt: "the fraction minus 3 over the square root of 2 minus the fraction 3 over the square root of 2 i")[$" " display(frac(− 3, sqrt(2))) − display(frac(3, sqrt(2))) i$] #figure(figph[square roots], alt: "square roots", caption: none) The fourth roots of #math.equation(block: false, alt: "minus 81")[$− 81$] The fifth roots of #math.equation(block: false, alt: "32")[$32$] + #math.equation(block: false, alt: "2 , 2 open parenthesis cos ⁡ the fraction 2 π over 5 plus i sin ⁡ the fraction 2 π over 5 close parenthesis")[$2 , " " 2 ( cos display(frac(2 π, 5)) + i sin display(frac(2 π, 5)) )$], #math.equation(block: false, alt: "2 open parenthesis cos ⁡ the fraction 4 π over 5 plus i sin ⁡ the fraction 4 π over 5 close parenthesis")[$" " 2 ( cos display(frac(4 π, 5)) + i sin display(frac(4 π, 5)) )$] , #math.equation(block: false, alt: "2 open parenthesis cos ⁡ the fraction 6 π over 5 plus i sin ⁡ the fraction 6 π over 5 close parenthesis")[$" " 2 ( cos display(frac(6 π, 5)) + i sin display(frac(6 π, 5)) )$], #math.equation(block: false, alt: "2 open parenthesis cos ⁡ the fraction 8 π over 5 plus i sin ⁡ the fraction 8 π over 5 close parenthesis")[$2 ( cos display(frac(8 π, 5)) + i sin display(frac(8 π, 5)) )$] + #math.equation(block: false, alt: "2 , 0.618 plus 1.9 i")[$2 , " " 0.618 + 1.9 i$], #math.equation(block: false, alt: "minus 1.618 plus 1.176 i")[$− 1.618 + 1.176 i$], #math.equation(block: false, alt: "minus 1.618 minus 1.176 i")[$− 1.618 − 1.176 i$], #math.equation(block: false, alt: "0.618 minus 1.902 i")[$0.618 − 1.902 i$] #figure(figph[fifth roots of complex number], alt: "fifth roots of complex number", caption: none) The cube roots of #math.equation(block: false, alt: "i")[$i$] The cube roots of #math.equation(block: false, alt: "4 the square root of 3 plus 4 i")[$4 sqrt(3) + 4 i$] + #math.equation(block: false, alt: "4 open parenthesis cos ⁡ the fraction π over 18 plus i sin ⁡ the fraction π over 18 close parenthesis")[$4 ( cos display(frac(π, 18)) + i sin display(frac(π, 18)) )$], #math.equation(block: false, alt: "4 open parenthesis cos ⁡ the fraction 13 π over 18 plus i sin ⁡ the fraction 13 π over 18 close parenthesis")[$" " 4 ( cos display(frac(13 π, 18)) + i sin display(frac(13 π, 18)) )$], #math.equation(block: false, alt: "4 open parenthesis cos ⁡ the fraction 25 π over 18 plus i sin ⁡ the fraction 25 π over 18 close parenthesis")[$" " 4 ( cos display(frac(25 π, 18)) + i sin display(frac(25 π, 18)) )$] + #math.equation(block: false, alt: "1.97 plus 0.347 i")[$1.97 + 0.347 i$], #math.equation(block: false, alt: "minus 1.286 plus 1.532 i")[$" " − 1.286 + 1.532 i$], #math.equation(block: false, alt: "minus 0.684 minus 1.879 i")[$" " − 0.684 − 1.879 i$] #figure(figph[cube roots of complex number], alt: "cube roots of complex number", caption: none) The square roots of #math.equation(block: false, alt: "minus 2 plus 2 the square root of 3 i")[$− 2 + 2 sqrt(3) i$] Show that any complex number of the form #math.equation(block: false, alt: "z equals cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis")[$z = cos ( θ ) + i sin ( θ )$] lies on the unit circle in the complex plane. #math.equation(block: true, alt: "vertical bar z vertical bar equals vertical bar cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis vertical bar equals the square root of cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$| z | = | cos ( θ ) + i sin ( θ ) | = sqrt(cos^(2) ( θ ) + sin^(2) ( θ )) = 1$] Show that if #math.equation(block: false, alt: "vertical bar z vertical bar equals 1")[$| z | = 1$], then #math.equation(block: false, alt: "the fraction 1 over z equals z bar")[$display(frac(1, z)) = overline(z)$]. + Find three distinct cube roots of 1. + Find four distinct fourth roots of 1. + Find five distinct fifth roots of 1. + Find six distinct sixth roots of 1. + #math.equation(block: false, alt: "1 , open parenthesis cos ⁡ the fraction 2 π over 3 plus i sin ⁡ the fraction 2 π over 3 close parenthesis , open parenthesis cos ⁡ the fraction 4 π over 3 plus i sin ⁡ the fraction 4 π over 3 close parenthesis")[$1 , " " ( cos display(frac(2 π, 3)) + i sin display(frac(2 π, 3)) ) , " " ( cos display(frac(4 π, 3)) + i sin display(frac(4 π, 3)) )$] + #math.equation(block: false, alt: "1 , i , minus 1 , minus i")[$1 , " " i , " " − 1 , " " − i$] + #math.equation(block: false, alt: "1 , open parenthesis cos ⁡ the fraction 2 π over 5 plus i sin ⁡ the fraction 2 π over 5 close parenthesis , open parenthesis cos ⁡ the fraction 4 π over 5 plus i sin ⁡ the fraction 4 π over 5 close parenthesis ,")[$1 , " " ( cos display(frac(2 π, 5)) + i sin display(frac(2 π, 5)) ) , " " ( cos display(frac(4 π, 5)) + i sin display(frac(4 π, 5)) ) ,$] #math.equation(block: false, alt: "open parenthesis cos ⁡ the fraction 6 π over 5 plus i sin ⁡ the fraction 6 π over 5 close parenthesis , open parenthesis cos ⁡ the fraction 8 π over 5 plus i sin ⁡ the fraction 8 π over 5 close parenthesis")[$( cos display(frac(6 π, 5)) + i sin display(frac(6 π, 5)) ) , " " ( cos display(frac(8 π, 5)) + i sin display(frac(8 π, 5)) )$] + #math.equation(block: false, alt: "1 , open parenthesis cos ⁡ the fraction π over 3 plus i sin ⁡ the fraction π over 3 close parenthesis , open parenthesis cos ⁡ the fraction 2 π over 3 plus i sin ⁡ the fraction 2 π over 3 close parenthesis , minus 1 ,")[$1 , " " ( cos display(frac(π, 3)) + i sin display(frac(π, 3)) ) , " " ( cos display(frac(2 π, 3)) + i sin display(frac(2 π, 3)) ) , " " − 1 ,$] #math.equation(block: false, alt: "open parenthesis cos ⁡ the fraction 4 π over 3 plus i sin ⁡ the fraction 4 π over 3 close parenthesis , open parenthesis cos ⁡ the fraction 5 π over 3 plus i sin ⁡ the fraction 5 π over 3 close parenthesis")[$( cos display(frac(4 π, 3)) + i sin display(frac(4 π, 3)) ) , " " ( cos display(frac(5 π, 3)) + i sin display(frac(5 π, 3)) )$] + Find the sum of the three distinct cube roots of 1. (Hint: Plot the roots.) + Find the sum of the four distinct fourth roots of 1. + Find the sum of the five distinct fifth roots of 1. + Find the sum of the six distinct sixth roots of 1. If #math.equation(block: false, alt: "n")[$n$] is a positive integer, define #math.equation(block: true, alt: "ω sub k equals cos ⁡ open parenthesis the fraction 2 π k over n close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π k over n close parenthesis")[$ω_(k) = cos ( display(frac(2 π k, n)) ) + i sin ( display(frac(2 π k, n)) )$] for #math.equation(block: false, alt: "k equals 0 , 1 , 2 , ⋯ , n minus 1.")[$k = 0 , " " 1 , " " 2 , " " ⋯ , " " n − 1.$] Show that #math.equation(block: false, alt: "open parenthesis ω sub k close parenthesis to the power n equals 1")[$( ω_(k) )^(n) = 1$]. (We call #math.equation(block: false, alt: "ω sub k")[$ω_(k)$] an #strong[#math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] root of unity].) #math.equation(block: true, alt: "open parenthesis ω sub k close parenthesis to the power n equals 1 to the power n open parenthesis cos ⁡ open parenthesis n times the fraction 2 π k over n close parenthesis plus i sin ⁡ open parenthesis n times the fraction 2 π k over n close parenthesis close parenthesis equals 1 open parenthesis cos ⁡ 2 π k plus i sin ⁡ 2 π k close parenthesis equals 1")[$( ω_(k) )^(n) = 1^(n) ( cos ( n ⋅ display(frac(2 π k, n)) ) + i sin ( n ⋅ display(frac(2 π k, n)) ) ) = 1 ( cos 2 π k + i sin 2 π k ) = 1$] Let #math.equation(block: false, alt: "ω equals cos ⁡ open parenthesis the fraction 2 π over n close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over n close parenthesis")[$ω = cos ( display(frac(2 π, n)) ) + i sin ( display(frac(2 π, n)) )$], where #math.equation(block: false, alt: "n")[$n$] is a positive integer. Show that the #math.equation(block: false, alt: "n")[$n$] distinct #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] roots of unity are #math.equation(block: false, alt: "ω , ω squared , ω cubed , ⋯ , ω to the power n minus 1")[$ω , " " ω^(2) , " " ω^(3) , " " ⋯ , " " ω^(n − 1)$]. For Problems 55-60, solve the equation. #math.equation(block: true, alt: "z to the power 4 plus 4 z squared plus 8 equals 0")[$z^(4) + 4 z^(2) + 8 = 0$] #math.equation(block: true, alt: "8 to the power 1 / 4 open parenthesis cos ⁡ open parenthesis the fraction 3 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 3 π over 8 close parenthesis close parenthesis")[$8^(1 / 4) ( cos ( display(frac(3 π, 8)) ) + i sin ( display(frac(3 π, 8)) ) )$], #math.equation(block: true, alt: "8 to the power 1 / 4 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 8 close parenthesis close parenthesis")[$" " 8^(1 / 4) ( cos ( display(frac(5 π, 8)) ) + i sin ( display(frac(5 π, 8)) ) )$], #math.equation(block: true, alt: "8 to the power 1 / 4 open parenthesis cos ⁡ open parenthesis the fraction 11 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 11 π over 8 close parenthesis close parenthesis")[$" " 8^(1 / 4) ( cos ( display(frac(11 π, 8)) ) + i sin ( display(frac(11 π, 8)) ) )$], #math.equation(block: true, alt: "8 to the power 1 / 4 open parenthesis cos ⁡ open parenthesis the fraction 13 π over 8 close parenthesis plus i sin ⁡ open parenthesis the fraction 13 π over 8 close parenthesis close parenthesis")[$" " 8^(1 / 4) ( cos ( display(frac(13 π, 8)) ) + i sin ( display(frac(13 π, 8)) ) )$] #math.equation(block: true, alt: "z to the power 6 plus 4 z cubed plus 8 equals 0")[$z^(6) + 4 z^(3) + 8 = 0$] #math.equation(block: true, alt: "z to the power 6 minus 8 equals 0")[$z^(6) − 8 = 0$] #math.equation(block: true, alt: "the square root of 2")[$sqrt(2)$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(π, 3)) ) + i sin ( display(frac(π, 3)) ) )$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(2 π, 3)) ) + i sin ( display(frac(2 π, 3)) ) )$], #math.equation(block: true, alt: "minus the square root of 2")[$" " − sqrt(2)$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 4 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 4 π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(4 π, 3)) ) + i sin ( display(frac(4 π, 3)) ) )$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(5 π, 3)) ) + i sin ( display(frac(5 π, 3)) ) )$] #math.equation(block: true, alt: "z to the power 4 minus 9 i equals 0")[$z^(4) − 9 i = 0$] #math.equation(block: true, alt: "z to the power 4 plus 2 z squared plus 4 equals 0")[$z^(4) + 2 z^(2) + 4 = 0$] #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction π over 3 close parenthesis close parenthesis")[$sqrt(2) ( cos ( display(frac(π, 3)) ) + i sin ( display(frac(π, 3)) ) )$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(2 π, 3)) ) + i sin ( display(frac(2 π, 3)) ) )$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 4 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 4 π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(4 π, 3)) ) + i sin ( display(frac(4 π, 3)) ) )$], #math.equation(block: true, alt: "the square root of 2 open parenthesis cos ⁡ open parenthesis the fraction 5 π over 3 close parenthesis plus i sin ⁡ open parenthesis the fraction 5 π over 3 close parenthesis close parenthesis")[$" " sqrt(2) ( cos ( display(frac(5 π, 3)) ) + i sin ( display(frac(5 π, 3)) ) )$] #math.equation(block: true, alt: "z to the power 4 minus 2 z squared plus 4 equals 0")[$z^(4) − 2 z^(2) + 4 = 0$] + Let #math.equation(block: false, alt: "z equals cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis")[$z = cos ( θ ) + i sin ( θ )$]. Compute #math.equation(block: false, alt: "z squared")[$z^(2)$] by expanding the product. + Use De Moivre's theorem to compute #math.equation(block: false, alt: "z squared")[$z^(2)$]. + Compare your answers to (a) and (b) to write identities for #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$]. + #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis plus open parenthesis 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis close parenthesis i")[$cos^(2) ( θ ) − sin^(2) ( θ ) + ( 2 sin ( θ ) cos ( θ ) ) i$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis plus i sin ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ ) + i sin ( 2 θ )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis ; cos ⁡ open parenthesis 2 θ close parenthesis equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis")[$sin ( 2 θ ) = 2 sin ( θ ) cos ( θ ) ; " " " " cos ( 2 θ ) = cos^(2) ( θ ) − sin^(2) ( θ )$] + Let #math.equation(block: false, alt: "z equals cos ⁡ open parenthesis θ close parenthesis plus i sin ⁡ open parenthesis θ close parenthesis")[$z = cos ( θ ) + i sin ( θ )$]. Compute #math.equation(block: false, alt: "z cubed")[$z^(3)$] by expanding the product. + Use De Moivre's theorem to compute #math.equation(block: false, alt: "z cubed")[$z^(3)$]. + Compare your answers to (a) and (b) to write identities for #math.equation(block: false, alt: "sin ⁡ open parenthesis 3 θ close parenthesis")[$sin ( 3 θ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 3 θ close parenthesis")[$cos ( 3 θ )$]. Problems 63 and 64 show that multiplication by #math.equation(block: false, alt: "i")[$i$] results in a rotation of #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. Suppose that #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] and that the real numbers #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are both nonzero. + What is the slope of the segment in the complex plane joining the origin to #math.equation(block: false, alt: "z")[$z$]? + What is the slope of the segment in the complex plane joining the origin to #math.equation(block: false, alt: "z i")[$z i$]? + What is the product of the slopes of the two segments from parts (a) and (b)? What can you conclude about the angle between the two segments? + #math.equation(block: false, alt: "the fraction b over a")[$display(frac(b, a))$] + #math.equation(block: false, alt: "the fraction minus a over b")[$display(frac(− a, b))$] + #math.equation(block: false, alt: "minus 1")[$− 1$], #math.equation(block: false, alt: "the fraction π over 2")[$" " display(frac(π, 2))$] Suppose that #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] and that #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are both real numbers. + If #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$] and #math.equation(block: false, alt: "b equals 0")[$b = 0$], then what is the slope of the segment in the complex plane joining the origin to #math.equation(block: false, alt: "z")[$z$]? What is the slope of the segment joining the origin to #math.equation(block: false, alt: "z i")[$z i$]? + If #math.equation(block: false, alt: "a equals 0")[$a = 0$] and #math.equation(block: false, alt: "b not equal to 0")[$b ≠ 0$], then what is the slope of the segment in the complex plane joining the origin to #math.equation(block: false, alt: "z")[$z$]? What is the slope of the segment joining the origin to #math.equation(block: false, alt: "z i")[$z i$]? + What can you conclude about the angle between the two segments from parts (a) and (b)? Prove the product rule by following the steps. + Suppose #math.equation(block: false, alt: "z sub 1 equals a plus b i")[$z_(1) = a + b i$] and #math.equation(block: false, alt: "z sub 2 equals c plus d i")[$z_(2) = c + d i$]. Compute #math.equation(block: false, alt: "z sub 1 z sub 2")[$z_(1) z_(2)$]. + Now suppose that #math.equation(block: false, alt: "z sub 1 equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z_(1) = r ( cos ( α ) + i sin ( α ) )$] and #math.equation(block: false, alt: "z sub 2 equals R open parenthesis cos ⁡ open parenthesis β close parenthesis plus i sin ⁡ open parenthesis β close parenthesis close parenthesis")[$z_(2) = R ( cos ( β ) + i sin ( β ) )$]. Write #math.equation(block: false, alt: "a , b , c")[$a , " " b , " " c$] and #math.equation(block: false, alt: "d")[$d$] in terms of #math.equation(block: false, alt: "r , R , α")[$r , " " R , " " α$] and #math.equation(block: false, alt: "β")[$β$]. + Substitute your expressions for #math.equation(block: false, alt: "a , b , c")[$a , " " b , " " c$] and #math.equation(block: false, alt: "d")[$d$] into your formula for #math.equation(block: false, alt: "z sub 1 z sub 2")[$z_(1) z_(2)$]. + Use the laws of sines and cosines to simplify your answer to part (c). + #math.equation(block: false, alt: "z sub 1 z sub 2 equals open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i")[$z_(1) z_(2) = ( a c − b d ) + ( a d + b c ) i$] + #math.equation(block: false, alt: "a equals r cos ⁡ open parenthesis α close parenthesis , b equals r sin ⁡ open parenthesis α close parenthesis , c equals R cos ⁡ open parenthesis β close parenthesis , d equals R sin ⁡ open parenthesis β close parenthesis")[$a = r cos ( α ) , " " b = r sin ( α ) , " " c = R cos ( β ) , " " d = R sin ( β )$] + #math.equation(block: false, alt: "open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i equals open parenthesis r R cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus r R sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis close parenthesis")[$( a c − b d ) + ( a d + b c ) i = ( r R cos ( α ) cos ( β ) − r R sin ( α ) sin ( β ) )$] #math.equation(block: false, alt: "plus open parenthesis r R cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis plus r R sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis close parenthesis i")[$+ ( r R cos ( α ) sin ( β ) + r R sin ( α ) cos ( β ) ) i$] + #math.equation(block: false, alt: "r R open parenthesis cos ⁡ open parenthesis α plus β close parenthesis plus i sin ⁡ open parenthesis α plus β close parenthesis close parenthesis")[$r R ( cos ( α + β ) + i sin ( α + β ) )$] Let #math.equation(block: false, alt: "z sub 1 equals r open parenthesis cos ⁡ open parenthesis α close parenthesis plus i sin ⁡ open parenthesis α close parenthesis close parenthesis")[$z_(1) = r ( cos ( α ) + i sin ( α ) )$] and #math.equation(block: false, alt: "z sub 2 equals R open parenthesis cos ⁡ open parenthesis β close parenthesis plus i sin ⁡ open parenthesis β close parenthesis close parenthesis")[$z_(2) = R ( cos ( β ) + i sin ( β ) )$]. Prove the quotient rule as follows: Set #math.equation(block: false, alt: "w equals the fraction r over R open parenthesis cos ⁡ open parenthesis α minus β close parenthesis plus i sin ⁡ open parenthesis α minus β close parenthesis close parenthesis ,")[$w = display(frac(r, R)) ( cos ( α − β ) + i sin ( α − β ) ) ,$] and show that #math.equation(block: false, alt: "z sub 1 equals w z sub 2")[$z_(1) = w z_(2)$].