#set document(title: "10.3 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")) == 10.3#h(0.6em)Complex Numbers So far most of your work in mathematics has been done using the set of #strong[real numbers]. We often represent the real numbers by a number line, because they can be matched up one-for-one with the points on the line. Every real number is either rational or irrational, and can be expressed as a decimal number, although irrational numbers are non-repeating, non-terminating decimals. However, the real numbers are actually a subset of a larger set of numbers called the #strong[complex numbers]. You may have first encountered complex numbers as solutions of certain quadratic equations. For example, the graph of #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 2 x plus 2")[$f ( x ) = x^(2) − 2 x + 2$] has no #math.equation(block: false, alt: "x")[$x$]-intercepts (as shown at right), because the equation #math.equation(block: false, alt: "x squared minus 2 x plus 2 minus 0")[$" " " " x^(2) − 2 x + 2 − 0 " " " "$] has no real-valued solutions. #figure(figph[triangle], alt: "triangle", caption: none) Applying the quadratic formula, we find #math.equation(block: true, alt: "x equals the fraction minus open parenthesis minus 2 close parenthesis plus or minus the square root of open parenthesis minus 2 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis 2 close parenthesis over 2 equals the fraction 2 plus or minus the square root of minus 4 over 2")[$x = display(frac(− ( − 2 ) ± sqrt(( − 2 )^(2) − 4 ( 1 ) ( 2 )), 2)) = display(frac(2 ± sqrt(− 4), 2))$] The solutions of the equation are #math.equation(block: false, alt: "the fraction 2 plus the square root of minus 4 over 2")[$display(frac(2 + sqrt(− 4), 2))$] and #math.equation(block: false, alt: "the fraction 2 minus the square root of minus 4 over 2")[$display(frac(2 − sqrt(− 4), 2))$], but they are not real numbers. Because #math.equation(block: false, alt: "the square root of minus 4")[$sqrt(− 4)$] is not a real number, the equation #math.equation(block: false, alt: "x squared minus 2 x plus 2 minus 0")[$x^(2) − 2 x + 2 − 0$] has no real solutions. === Imaginary Numbers Although square roots of negative numbers such as #math.equation(block: false, alt: "the square root of minus 4")[$sqrt(− 4)$] are not real numbers, they occur often in mathematics and its applications. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. René Descartes gave them the name #strong[imaginary numbers], which reflected the mistrust with which mathematicians regarded them at the time. Today, however, such numbers are well understood and used routinely by scientists and engineers. We begin by defining a new number, called #strong[#math.equation(block: false, alt: "i")[$i$]], whose square is #math.equation(block: false, alt: "minus 1")[$− 1$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Imaginary Unit] We define the imaginary unit, #strong[#math.equation(block: false, alt: "i")[$i$]], by #math.equation(block: true, alt: "i squared equals minus 1 or i equals the square root of minus 1")[$bold(italic(i))^(2) = − 1 " " " " " " " " " " " " "or" " " " " " " " " " " " " bold(italic(i)) = sqrt(− 1)$] ] The letter #strong[#math.equation(block: false, alt: "i")[$i$]] used in this way is not a variable, it is the name of a specific number, and hence is a constant. The square root of any negative number can be written as the product of a real number and #strong[#math.equation(block: false, alt: "i")[$i$]]. For example, #math.equation(block: true, alt: "the square root of minus 4, equals the square root of minus 1 times 4; equals the square root of minus 1 the square root of 4 equals i times 2")[$sqrt(− 4) & = sqrt(− 1 ⋅ 4) \ & = sqrt(− 1) sqrt(4) = i ⋅ 2$] or #math.equation(block: false, alt: "the square root of minus 4 equals 2 i")[$sqrt(− 4) = 2 i$]. Any number that is the product of #strong[#math.equation(block: false, alt: "i")[$i$]] and a real number is called an #strong[imaginary number]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Imaginary Numbers] For any real number #math.equation(block: false, alt: "a greater than 0")[$a > 0$], #math.equation(block: true, alt: "the square root of minus a equals the square root of minus 1 the square root of a equals i the square root of a")[$sqrt(− bold(italic(a))) = sqrt(− 1) sqrt(bold(italic(a))) = bold(italic(i)) sqrt(bold(italic(a)))$] ] Examples of imaginary numbers are #math.equation(block: false, alt: "3 i , the fraction 7 over 8 i , minus 38 i ,")[$3 i , " " display(frac(7, 8)) i , " " − 38 i ,$] and #math.equation(block: false, alt: "i the square root of 5")[$i sqrt(5)$]. #examplebox("Example 1")[][ Write each radical as an imaginary number. + #math.equation(block: false, alt: "the square root of minus 25")[$sqrt(− 25)$] + #math.equation(block: false, alt: "2 the square root of minus 3")[$2 sqrt(− 3)$] #solutionbox[ + #math.equation(block: false, alt: "the square root of minus 25 equals the square root of minus 1 the square root of 25 equals i the square root of 25 equals 5 i")[$sqrt(− 25) = sqrt(− 1) sqrt(25) = i sqrt(25) = 5 i$] + #math.equation(block: false, alt: "2 the square root of minus 3 equals the square root of minus 1 the square root of 3 equals 2 i the square root of 3")[$2 sqrt(− 3) = sqrt(− 1) sqrt(3) = 2 i sqrt(3)$] ] ] Write each radical as an imaginary number. + #math.equation(block: false, alt: "the square root of minus 18")[$sqrt(− 18)$] + #math.equation(block: false, alt: "minus 6 the square root of minus 5")[$− 6 sqrt(− 5)$] + #math.equation(block: false, alt: "3 i the square root of 2")[$3 i sqrt(2)$] + #math.equation(block: false, alt: "minus 6 i the square root of 5")[$− 6 i sqrt(5)$] Just as each positive number has two real-valued square roots, every negative number has two imaginary square roots. For example, the two square roots of #math.equation(block: false, alt: "minus 9")[$− 9$] are #math.equation(block: false, alt: "3 i")[$3 i$] and #math.equation(block: false, alt: "minus 3 i")[$− 3 i$]. === Complex Numbers Consider the quadratic equation #math.equation(block: true, alt: "x squared minus 2 x plus 5 equals 0")[$x^(2) − 2 x + 5 = 0$] Using the quadratic formula to solve the equation, we find #math.equation(block: true, alt: "x equals the fraction minus open parenthesis minus 2 close parenthesis plus or minus the square root of open parenthesis minus 2 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis 5 close parenthesis over 2 equals the fraction 2 plus or minus the square root of minus 16 over 2")[$x = display(frac(− ( − 2 ) ± sqrt(( − 2 )^(2) − 4 ( 1 ) ( 5 )), 2)) = display(frac(2 ± sqrt(− 16), 2))$] If we now replace #math.equation(block: false, alt: "the square root of minus 16")[$sqrt(− 16)$] by #math.equation(block: false, alt: "4 i")[$4 i$], we have #math.equation(block: true, alt: "x equals the fraction 2 plus or minus 4 i over 2 equals 1 plus or minus 2 i")[$x = display(frac(2 ± 4 i, 2)) = 1 ± 2 i$] The two solutions are #math.equation(block: false, alt: "1 plus 2 i")[$1 + 2 i$] and #math.equation(block: false, alt: "1 minus 2 i")[$1 − 2 i$]. These are examples of #strong[complex numbers]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Complex Numbers] A #strong[complex number] can be written in the form #math.equation(block: true, alt: "z equals a plus b i")[$bold(italic(z)) = bold(italic(a)) + bold(italic(b)) bold(italic(i))$] where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are real numbers. ] Examples of complex numbers are #math.equation(block: true, alt: "3 minus 5 i , 2 plus the square root of 7 i , the fraction 4 minus i over 3 , 6 i , and minus 9")[$3 − 5 i , " " 2 + sqrt(7) i , " " display(frac(4 − i, 3)) , " " 6 i , " " " " "and" " " " " − 9$] In a complex number #math.equation(block: false, alt: "a plus b i")[$a + b i$], #math.equation(block: false, alt: "a")[$a$] is called the #strong[real part], and #math.equation(block: false, alt: "b")[$b$] is called the #strong[imaginary part]. All real numbers are also complex numbers (with imaginary part equal to zero). A complex number whose real part equals 0 is called a #strong[pure imaginary number]. The real and imaginary parts of a complex number cannot be combined. Thus, two complex numbers #math.equation(block: false, alt: "z sub 1")[$z_(1)$] and #math.equation(block: false, alt: "z sub 2")[$z_(2)$] are equal if and only if their real parts are equal and their imaginary parts are equal. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Equality of Complex Numbers] Two complex numbers #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")[$" " z_(2) = c + d i " "$] are equal if and only if #math.equation(block: true, alt: "a equals c and c equals d")[$a = c " " " " "and" " " " " c = d$] ] #examplebox("Example 2")[][ The solutions of the equation #math.equation(block: false, alt: "x squared minus 2 x plus 2 equals 0")[$" " x^(2) − 2 x + 2 = 0 " "$] are #math.equation(block: false, alt: "the fraction 2 plus or minus the square root of minus 4 over 2")[$display(frac(2 ± sqrt(− 4), 2))$]. Write these solutions as complex numbers, and simplify. #solutionbox[ Because #math.equation(block: false, alt: "the square root of minus 4 equals the square root of minus 1 the square root of 4 equals 2 i ,")[$" " sqrt(− 4) = sqrt(− 1) sqrt(4) = 2 i ,$] we have #math.equation(block: false, alt: "the fraction 2 plus or minus the square root of minus 4 over 2 equals the fraction 2 plus or minus 2 i over 2")[$" " display(frac(2 ± sqrt(− 4), 2)) = display(frac(2 ± 2 i, 2))$], or #math.equation(block: false, alt: "1 plus or minus i")[$1 ± i$]. The solutions are #math.equation(block: false, alt: "1 plus i")[$1 + i$] and #math.equation(block: false, alt: "1 minus i")[$1 − i$]. ] ] Use extraction of roots to solve #math.equation(block: false, alt: "open parenthesis 2 x plus 1 close parenthesis squared plus 9 equals 0")[$" " ( 2 x + 1 )^(2) + 9 = 0$]. Write your answers as complex numbers. #math.equation(block: true, alt: "x equals the fraction minus 1 over 2 plus or minus the fraction 3 over 2 i")[$x = display(frac(− 1, 2)) ± display(frac(3, 2)) i$] === Arithmetic of Complex Numbers We add and subtract complex numbers by combining their real and imaginary parts separately. For example, #math.equation(block: true, alt: "open parenthesis 4 plus 5 i close parenthesis plus open parenthesis 2 minus 3 i close parenthesis, equals open parenthesis 4 plus 2 close parenthesis plus open parenthesis 5 minus 3 close parenthesis i; equals 6 plus 2 i")[$( 4 + 5 i ) + ( 2 − 3 i ) & = ( 4 + 2 ) + ( 5 − 3 ) i \ & = 6 + 2 i$] The algebraic form of this rule can be stated as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sum and Difference of Complex Numbers] The sum of two complex numbers #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")[$z_(2) = c + d i$] is #math.equation(block: true, alt: "z sub 1 plus z sub 2 equals open parenthesis a plus b i close parenthesis plus open parenthesis c plus d i close parenthesis equals open parenthesis a plus c close parenthesis plus open parenthesis b plus d close parenthesis i")[$bold(italic(z))_(1) + bold(italic(z))_(2) = ( bold(italic(a)) + bold(italic(b)) bold(italic(i)) ) + ( bold(italic(c)) + bold(italic(d)) bold(italic(i)) ) = ( bold(italic(a)) + bold(italic(c)) ) + ( bold(italic(b)) + bold(italic(d)) ) bold(italic(i))$] Their difference is #math.equation(block: true, alt: "z sub 1 minus z sub 2 equals open parenthesis a plus b i close parenthesis minus open parenthesis c plus d i close parenthesis equals open parenthesis a minus c close parenthesis plus open parenthesis b minus d close parenthesis i")[$bold(italic(z))_(1) − bold(italic(z))_(2) = ( bold(italic(a)) + bold(italic(b)) bold(italic(i)) ) − ( bold(italic(c)) + bold(italic(d)) bold(italic(i)) ) = ( bold(italic(a)) − bold(italic(c)) ) + ( bold(italic(b)) − bold(italic(d)) ) bold(italic(i))$] ] #examplebox("Example 3")[][ Subtract: #math.equation(block: false, alt: "open parenthesis 8 minus 6 i close parenthesis minus open parenthesis 5 plus 2 i close parenthesis")[$" " ( 8 − 6 i ) − ( 5 + 2 i )$]. #solutionbox[ Combine the real and imaginary parts. #math.equation(block: true, alt: "open parenthesis 8 minus 6 i close parenthesis minus open parenthesis 5 plus 2 i close parenthesis, equals open parenthesis 8 minus 5 close parenthesis plus open parenthesis minus 6 minus 2 close parenthesis i; equals 3 plus open parenthesis minus 8 close parenthesis i equals 3 minus 8 i")[$( 8 − 6 i ) − ( 5 + 2 i ) & = ( 8 − 5 ) + ( − 6 − 2 ) i \ & = 3 + ( − 8 ) i = 3 − 8 i$] ] ] Subtract: #math.equation(block: false, alt: "open parenthesis minus 3 plus 2 i close parenthesis minus open parenthesis minus 3 minus 2 i close parenthesis")[$" " ( − 3 + 2 i ) − ( − 3 − 2 i )$] #math.equation(block: true, alt: "4 i")[$4 i$] === Products of Complex Numbers To find the product of two imaginary numbers, we use the fact that #math.equation(block: false, alt: "i squared equals minus 1")[$i^(2) = − 1$]. For example, #math.equation(block: true, alt: "open parenthesis 3 i close parenthesis open parenthesis 4 i close parenthesis, equals 3 times 4 i squared; equals 12 open parenthesis minus 1 close parenthesis equals minus 12")[$( 3 i ) ( 4 i ) & = 3 ⋅ 4 i^(2) \ & = 12 ( − 1 ) = − 12$] To find the product of two complex numbers, we use the distributive law, as if the numbers were binomials. #examplebox("Example 4")[][ Multiply #math.equation(block: false, alt: "open parenthesis 2 plus 3 i close parenthesis open parenthesis 3 minus 5 i close parenthesis")[$" " ( 2 + 3 i ) ( 3 − 5 i )$]. #solutionbox[ We multiply each term of the first complex number by each term of the second one. #math.equation(block: true, alt: "open parenthesis 2 plus 3 i close parenthesis open parenthesis 3 minus 5 i close parenthesis equals 6 minus 10 i plus 9 i minus 15 i squared")[$( 2 + 3 i ) ( 3 − 5 i ) = 6 − 10 i + 9 i − 15 i^(2)$] Because #math.equation(block: false, alt: "i squared equals minus 1")[$i^(2) = − 1$], the last term, #math.equation(block: false, alt: "minus 15 i squared")[$− 15 i^(2)$], can be replaced by #math.equation(block: false, alt: "minus 15 open parenthesis minus 1 close parenthesis")[$− 15 ( − 1 )$], or #math.equation(block: false, alt: "15")[$15$], to obtain #math.equation(block: true, alt: "6 minus 10 i plus 9 i plus 15")[$6 − 10 i + 9 i + 15$] Finally, we combine the real parts and imaginary parts to obtain #math.equation(block: true, alt: "open parenthesis 6 plus 15 close parenthesis plus open parenthesis minus 10 i plus 9 i close parenthesis equals 21 minus i")[$( 6 + 15 ) + ( − 10 i + 9 i ) = 21 − i$] ] ] Multiply #math.equation(block: false, alt: "open parenthesis minus 3 plus 2 i close parenthesis open parenthesis minus 3 minus 2 i close parenthesis")[$" " ( − 3 + 2 i ) ( − 3 − 2 i )$]. #math.equation(block: true, alt: "13")[$13$] In the Homework Probems, you will verify that the following rule holds. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Product of Complex Numbers] The product of two complex numbers #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")[$" " z_(2) = c + d i " "$] is #math.equation(block: true, alt: "z sub 1 z sub 2 equals open parenthesis a plus b i close parenthesis open parenthesis c plus d i close parenthesis equals open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i")[$bold(italic(z))_(1) bold(italic(z))_(2) = ( bold(italic(a)) + bold(italic(b)) bold(italic(i)) ) ( bold(italic(c)) + bold(italic(d)) bold(italic(i)) ) = ( bold(italic(a)) bold(italic(c)) − bold(italic(b)) bold(italic(d)) ) + ( bold(italic(a)) bold(italic(d)) + bold(italic(b)) bold(italic(c)) ) bold(italic(i))$] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ One property of real numbers that is #emph[not] true of complex numbers is #math.equation(block: false, alt: "the square root of a b equals the square root of a the square root of b")[$sqrt(a b) = sqrt(a) sqrt(b)$]. This identity fails when #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are both negative. For example, if #math.equation(block: false, alt: "a equals b equals minus 2")[$a = b = − 2$], we have #math.equation(block: true, alt: "the square root of a b equals the square root of open parenthesis minus 2 close parenthesis open parenthesis minus 2 close parenthesis equals the square root of 4 equals 2")[$sqrt(a b) = sqrt(( − 2 ) ( − 2 )) = sqrt(4) = 2$] but #math.equation(block: true, alt: "the square root of a times the square root of b, equals the square root of open parenthesis minus 2 close parenthesis times the square root of open parenthesis minus 2 close parenthesis; equals the square root of open parenthesis minus 1 times 2 close parenthesis times the square root of open parenthesis minus 1 times 2 close parenthesis; equals i the square root of 2 times i the square root of 2 equals i squared open parenthesis the square root of 2 close parenthesis squared equals minus 2")[$sqrt(a) ⋅ sqrt(b) & = sqrt(( − 2 )) ⋅ sqrt(( − 2 )) \ & = sqrt(( − 1 ⋅ 2 )) ⋅ sqrt(( − 1 ⋅ 2 )) \ & = i sqrt(2) ⋅ i sqrt(2) = i^(2) ( sqrt(2) )^(2) = − 2$] so #math.equation(block: false, alt: "the square root of a b not equal to the square root of a the square root of b")[$" " sqrt(a b) ≠ sqrt(a) sqrt(b)$]. We can avoid possible errors by writing square roots of negative numbers as imaginary numbers before attempting any calculations. ] === Quotients of Complex Numbers To find the quotient of two complex numbers, we use the technique of rationalizing the denominator. First consider division by a pure imaginary number. #examplebox("Example 5")[][ Divide #math.equation(block: false, alt: "the fraction 3 plus 4 i over 2 i")[$" " " " display(frac(3 + 4 i, 2 i))$]. #solutionbox[ Because #math.equation(block: false, alt: "i")[$i$] is really a radical (remember that #math.equation(block: false, alt: "i equals the square root of minus 1")[$i = sqrt(− 1)$]), we multiply the numerator and denominator of the quotient by #math.equation(block: false, alt: "i")[$i$] to obtain #math.equation(block: true, alt: "the fraction open parenthesis 3 plus 4 i close parenthesis times i over 2 i times i, equals the fraction 3 i plus 4 i squared over 2 i squared, Apply the distributive law.; equals the fraction 3 i minus 4 over minus 2, Recall that i squared equals minus 1.")[$display(frac(( 3 + 4 i ) ⋅ i, 2 i ⋅ i)) & = display(frac(3 i + 4 i^(2), 2 i^(2))) & & "Apply the distributive law." \ & = display(frac(3 i − 4, − 2)) & & "Recall that " bold(italic(i))^(2) = − 1.$] To write the quotient in the form #math.equation(block: false, alt: "a plus b i")[$a + b i$], we divide #math.equation(block: false, alt: "minus 2")[$− 2$] into each term of the numerator. #math.equation(block: true, alt: "the fraction 3 i over minus 2 minus the fraction 4 over minus 2 equals the fraction minus 3 over 2 i plus 2")[$display(frac(3 i, − 2)) − display(frac(4, − 2)) = display(frac(− 3, 2)) i + 2$] The quotient is #math.equation(block: false, alt: "2 minus the fraction 3 over 2 i")[$2 − display(frac(3, 2)) i$]. ] ] Divide #math.equation(block: false, alt: "the fraction 8 plus 9 i over 3 i")[$" " " " display(frac(8 + 9 i, 3 i))$]. #math.equation(block: true, alt: "3 minus the fraction 8 over 3 i")[$3 − display(frac(8, 3)) i$] Perhaps you recall that to rationalize a binomial denominator, we multiply by its conjugate. For example, to rationalize the denominator of #math.equation(block: false, alt: "the fraction 5 over 2 plus the square root of 3")[$" " display(frac(5, 2 + sqrt(3)))$], we multiply numerator and denominator by #math.equation(block: false, alt: "2 minus the square root of 3")[$" " 2 − sqrt(3)$]. A similar technique works for dividing complex numbers. We first define the #strong[conjugate] of a complex number. #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Complex Conjugate] For any complex number #math.equation(block: false, alt: "z equals a plus b i")[$" " z = a + b i$], the number #math.equation(block: true, alt: "z bar equals a minus b i")[$overline(bold(italic(z))) = bold(italic(a)) − bold(italic(b)) bold(italic(i))$] is called the #strong[complex conjugate] of #math.equation(block: false, alt: "z")[$z$]. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The conjugate of a complex number has several useful properties. In particular, the product of a nonzero complex number and its conjugate is always a positive real number. #math.equation(block: true, alt: "z z bar, equals open parenthesis a plus b i close parenthesis open parenthesis a minus b i close parenthesis; equals a squared minus b squared i squared equals a squared minus b squared open parenthesis minus 1 close parenthesis equals a squared plus b squared")[$z overline(z) & = ( a + b i ) ( a − b i ) \ & = a^(2) − b^(2) i^(2) = a^(2) − b^(2) ( − 1 ) = a^(2) + b^(2)$] ] To illustrate, we calculate the quotient #math.equation(block: false, alt: "the fraction 5 over 2 plus 3 i")[$" " " " display(frac(5, 2 + 3 i))$]. We multiply numerator and denominator by the complex conjugate of the denominator, #math.equation(block: false, alt: "2 minus 3 i")[$2 − 3 i$], to obtain #math.equation(block: true, alt: "the fraction 5 over 2 plus 3 i times the fraction 2 minus 3 i over 2 minus 3 i")[$display(frac(5, 2 + 3 i)) ⋅ display(frac(2 − 3 i, 2 − 3 i))$] The denominator is then a real number, because #math.equation(block: true, alt: "open parenthesis 2 plus 3 i close parenthesis open parenthesis 2 minus 3 i close parenthesis equals 4 minus 6 i plus 6 i minus 9 i squared equals 4 plus 9 equals 13")[$( 2 + 3 i ) ( 2 − 3 i ) = 4 − 6 i + 6 i − 9 i^(2) = 4 + 9 = 13$] and the quotient is #math.equation(block: true, alt: "the fraction 5 over 2 plus 3 i times the fraction 2 minus 3 i over 2 minus 3 i equals the fraction 10 minus 15 i over 13 equals the fraction 10 over 13 minus the fraction 15 over 13 i")[$display(frac(5, 2 + 3 i)) ⋅ display(frac(2 − 3 i, 2 − 3 i)) = display(frac(10 − 15 i, 13)) = display(frac(10, 13)) − display(frac(15, 13)) i$] #examplebox("Example 6")[][ Divide #math.equation(block: false, alt: "the fraction 2 plus 3 i over 4 minus 2 i")[$" " " " display(frac(2 + 3 i, 4 − 2 i))$]. #solutionbox[ We multiply numerator and denominator by #math.equation(block: false, alt: "4 plus 2 i")[$4 + 2 i$], the complex conjugate of the denominator. #math.equation(block: true, alt: "the fraction 2 plus 3 i over 4 minus 2 i, equals the fraction open parenthesis 2 plus 3 i close parenthesis open parenthesis 4 plus 2 i close parenthesis over open parenthesis 4 minus 2 i close parenthesis open parenthesis 4 plus 2 i close parenthesis, Expand numerator; and denominator.; equals the fraction 8 plus 4 i plus 12 i plus 6 i squared over 16 plus 8 i minus 8 i minus 4 i squared, Replace i squared by minus 1.; equals the fraction 8 plus 16 i minus 6 over 16 minus open parenthesis minus 4 close parenthesis, Combine like terms.; equals the fraction 2 plus 16 i over 20, Divide 20 into each term; of numerator.; equals the fraction 2 over 20 plus the fraction 16 i over 20 equals the fraction 1 over 10 plus the fraction 4 over 5 i")[$display(frac(2 + 3 i, 4 − 2 i)) & = display(frac(( 2 + 3 i ) ( 4 + 2 bold(italic(i)) ), ( 4 − 2 i ) ( 4 + 2 bold(italic(i)) ))) & & " " "Expand numerator " \ #h(1em) "and denominator." \ & = display(frac(8 + 4 i + 12 i + 6 i^(2), 16 + 8 i − 8 i − 4 i^(2))) & & " Replace " " " bold(italic(i))^(2) " " "by" " " − 1 . \ & = display(frac(8 + 16 i − 6, 16 − ( − 4 ))) & & " Combine like terms." \ & = display(frac(2 + 16 i, 20)) & & " " "Divide 20 into each term" \ #h(1em) " of numerator." \ & = display(frac(2, 20)) + display(frac(16 i, 20)) = display(frac(1, 10)) + display(frac(4, 5)) i$] ] ] Write the quotient #math.equation(block: false, alt: "the fraction 4 minus 2 i over 1 plus i")[$" " " " display(frac(4 − 2 i, 1 + i))$] in the form #math.equation(block: false, alt: "a plus b i")[$a + b i$]. #math.equation(block: true, alt: "1 minus 3 i")[$1 − 3 i$] In the Homework Problems you will verify the rule for dividing complex numbers. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quotient of Complex Numbers] The quotient of two complex numbers #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")[$" " z_(2) = c + d i " "$] is #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2, equals the fraction a plus b i over c plus d i; equals the fraction a plus b i over c plus d i times the fraction c minus d i over c minus d i equals the fraction a c plus b d over c squared plus d squared plus the fraction b c minus a d over c squared plus d squared i")[$display(frac(bold(italic(z))_(1), bold(italic(z))_(2))) & = display(frac(bold(italic(a)) + bold(italic(b)) bold(italic(i)), bold(italic(c)) + bold(italic(d)) bold(italic(i)))) \ & = display(frac(bold(italic(a)) + bold(italic(b)) bold(italic(i)), bold(italic(c)) + bold(italic(d)) bold(italic(i)))) ⋅ display(frac(bold(italic(c)) − bold(italic(d)) bold(italic(i)), bold(italic(c)) − bold(italic(d)) bold(italic(i)))) = display(frac(bold(italic(a)) bold(italic(c)) + bold(italic(b)) bold(italic(d)), bold(italic(c))^(2) + bold(italic(d))^(2))) + display(frac(bold(italic(b)) bold(italic(c)) − bold(italic(a)) bold(italic(d)), bold(italic(c))^(2) + bold(italic(d))^(2))) bold(italic(i))$] ] === Graphing Complex Numbers Real numbers can be plotted on a number line, but to graph a complex number we use a plane, called the #strong[complex plane]. In the complex plane, the real numbers lie on the horizontal or #strong[real axis], and pure imaginary numbers lie on the vertical or #strong[imaginary axis]. To plot a complex number #math.equation(block: false, alt: "a plus b i")[$a + b i$] we move #math.equation(block: false, alt: "a")[$a$] units from the origin in the horizontal direction and #math.equation(block: false, alt: "b")[$b$] units in the vertical direction. #examplebox("Example 7")[][ Plot the numbers #math.equation(block: false, alt: "z equals 2 plus 3 i")[$" " z = 2 + 3 i " "$] and #math.equation(block: false, alt: "z bar equals 2 minus 3 i")[$" " overline(z) = 2 − 3 i " "$] in the complex plane. #solutionbox[ To plot #math.equation(block: false, alt: "z equals 2 plus 3 i")[$" " z = 2 + 3 i$], we move from the origin 2 units to the right and 3 units up. To plot #math.equation(block: false, alt: "z bar equals 2 minus 3 i")[$" " overline(z) = 2 − 3 i$] we move from the origin 2 units to the right and 3 units down. The numbers #math.equation(block: false, alt: "2 plus 3 i")[$2 + 3 i$] and #math.equation(block: false, alt: "2 minus 3 i")[$2 − 3 i$] are plotted in the figure at right. #figure(figph[complex conjugates], alt: "complex conjugates", caption: none) ] ] Plot #math.equation(block: false, alt: "z equals minus 3 plus 5 i")[$" " z = − 3 + 5 i " "$] and its conjugate in the complex plane. #figure(figph[complex conjugates], alt: "complex conjugates", caption: none) #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ You are used to plotting ordered pairs #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] in the Cartesian plane, so that one point represents two real numbers, given by its coordinates. However, in the complex plane one point represents a single complex number #math.equation(block: false, alt: "z equals a plus b i")[$" " z = a + b i$], and the coordinates of the point are the real and imaginary parts of #math.equation(block: false, alt: "z")[$z$]. For example, the point #math.equation(block: false, alt: "open parenthesis 2 , minus 6 close parenthesis")[$( 2 , − 6 )$] in the real plane represents the pair of real numbers #math.equation(block: false, alt: "x equals 2")[$x = 2$] and #math.equation(block: false, alt: "y equals minus 6")[$y = − 6$], but the point #math.equation(block: false, alt: "open parenthesis 2 , minus 6 close parenthesis")[$( 2 , − 6 )$] in the complex plane represents the complex number #math.equation(block: false, alt: "z equals 2 minus 6 i")[$z = 2 − 6 i$]. ] The #strong[modulus], or length, of a complex number is its distance from the origin in the complex plane. The modulus of a complex number is analogous to the absolute value of a real number, and is denoted by #math.equation(block: false, alt: "vertical bar z vertical bar")[$| z |$]. If #math.equation(block: false, alt: "z equals a plus b i")[$" " z = a + b i$], we can use the Pythagorean theorem to compute its modulus. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Modulus of a Complex Number] The #strong[modulus] of a complex number #math.equation(block: false, alt: "z equals a plus b i")[$" " z = a + b i " "$] is given by #math.equation(block: true, alt: "vertical bar z vertical bar equals the square root of a squared plus b squared")[$| bold(italic(z)) | = sqrt(bold(italic(a))^(2) + bold(italic(b))^(2))$] ] For example, the modulus of #math.equation(block: false, alt: "z equals 2 plus 3 i")[$" " z = 2 + 3 i " "$] is #math.equation(block: true, alt: "vertical bar z vertical bar equals the square root of 2 squared plus 3 squared equals the square root of 13")[$| z | = sqrt(2^(2) + 3^(2)) = sqrt(13)$] #examplebox("Example 8")[][ Plot all points in the complex plane that satisfy the equation #math.equation(block: false, alt: "vertical bar z vertical bar equals 4")[$| z | = 4$]. #solutionbox[ The points that satisfy #math.equation(block: false, alt: "vertical bar z vertical bar equals 4")[$| z | = 4$] all lie at a distance of 4 units from the origin. In other words, they lie on a circle of radius 4 centered at the origin, as shown at right. #figure(figph[circle], alt: "circle", caption: none) ] ] Write an equation for the circle of radius 1 centered at the origin in the complex plane. #math.equation(block: true, alt: "vertical bar z vertical bar equals 1")[$| z | = 1$] We can think of the graph of a complex number #math.equation(block: false, alt: "z equals a plus b i")[$" " z = a + b i " "$] as the vector (or arrow) that starts at the origin and ends at the point #math.equation(block: false, alt: "open parenthesis a , b close parenthesis")[$( a , b )$] in the complex plane, as shown in figure (a). #figure(figph[complex numbers as vectors], alt: "complex numbers as vectors", caption: none) Then the sum of two complex numbers corresponds to the sum of the two vectors representing them. Figure (b) illustrates the sum of two complex numbers, #math.equation(block: false, alt: "z sub 1")[$z_(1)$] and #math.equation(block: false, alt: "z sub 2")[$z_(2)$], by vector addition using the parallelogram law: we form a parallelogram with the vectors #math.equation(block: false, alt: "z sub 1")[$z_(1)$] and #math.equation(block: false, alt: "z sub 2")[$z_(2)$] and as adjacent sides. Their sum is the vector that forms the diagonal of the parallelogram, starting at the origin. #examplebox("Example 9")[][ Illustrate the sum of #math.equation(block: false, alt: "z sub 1 equals 5 minus i")[$" " z_(1) = 5 − i " "$] and #math.equation(block: false, alt: "z sub 2 equals minus 7 plus 4 i")[$" " z_(2) = − 7 + 4 i " "$] by vector addition. #solutionbox[ Draw a vector starting at the origin and ending at the point #math.equation(block: false, alt: "open parenthesis 5 , minus 1 close parenthesis")[$( 5 , − 1 )$] in the complex plane to represent #math.equation(block: false, alt: "z sub 1")[$z_(1)$], and another vector ending at #math.equation(block: false, alt: "open parenthesis minus 7 , 4 close parenthesis")[$( − 7 , 4 )$] to represent #math.equation(block: false, alt: "z sub 2")[$z_(2)$]. Now sketch the opposite sides of a parallelogram, and draw the diagonal starting at the origin, as shown at right. #figure(figph[complex numbers as vectors], alt: "complex numbers as vectors", caption: none) This vector represents the sum #math.equation(block: false, alt: "z sub 1 plus z sub 2")[$z_(1) + z_(2)$]. You can check that the vector ends at the point #math.equation(block: true, alt: "z sub 1 plus z sub 2 equals open parenthesis 5 minus 7 close parenthesis plus open parenthesis minus 1 plus 4 close parenthesis i equals minus 2 plus 3 i")[$z_(1) + z_(2) = ( 5 − 7 ) + ( − 1 + 4 ) i = − 2 + 3 i$] ] ] To visualize subtraction, we add the opposite of the second vector, because #math.equation(block: true, alt: "z sub 1 minus z sub 2 equals z sub 1 plus open parenthesis minus z sub 2 close parenthesis")[$z_(1) − z_(2) = z_(1) + ( − z_(2) )$] Recall that the opposite of a vector has the same length, but it points in the opposite direction. Illustrate #math.equation(block: false, alt: "z sub 1 minus z sub 2")[$z_(1) − z_(2)$], for #math.equation(block: false, alt: "z sub 1 equals 3 minus 2 i")[$" " z_(1) = 3 − 2 i " "$] and #math.equation(block: false, alt: "z sub 2 equals minus 4 plus 5 i")[$" " z_(2) = − 4 + 5 i " "$]. #figure(figph[complex numbers as vectors], alt: "complex numbers as vectors", caption: none) === Zeros of Polynomials A polynomial with real-number coefficients may or may not have real-valued zeros. For example, the polynomial #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power 4 plus 4")[$" " f ( x ) = x^(4) + 4 " "$] has no real-valued zeros. But a polynomial always has a zero if we allow complex numbers as inputs. Because we can add, subtract, and multiply any two complex numbers, we can evaluate a polynomial function at a complex number. Thus, we can extend the domain of any polynomial to include all complex numbers. #examplebox("Example 10")[][ Evaluate the polynomial #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 2 x plus 2")[$" " f ( x ) = x^(2) − 2 x + 2 " "$] for #math.equation(block: false, alt: "x equals 1 plus i")[$x = 1 + i$], and simplify. #solutionbox[ We substitute #math.equation(block: false, alt: "x equals 1 plus i")[$x = 1 + i$], to find #math.equation(block: true, alt: "f open parenthesis 1 plus i close parenthesis, equals open parenthesis 1 plus i close parenthesis squared minus 2 open parenthesis 1 plus i close parenthesis plus 2; equals 1 squared plus 2 i plus i squared minus 2 minus 2 i plus 2; equals 1 plus 2 i plus open parenthesis minus 1 close parenthesis minus 2 minus 2 i plus 2")[$f ( 1 + i ) & = ( 1 + i )^(2) − 2 ( 1 + i ) + 2 \ & = 1^(2) + 2 i + i^(2) − 2 − 2 i + 2 \ & = 1 + 2 i + ( − 1 ) − 2 − 2 i + 2$] Thus, #math.equation(block: false, alt: "f open parenthesis 1 plus i close parenthesis equals 0")[$" " f ( 1 + i ) = 0$], so #math.equation(block: false, alt: "1 plus i")[$1 + i$] is a zero of the polynomial. ] ] If #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 6 x plus 13")[$" " f ( x ) = x^(2) − 6 x + 13$], evaluate #math.equation(block: false, alt: "f open parenthesis 3 plus 2 i close parenthesis")[$f ( 3 + 2 i )$]. #math.equation(block: true, alt: "0")[$0$] The zeros of a quadratic polynomial #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$" " f ( x ) = a x^(2) + b x + c " "$], of course, are the solutions of the quadratic equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 ,")[$" " a x^(2) + b x + c = 0 " " ,$] and those solutions are given by the quadratic formula, #math.equation(block: true, alt: "x equals the fraction minus b over 2 a plus the fraction b squared minus 4 a c over 2 a and x equals the fraction minus b over 2 a minus the fraction b squared minus 4 a c over 2 a")[$x = display(frac(− b, 2 a)) + display(frac(b^(2) − 4 a c, 2 a)) " " " " " " "and" " " " " " " x = display(frac(− b, 2 a)) − display(frac(b^(2) − 4 a c, 2 a))$] If the discriminant #math.equation(block: false, alt: "D equals b squared minus 4 a c")[$" " D = b^(2) − 4 a c " "$] is negative, the two solutions are complex conjugates. For example, the solutions of the equation #math.equation(block: false, alt: "x squared minus 4 x plus 5 equals 0")[$" " x^(2) − 4 x + 5 = 0 " "$] are #math.equation(block: true, alt: "the fraction minus open parenthesis minus 4 close parenthesis plus or minus the square root of open parenthesis minus 4 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis 5 close parenthesis over 2 open parenthesis 1 close parenthesis equals the fraction 4 plus or minus the square root of minus 4 over 2")[$display(frac(− ( − 4 ) ± sqrt(( − 4 )^(2) − 4 ( 1 ) ( 5 )), 2 ( 1 ))) = display(frac(4 ± sqrt(− 4), 2))$] or #math.equation(block: false, alt: "z equals 2 plus i")[$z = 2 + i$] and #math.equation(block: false, alt: "z bar equals 2 minus i")[$overline(z) = 2 − i$]. Thus, if we know that #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] is one complex solution of a quadratic equation, we know that #math.equation(block: false, alt: "z bar equals a minus b i")[$overline(z) = a − b i$] is the other solution. We can now write a quadratic polynomial, with real coefficients, having any complex number as one of its zeros. The factored form of the quadratic polynomial with zeros #math.equation(block: false, alt: "z")[$z$] and #math.equation(block: false, alt: "z bar")[$overline(z)$] is #math.equation(block: true, alt: "p open parenthesis x close parenthesis equals open parenthesis x minus z close parenthesis open parenthesis x minus z bar close parenthesis")[$p ( x ) = ( x − z ) ( x − overline(z) )$] Expanding the right side, we find #math.equation(block: true, alt: "p open parenthesis x close parenthesis equals x squared minus open parenthesis z plus z bar close parenthesis x plus z z bar")[$p ( x ) = x^(2) − ( z + overline(z) ) x + z overline(z)$] Because #math.equation(block: false, alt: "open parenthesis z plus z bar close parenthesis")[$( z + overline(z) )$] and #math.equation(block: false, alt: "z z bar")[$z overline(z)$] are both real numbers, this polynomial has real-valued coefficients. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Complex Zeros of a Quadratic Polynomial] If #math.equation(block: false, alt: "z equals a plus b i , b not equal to 0")[$z = a + b i , " " b ≠ 0$], is a complex zero of a quadratic polynomial #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$] with real coefficients and lead coefficient 1, then #math.equation(block: false, alt: "z bar equals a minus b i")[$overline(z) = a − b i$] is the other zero, and #math.equation(block: true, alt: "p open parenthesis x close parenthesis equals open parenthesis x minus z close parenthesis open parenthesis x minus z bar close parenthesis equals x squared minus open parenthesis z plus z bar close parenthesis x plus z z bar")[$bold(italic(p)) ( bold(italic(x)) ) = ( bold(italic(x)) − bold(italic(z)) ) ( bold(italic(x)) − overline(bold(italic(z))) ) = bold(italic(x))^(2) − ( bold(italic(z)) + overline(bold(italic(z))) ) bold(italic(x)) + bold(italic(z)) overline(bold(italic(z)))$] ] #examplebox("Example 11")[][ + Let #math.equation(block: false, alt: "z equals 7 minus 5 i")[$z = 7 − 5 i$]. Compute #math.equation(block: false, alt: "open parenthesis z plus z bar close parenthesis")[$( z + overline(z) )$] and #math.equation(block: false, alt: "z z bar")[$z overline(z)$]. + Find a quadratic polynomial with one zero being #math.equation(block: false, alt: "z equals 7 minus 5 i")[$z = 7 − 5 i$]. #solutionbox[ + The conjugate of #math.equation(block: false, alt: "z equals 7 minus 5 i")[$z = 7 − 5 i$] is #math.equation(block: false, alt: "z bar equals 7 plus 5 i")[$overline(z) = 7 + 5 i$], so #math.equation(block: true, alt: "open parenthesis z plus z bar close parenthesis equals open parenthesis 7 minus 5 i close parenthesis plus open parenthesis 7 plus 5 i close parenthesis equals 14")[$( z + overline(z) ) = ( 7 − 5 i ) + ( 7 + 5 i ) = 14$] and #math.equation(block: true, alt: "z z bar, equals open parenthesis 7 minus 5 i close parenthesis open parenthesis 7 plus 5 i close parenthesis; equals 49 minus 25 i squared equals 49 plus 25 equals 74")[$z overline(z) & = ( 7 − 5 i ) ( 7 + 5 i ) \ & = 49 − 25 i^(2) = 49 + 25 = 74$] + The polynomial is #math.equation(block: true, alt: "p open parenthesis x close parenthesis, equals x squared plus open parenthesis z plus z bar close parenthesis x plus z z bar; equals x squared minus 14 x plus 74")[$p ( x ) & = x^(2) + ( z + overline(z) ) x + z overline(z) \ & = x^(2) − 14 x + 74$] ] ] + Let #math.equation(block: false, alt: "z equals minus 3 plus 4 i")[$z = − 3 + 4 i$]. Compute #math.equation(block: false, alt: "open parenthesis z plus z bar close parenthesis")[$( z + overline(z) )$] and #math.equation(block: false, alt: "z z bar")[$z overline(z)$]. + Find a quadratic polynomial with one zero being #math.equation(block: false, alt: "z equals minus 3 plus 4 i")[$z = − 3 + 4 i$]. + #math.equation(block: false, alt: "z plus z bar equals minus 6")[$z + overline(z) = − 6$], #math.equation(block: false, alt: "z z bar equals 25")[$" " z overline(z) = 25$] + #math.equation(block: false, alt: "x squared plus 6 x plus 25 equals 0")[$x^(2) + 6 x + 25 = 0$] One of the most important results in mathematics is the #strong[fundamental theorem of algebra], which says that if we allow complex numbers as inputs, then every polynomial #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$] of degree #math.equation(block: false, alt: "n greater than or equal to 1")[$n ≥ 1$] has exactly #math.equation(block: false, alt: "n")[$n$] complex number zeros. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Fundamental Theorem of Algebra] Let #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$] be a polynomial of degree #math.equation(block: false, alt: "n greater than or equal to 1")[$n ≥ 1$]. Then #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$] has exactly #math.equation(block: false, alt: "n")[$n$] complex zeros, counting multiplicities. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A polynomial has a zero "of multiplicity #math.equation(block: false, alt: "k")[$k$]" at #math.equation(block: false, alt: "x equals a")[$x = a$] if #math.equation(block: false, alt: "open parenthesis x minus a close parenthesis")[$( x − a )$] occurs #math.equation(block: false, alt: "k")[$k$] times as a factor of the polynomial. For example, the polynomial #math.equation(block: false, alt: "p open parenthesis x close parenthesis equals open parenthesis x minus 5 close parenthesis squared")[$p ( x ) = ( x − 5 )^(2)$] has a zero of multiplicity 2 at #math.equation(block: false, alt: "x equals 5")[$x = 5$]. ] As a result, every polynomial of degree can be factored as the product of linear terms. For example, although the graph of #math.equation(block: false, alt: "y equals x to the power 4 plus 4")[$" " " " y = x^(4) + 4 " " " "$]shown at right has no #math.equation(block: false, alt: "x")[$x$]-intercepts, the fundamental theorem tells us that there are four complex solutions to #math.equation(block: false, alt: "x to the power 4 plus 4 equals 0")[$" " x^(4) + 4 = 0 " "$], and that #math.equation(block: false, alt: "x to the power 4 plus 4")[$x^(4) + 4$] can be factored. You can check that the four solutions to #math.equation(block: false, alt: "x to the power 4 plus 4 equals 0")[$" " x^(4) + 4 = 0 " "$] are #math.equation(block: true, alt: "1 plus i , minus 1 plus i , minus 1 minus i , and 1 minus i")[$1 + i , " " − 1 + i , " " − 1 − i , " " "and" " " 1 − i$] #figure(figph[quartic], alt: "quartic", caption: none) For example, if #math.equation(block: false, alt: "x equals 1 plus i")[$x = 1 + i$], then #math.equation(block: true, alt: "x squared equals open parenthesis 1 plus i close parenthesis squared equals 1 plus 2 i plus i squared equals 2 i")[$x^(2) = ( 1 + i )^(2) = 1 + 2 i + i^(2) = 2 i$] and #math.equation(block: true, alt: "x to the power 4 equals open parenthesis x squared close parenthesis squared equals open parenthesis 2 i close parenthesis squared equals minus 4")[$x^(4) = ( x^(2) )^(2) = ( 2 i )^(2) = − 4$] Because each zero corresponds to a factor of the polynomial, the factored form of #math.equation(block: false, alt: "x to the power 4 plus 4")[$x^(4) + 4$] is #math.equation(block: true, alt: "x to the power 4 plus 4 equals open bracket x minus open parenthesis 1 plus i close parenthesis close bracket open bracket x minus open parenthesis minus 1 plus i close parenthesis close bracket open bracket x minus open parenthesis minus 1 minus i close parenthesis close bracket open bracket x minus open parenthesis 1 minus i close parenthesis close bracket")[$x^(4) + 4 = [ x − ( 1 + i ) ] [ x − ( − 1 + i ) ] [ x − ( − 1 − i ) ] [ x − ( 1 − i ) ]$] The four solutions to #math.equation(block: false, alt: "x to the power 4 plus 4 equals 0")[$" " x^(4) + 4 = 0 " "$] form two complex conjugate pairs, namely #math.equation(block: false, alt: "1 plus or minus i")[$1 ± i$] and #math.equation(block: false, alt: "minus 1 plus or minus i")[$− 1 ± i$]. We see again that, for every polynomial with real coefficients, the nonreal zeros always occur in complex conjugate pairs. #examplebox("Example 12")[][ Find a fourth degree polynomial with real coefficients, two of whose zeros are #math.equation(block: false, alt: "3 i")[$3 i$] and #math.equation(block: false, alt: "2 plus i")[$2 + i$]. #solutionbox[ The other two zeros are #math.equation(block: false, alt: "minus 3 i")[$− 3 i$] and #math.equation(block: false, alt: "2 minus i")[$2 − i$]. The factored form of the polynomial is #math.equation(block: true, alt: "open parenthesis x minus 3 i close parenthesis open parenthesis x plus 3 i close parenthesis open bracket x minus open parenthesis 2 plus i close parenthesis close bracket open bracket x minus open parenthesis 2 minus i close parenthesis close bracket")[$( x − 3 i ) ( x + 3 i ) [ x − ( 2 + i ) ] [ x − ( 2 − i ) ]$] We multiply together the factors to find the polynomial. The product of #math.equation(block: false, alt: "open parenthesis x minus 3 i close parenthesis open parenthesis x plus 3 i close parenthesis")[$( x − 3 i ) ( x + 3 i )$] is #math.equation(block: false, alt: "x squared plus 9")[$x^(2) + 9$], and #math.equation(block: true, alt: "open bracket x minus open parenthesis 2 plus i close parenthesis close bracket open bracket x minus open parenthesis 2 minus i close parenthesis close bracket, equals x squared minus open parenthesis 2 plus i plus 2 minus i close parenthesis x plus open parenthesis 2 plus i close parenthesis open parenthesis 2 minus i close parenthesis; equals x squared minus 4 x plus 5")[$[ x − ( 2 + i ) ] [ x − ( 2 − i ) ] & = x^(2) − ( 2 + i + 2 − i ) x + ( 2 + i ) ( 2 − i ) \ & = x^(2) − 4 x + 5$] Finally, we multiply these two partial products to find the polynomial we seek, #math.equation(block: true, alt: "open parenthesis x squared plus 9 close parenthesis open parenthesis x squared minus 4 x plus 5 close parenthesis equals x to the power 4 minus 4 x cubed plus 14 x squared minus 36 x plus 45")[$( x^(2) + 9 ) ( x^(2) − 4 x + 5 ) = x^(4) − 4 x^(3) + 14 x^(2) − 36 x + 45$] ] ] + Find the zeros of the polynomial #math.equation(block: false, alt: "x to the power 4 plus 15 x squared minus 16")[$x^(4) + 15 x^(2) − 16$]. + Write the polynomial in factored form. + #math.equation(block: false, alt: "plus or minus 1 , plus or minus 4 i")[$± 1 , " " ± 4 i$] + #math.equation(block: false, alt: "open parenthesis x minus 1 close parenthesis open parenthesis x plus 1 close parenthesis open parenthesis x minus 4 i close parenthesis open parenthesis x plus 4 i close parenthesis")[$( x − 1 ) ( x + 1 ) ( x − 4 i ) ( x + 4 i )$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve. + #math.equation(block: false, alt: "3 x squared equals 6 x")[$3 x^(2) = 6 x$] + #math.equation(block: false, alt: "16 x squared equals 9")[$16 x^(2) = 9$] + #math.equation(block: false, alt: "x squared plus 4 x equals 12")[$x^(2) + 4 x = 12$] + #math.equation(block: false, alt: "2 open parenthesis x minus 3 close parenthesis squared equals 8")[$2 ( x − 3 )^(2) = 8$] + #math.equation(block: false, alt: "3 x squared minus 8 equals 10 x")[$3 x^(2) − 8 = 10 x$] + #math.equation(block: false, alt: "x squared minus 3 x plus 1 equals 0")[$x^(2) − 3 x + 1 = 0$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "0 , 2")[$0 , " " 2$] + #math.equation(block: false, alt: "the fraction plus or minus 3 over 4")[$display(frac(± 3, 4))$] + #math.equation(block: false, alt: "minus 6 , 2")[$− 6 , " " 2$] + #math.equation(block: false, alt: "1 , 5")[$1 , " " 5$] + #math.equation(block: false, alt: "the fraction minus 2 over 3 , 4")[$display(frac(− 2, 3)) , " " 4$] + #math.equation(block: false, alt: "the fraction 3 plus or minus the square root of 5 over 2")[$display(frac(3 ± sqrt(5), 2))$] ] === Section 10.3 Summary ==== Vocabulary - Imaginary unit - Imaginary number - Complex number - Complex conjugate - Real part - Imaginary part - Complex plane - Real axis - Imaginary axis - Modulus ==== Concepts + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Imaginary Unit] We define the imaginary unit, #strong[#math.equation(block: false, alt: "i")[$i$]], by #math.equation(block: true, alt: "i squared equals minus 1 or i equals the square root of minus 1")[$i^(2) = − 1 " " " " " " " " " " " " "or" " " " " " " " " " " " " i = sqrt(− 1)$] ] + The square root of a negative number is an imaginary number: if #math.equation(block: false, alt: "a greater than 0 , the square root of minus a equals i the square root of a")[$a > 0 , " " sqrt(− a) = i sqrt(a)$] + A complex number #math.equation(block: false, alt: "z")[$z$] is the sum of a real number and an imaginary number, #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$]. + We can perform the four arithmetic operations on complex numbers. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Operations on Complex Numbers] #math.equation(block: true, alt: "z sub 1 plus z sub 2 equals open parenthesis a plus b i close parenthesis plus open parenthesis c plus d i close parenthesis equals open parenthesis a plus c close parenthesis plus open parenthesis b plus d close parenthesis i")[$z_(1) + z_(2) = ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i$]#math.equation(block: true, alt: "z sub 1 minus z sub 2 equals open parenthesis a plus b i close parenthesis minus open parenthesis c plus d i close parenthesis equals open parenthesis a minus c close parenthesis plus open parenthesis b minus d close parenthesis i")[$z_(1) − z_(2) = ( a + b i ) − ( c + d i ) = ( a − c ) + ( b − d ) i$]#math.equation(block: true, alt: "z sub 1 z sub 2 equals open parenthesis a plus b i close parenthesis open parenthesis c plus d i close parenthesis equals open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i")[$z_(1) z_(2) = ( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i$]#math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals the fraction a plus b i over c plus d i equals the fraction a plus b i over c plus d i times the fraction c minus d i over c minus d i equals the fraction a c plus b d over c squared plus d squared plus the fraction b c minus a d over c squared plus d squared i")[$display(frac(z_(1), z_(2))) = display(frac(a + b i, c + d i)) = display(frac(a + b i, c + d i)) ⋅ display(frac(c − d i, c − d i)) = display(frac(a c + b d, c^(2) + d^(2))) + display(frac(b c − a d, c^(2) + d^(2))) i$] ] + The product of a nonzero complex number and its conjugate is always a positive real number. #math.equation(block: true, alt: "z z bar equals open parenthesis a plus b i close parenthesis open parenthesis a minus b i close parenthesis equals a squared minus b squared i squared equals a squared minus b squared open parenthesis minus 1 close parenthesis equals a squared plus b squared")[$z overline(z) = ( a + b i ) ( a − b i ) = a^(2) − b^(2) i^(2) = a^(2) − b^(2) ( − 1 ) = a^(2) + b^(2)$]. + We can graph complex numbers in the complex plane. + We can visualize the sum of two complex numbers by vector addition in the complex plane. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Fundamental Theorem of Algebra] Let #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$] be a polynomial of degree #math.equation(block: false, alt: "n greater than or equal to 1")[$n ≥ 1$]. Then #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$] has exactly #math.equation(block: false, alt: "n")[$n$] complex zeros. ] + The nonreal zeros of a polynomial with real coefficients always occur in conjugate pairs. ==== Study Questions + What are imaginary numbers, and why were they invented? + Simplify the following powers of #math.equation(block: false, alt: "i")[$i$]: #math.equation(block: true, alt: "i squared , i cubed , i to the power 4 , i to the power 5 , i to the power 6 , i to the power 7 , i to the power 8")[$i^(2) , " " i^(3) , " " i^(4) , " " i^(5) , " " i^(6) , " " i^(7) , " " i^(8)$] What do you notice? + Explain how the complex conjugate is used in dividing complex numbers. + If one solution of a quadratic equation is #math.equation(block: false, alt: "3 plus i the square root of 2")[$3 + i sqrt(2)$], what is the other solution? + If #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] is a polynomial of degree #math.equation(block: false, alt: "7")[$7$], how many zeros does #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] have? How many #math.equation(block: false, alt: "x")[$x$]-intercepts could its graph have? How many complex zeros could #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] have? ==== Skills + Write and simplify complex numbers \#1–6 + Perform arithmetic operations on complex numbers \#7–34 + Evaluate polynomials at complex numbers, expand polynomials \#35–46 + Graph complex numbers \#47–58 + Find a polynomial with given zeros \#67–74 === Homework 10-3 For Problems 1–2, write the complex number in the form #math.equation(block: false, alt: "a plus b i")[$a + b i$], where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are real numbers. + #math.equation(block: false, alt: "the square root of minus 25 minus 4")[$sqrt(− 25) − 4$] + #math.equation(block: false, alt: "the fraction minus 8 plus the square root of minus 4 over 2")[$display(frac(− 8 + sqrt(− 4), 2))$] + #math.equation(block: false, alt: "the fraction minus 5 minus the square root of minus 2 over 6")[$display(frac(− 5 − sqrt(− 2), 6))$] + #math.equation(block: false, alt: "5 i minus 4")[$5 i − 4$] + #math.equation(block: false, alt: "minus 4 plus i")[$− 4 + i$] + #math.equation(block: false, alt: "the fraction minus 5 over 6 minus the fraction the square root of 2 over 6 i")[$display(frac(− 5, 6)) − display(frac(sqrt(2), 6)) i$] + #math.equation(block: false, alt: "the square root of minus 9 plus 3")[$sqrt(− 9) + 3$] + #math.equation(block: false, alt: "the fraction 6 minus the square root of minus 36 over 2")[$display(frac(6 − sqrt(− 36), 2))$] + #math.equation(block: false, alt: "the fraction 7 plus the square root of minus 3 over 4")[$display(frac(7 + sqrt(− 3), 4))$] For Problems 3–6, find the zeros of the quadratic polynomial. Write each zero in the form #math.equation(block: false, alt: "a plus b i")[$a + b i$], where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are real numbers. #math.equation(block: true, alt: "x squared plus 6 x plus 13")[$x^(2) + 6 x + 13$] #math.equation(block: true, alt: "minus 3 plus or minus 2 i")[$− 3 ± 2 i$] #math.equation(block: true, alt: "x squared minus 2 x plus 10")[$x^(2) − 2 x + 10$] #math.equation(block: true, alt: "3 x squared minus x plus 1")[$3 x^(2) − x + 1$] #math.equation(block: true, alt: "the fraction 1 over 6 plus or minus the fraction the square root of 11 over 6 i")[$display(frac(1, 6)) ± display(frac(sqrt(11), 6)) i$] #math.equation(block: true, alt: "5 x squared plus 2 x plus 2")[$5 x^(2) + 2 x + 2$] For Problems 7–10, add or subtract. #math.equation(block: true, alt: "open parenthesis 11 minus 4 i close parenthesis minus open parenthesis minus 2 minus 8 i close parenthesis")[$( 11 − 4 i ) − ( − 2 − 8 i )$] #math.equation(block: true, alt: "13 plus 4 i")[$13 + 4 i$] #math.equation(block: true, alt: "open parenthesis 7 i minus 2 close parenthesis plus open parenthesis 6 minus 4 i close parenthesis")[$( 7 i − 2 ) + ( 6 − 4 i )$] #math.equation(block: true, alt: "open parenthesis 2.1 plus 5.6 i close parenthesis plus open parenthesis minus 1.8 i minus 2.9 close parenthesis")[$( 2.1 + 5.6 i ) + ( − 1.8 i − 2.9 )$] #math.equation(block: true, alt: "minus 0.8 plus 3.8 i")[$− 0.8 + 3.8 i$] #math.equation(block: true, alt: "open parenthesis the fraction 1 over 5 i minus the fraction 2 over 5 close parenthesis minus open parenthesis the fraction 4 over 5 minus the fraction 3 over 5 i close parenthesis")[$( display(frac(1, 5)) i − display(frac(2, 5)) ) − ( display(frac(4, 5)) − display(frac(3, 5)) i )$] For Problems 11–20, multiply. #math.equation(block: true, alt: "5 i open parenthesis 2 minus 4 i close parenthesis")[$5 i ( 2 − 4 i )$] #math.equation(block: true, alt: "20 minus 10 i")[$20 − 10 i$] #math.equation(block: true, alt: "minus 7 i open parenthesis minus 1 plus 4 i close parenthesis")[$− 7 i ( − 1 + 4 i )$] #math.equation(block: true, alt: "open parenthesis 4 minus i close parenthesis open parenthesis minus 6 plus 7 i close parenthesis")[$( 4 − i ) ( − 6 + 7 i )$] #math.equation(block: true, alt: "minus 14 plus 34 i")[$− 14 + 34 i$] #math.equation(block: true, alt: "open parenthesis 2 minus 3 i close parenthesis open parenthesis 2 minus 3 i close parenthesis")[$( 2 − 3 i ) ( 2 − 3 i )$] #math.equation(block: true, alt: "open parenthesis 7 plus i the square root of 3 close parenthesis squared")[$( 7 + i sqrt(3) )^(2)$] #math.equation(block: true, alt: "46 plus 14 i the square root of 3")[$46 + 14 i sqrt(3)$] #math.equation(block: true, alt: "open parenthesis 5 minus i the square root of 2 close parenthesis squared")[$( 5 − i sqrt(2) )^(2)$] #math.equation(block: true, alt: "open parenthesis 7 plus i the square root of 3 close parenthesis open parenthesis 7 minus i the square root of 3 close parenthesis")[$( 7 + i sqrt(3) ) ( 7 − i sqrt(3) )$] #math.equation(block: true, alt: "52")[$52$] #math.equation(block: true, alt: "open parenthesis 5 minus i the square root of 2 close parenthesis open parenthesis 5 plus i the square root of 2 close parenthesis")[$( 5 − i sqrt(2) ) ( 5 + i sqrt(2) )$] #math.equation(block: true, alt: "open parenthesis 1 minus i close parenthesis cubed")[$( 1 − i )^(3)$] #math.equation(block: true, alt: "minus 2 minus 2 i")[$− 2 − 2 i$] #math.equation(block: true, alt: "open parenthesis 2 plus i close parenthesis cubed")[$( 2 + i )^(3)$] For Problems 21–32, divide. #math.equation(block: true, alt: "the fraction 12 plus 3 i over minus 3 i")[$display(frac(12 + 3 i, − 3 i))$] #math.equation(block: true, alt: "minus 1 plus 4 i")[$− 1 + 4 i$] #math.equation(block: true, alt: "the fraction 12 plus 4 i over 8 i")[$display(frac(12 + 4 i, 8 i))$] #math.equation(block: true, alt: "the fraction 10 plus 15 i over 2 plus i")[$display(frac(10 + 15 i, 2 + i))$] #math.equation(block: true, alt: "the fraction 35 over 3 plus the fraction 20 over 3 i")[$display(frac(35, 3)) + display(frac(20, 3)) i$] #math.equation(block: true, alt: "the fraction 4 minus 6 i over 1 minus i")[$display(frac(4 − 6 i, 1 − i))$] #math.equation(block: true, alt: "the fraction 5 i over 2 minus 5 i")[$display(frac(5 i, 2 − 5 i))$] #math.equation(block: true, alt: "the fraction minus 25 over 29 plus the fraction 10 over 29 i")[$display(frac(− 25, 29)) + display(frac(10, 29)) i$] #math.equation(block: true, alt: "the fraction minus 2 i over 7 plus 2 i")[$display(frac(− 2 i, 7 + 2 i))$] #math.equation(block: true, alt: "the fraction the square root of 3 over the square root of 3 plus i")[$display(frac(sqrt(3), sqrt(3) + i))$] #math.equation(block: true, alt: "the fraction 3 over 4 minus the fraction the square root of 3 over 4 i")[$display(frac(3, 4)) − display(frac(sqrt(3), 4)) i$] #math.equation(block: true, alt: "the fraction 2 the square root of 2 over 1 minus i the square root of 2")[$display(frac(2 sqrt(2), 1 − i sqrt(2)))$] #math.equation(block: true, alt: "the fraction 1 plus i the square root of 5 over 1 minus i the square root of 5")[$display(frac(1 + i sqrt(5), 1 − i sqrt(5)))$] #math.equation(block: true, alt: "the fraction minus 2 over 3 plus the fraction the square root of 5 over 3 i")[$display(frac(− 2, 3)) + display(frac(sqrt(5), 3)) i$] #math.equation(block: true, alt: "the fraction the square root of 2 minus i over the square root of 2 plus i")[$display(frac(sqrt(2) − i, sqrt(2) + i))$] #math.equation(block: true, alt: "the fraction 3 plus 2 i over 2 minus 3 i")[$display(frac(3 + 2 i, 2 − 3 i))$] #math.equation(block: true, alt: "i")[$i$] #math.equation(block: true, alt: "the fraction 4 minus 6 i over minus 3 minus 2 i")[$display(frac(4 − 6 i, − 3 − 2 i))$] Simplify. + #math.equation(block: false, alt: "i to the power 6")[$i^(6)$] + #math.equation(block: false, alt: "i to the power 12")[$i^(12)$] + #math.equation(block: false, alt: "i to the power 15")[$i^(15)$] + #math.equation(block: false, alt: "i to the power 102")[$i^(102)$] + #math.equation(block: false, alt: "minus 1")[$− 1$] + #math.equation(block: false, alt: "1")[$1$] + #math.equation(block: false, alt: "minus i")[$− i$] + #math.equation(block: false, alt: "minus 1")[$− 1$] Express with a positive exponent and simplify. + #math.equation(block: false, alt: "i to the power minus 1")[$i^(− 1)$] + #math.equation(block: false, alt: "i to the power minus 2")[$i^(− 2)$] + #math.equation(block: false, alt: "i to the power minus 3")[$i^(− 3)$] + #math.equation(block: false, alt: "i to the power minus 6")[$i^(− 6)$] For Problems 35-40, evaluate the polynomial for the given values of the variable. #math.equation(block: true, alt: "z squared plus 9")[$z^(2) + 9$] + #math.equation(block: false, alt: "z equals 3 i")[$z = 3 i$] + #math.equation(block: false, alt: "z equals minus 3 i")[$z = − 3 i$] + #math.equation(block: false, alt: "0")[$0$] + #math.equation(block: false, alt: "0")[$0$] #math.equation(block: true, alt: "2 y squared minus y minus 2")[$2 y^(2) − y − 2$] + #math.equation(block: false, alt: "y equals 2 minus i")[$y = 2 − i$] + #math.equation(block: false, alt: "y equals minus 2 minus i")[$y = − 2 − i$] #math.equation(block: true, alt: "x squared minus 2 x plus 2")[$x^(2) − 2 x + 2$] + #math.equation(block: false, alt: "x equals 1 minus i")[$x = 1 − i$] + #math.equation(block: false, alt: "x equals 1 plus i")[$x = 1 + i$] + #math.equation(block: false, alt: "0")[$0$] + #math.equation(block: false, alt: "0")[$0$] #math.equation(block: true, alt: "3 w squared plus 5")[$3 w^(2) + 5$] + #math.equation(block: false, alt: "w equals 2 i")[$w = 2 i$] + #math.equation(block: false, alt: "w equals minus 2 i")[$w = − 2 i$] #math.equation(block: true, alt: "q squared plus 4 q plus 13")[$q^(2) + 4 q + 13$] + #math.equation(block: false, alt: "q equals minus 2 plus 3 i")[$q = − 2 + 3 i$] + #math.equation(block: false, alt: "q equals minus 2 minus 3 i")[$q = − 2 − 3 i$] + #math.equation(block: false, alt: "0")[$0$] + #math.equation(block: false, alt: "0")[$0$] #math.equation(block: true, alt: "v squared plus 2 v plus 3")[$v^(2) + 2 v + 3$] + #math.equation(block: false, alt: "v equals 1 plus i")[$v = 1 + i$] + #math.equation(block: false, alt: "v equals minus 1 plus i")[$v = − 1 + i$] For Problems 41–46, expand the product of polynomials. #math.equation(block: true, alt: "open parenthesis 2 z plus 7 i close parenthesis open parenthesis 2 z minus 7 i close parenthesis")[$( 2 z + 7 i ) ( 2 z − 7 i )$] #math.equation(block: true, alt: "4 z squared plus 49")[$4 z^(2) + 49$] #math.equation(block: true, alt: "open parenthesis 5 w plus 3 i close parenthesis open parenthesis 5 w minus 3 i close parenthesis")[$( 5 w + 3 i ) ( 5 w − 3 i )$] #math.equation(block: true, alt: "open bracket x plus open parenthesis 3 plus i close parenthesis close bracket open bracket x plus open parenthesis 3 minus i close parenthesis close bracket")[$[ x + ( 3 + i ) ] [ x + ( 3 − i ) ]$] #math.equation(block: true, alt: "x squared plus 6 x plus 10")[$x^(2) + 6 x + 10$] #math.equation(block: true, alt: "open bracket s minus open parenthesis 1 plus 2 i close parenthesis close bracket open bracket s minus open parenthesis 1 minus 2 i close parenthesis close bracket")[$[ s − ( 1 + 2 i ) ] [ s − ( 1 − 2 i ) ]$] #math.equation(block: true, alt: "open bracket v minus open parenthesis 4 plus i close parenthesis close bracket open bracket v minus open parenthesis 4 minus i close parenthesis close bracket")[$[ v − ( 4 + i ) ] [ v − ( 4 − i ) ]$] #math.equation(block: true, alt: "v squared minus 8 v plus 17")[$v^(2) − 8 v + 17$] #math.equation(block: true, alt: "open bracket Z plus open parenthesis 2 plus i close parenthesis close bracket open bracket Z plus open parenthesis 2 minus i close parenthesis close bracket")[$[ Z + ( 2 + i ) ] [ Z + ( 2 − i ) ]$] For Problems 47–50, plot the number and its complex conjugate in the complex plane. What is the geometric relationship between complex conjugates? #math.equation(block: true, alt: "z equals minus 3 plus 2 i")[$z = − 3 + 2 i$] #figure(figph[complex conjugates], alt: "complex conjugates", caption: none) #math.equation(block: true, alt: "z equals 4 minus 3 i")[$z = 4 − 3 i$] #math.equation(block: true, alt: "z equals the fraction the square root of 3 over 2 minus the fraction 1 over 2 i")[$z = display(frac(sqrt(3), 2)) − display(frac(1, 2)) i$] #figure(figph[complex conjugates], alt: "complex conjugates", caption: none) #math.equation(block: true, alt: "z equals minus the fraction the square root of 2 over 2 minus the fraction the square root of 2 over 2 i")[$z = − display(frac(sqrt(2), 2)) − display(frac(sqrt(2), 2)) i$] For Problems 51–54, sketch the set of points in the complex plane. #math.equation(block: true, alt: "vertical bar z vertical bar less than or equal to 2")[$| z | ≤ 2$] #figure(figph[disk], alt: "disk", caption: none) #math.equation(block: true, alt: "1 less than vertical bar z vertical bar less than 3")[$1 < | z | < 3$] The set of all #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] for which #math.equation(block: false, alt: "a greater than or equal to b")[$a ≥ b$] #figure(figph[inequality], alt: "inequality", caption: none) The set of all #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] for which #math.equation(block: false, alt: "a plus b less than 2")[$a + b < 2$] For Problems 55–58, illustrate addition using the parallelogram rule in the complex plane. #math.equation(block: true, alt: "open parenthesis 1 minus 4 i close parenthesis plus open parenthesis minus 3 plus 2 i close parenthesis")[$( 1 − 4 i ) + ( − 3 + 2 i )$] #figure(figph[complex numbers as vectors], alt: "complex numbers as vectors", caption: none) #math.equation(block: true, alt: "open parenthesis minus 4 plus 2 i close parenthesis plus open parenthesis minus 2 plus i close parenthesis")[$( − 4 + 2 i ) + ( − 2 + i )$] #math.equation(block: true, alt: "open parenthesis 2 plus 6 i close parenthesis minus open parenthesis 3 plus 3 i close parenthesis")[$( 2 + 6 i ) − ( 3 + 3 i )$] #figure(figph[complex numbers as vectors], alt: "complex numbers as vectors", caption: none) #math.equation(block: true, alt: "open parenthesis minus 5 minus 2 i close parenthesis minus open parenthesis minus 3 plus 2 i close parenthesis")[$( − 5 − 2 i ) − ( − 3 + 2 i )$] Prove that the product of two complex numbers #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")[$z_(2) = c + d i$] is #math.equation(block: true, alt: "z sub 1 z sub 2 equals open parenthesis a plus b i close parenthesis open parenthesis c plus d i close parenthesis equals open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i")[$z_(1) z_(2) = ( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i$] #math.equation(block: true, alt: "open parenthesis a plus b i close parenthesis open parenthesis c plus d i close parenthesis equals a c plus a d i plus b c i plus b d i squared equals open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i")[$( a + b i ) ( c + d i ) = a c + a d i + b c i + b d i^(2) = ( a c − b d ) + ( a d + b c ) i$] Prove that the quotient of two complex numbers #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")[$z_(2) = c + d i$] is #math.equation(block: true, alt: "the fraction z sub 1 over z sub 2 equals the fraction a plus b i over c plus d i equals the fraction a plus b i over c plus d i times the fraction c minus d i over c minus d i equals the fraction a c plus b d over c squared plus d squared plus the fraction b c minus a d over c squared plus d squared i")[$display(frac(z_(1), z_(2))) = display(frac(a + b i, c + d i)) = display(frac(a + b i, c + d i)) ⋅ display(frac(c − d i, c − d i)) = display(frac(a c + b d, c^(2) + d^(2))) + display(frac(b c − a d, c^(2) + d^(2))) i$] Prove the commutative laws for addition and multiplication of complex numbers: #math.equation(block: true, alt: "z sub 1 plus z sub 2 equals z sub 2 plus z sub 1")[$z_(1) + z_(2) = z_(2) + z_(1)$] #math.equation(block: true, alt: "z sub 1 z sub 2 equals z sub 2 z sub 1")[$z_(1) z_(2) = z_(2) z_(1)$] #math.equation(block: true, alt: "z sub 1 plus z sub 2, equals open parenthesis a plus b i close parenthesis plus open parenthesis c plus d i close parenthesis equals open parenthesis a plus c close parenthesis plus open parenthesis b plus d close parenthesis i; equals open parenthesis c plus a close parenthesis plus open parenthesis d plus b close parenthesis i equals open parenthesis c plus d i close parenthesis plus open parenthesis a plus b i close parenthesis equals z sub 2 plus z sub 1")[$z_(1) + z_(2) & = ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i \ & = ( c + a ) + ( d + b ) i = ( c + d i ) + ( a + b i ) = z_(2) + z_(1)$] #math.equation(block: true, alt: "z sub 1 z sub 2, equals open parenthesis a plus b i close parenthesis open parenthesis c plus d i close parenthesis equals open parenthesis a c minus b d close parenthesis plus open parenthesis a d plus b c close parenthesis i; equals open parenthesis c a minus d b close parenthesis plus open parenthesis d a plus c b close parenthesis i equals z sub 2 z sub 1")[$z_(1) z_(2) & = ( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i \ & = ( c a − d b ) + ( d a + c b ) i = z_(2) z_(1)$] Prove the distributive law for complex numbers: #math.equation(block: true, alt: "z sub 1 open parenthesis z sub 2 plus z sub 3 close parenthesis equals z sub 1 z sub 2 plus z sub 1 z sub 3")[$z_(1) ( z_(2) + z_(3) ) = z_(1) z_(2) + z_(1) z_(3)$] + Show that #math.equation(block: false, alt: "z plus z bar")[$z + overline(z)$] is a real number, and #math.equation(block: false, alt: "z minus z bar")[$z − overline(z)$] is an imaginary number. + Show that #math.equation(block: false, alt: "z z bar equals vertical bar z vertical bar squared")[$z overline(z) = attach(| z |, t: 2)$] + #math.equation(block: false, alt: "z plus z bar equals open parenthesis a plus b i close parenthesis plus open parenthesis a minus b i close parenthesis equals 2 a ;")[$z + overline(z) = ( a + b i ) + ( a − b i ) = 2 a ; " " " " " "$] #math.equation(block: false, alt: "z minus z bar equals open parenthesis a plus b i close parenthesis minus open parenthesis a minus b i close parenthesis equals minus 2 b i")[$z − overline(z) = ( a + b i ) − ( a − b i ) = − 2 b i$] + #math.equation(block: false, alt: "z z bar equals open parenthesis a plus b i close parenthesis open parenthesis a minus b i close parenthesis equals a squared plus b squared equals vertical bar z vertical bar squared")[$z overline(z) = ( a + b i ) ( a − b i ) = a^(2) + b^(2) = attach(| z |, t: 2)$] Show that #math.equation(block: false, alt: "z w bar equals z bar times w bar")[$limits(z w)^(―) = overline(z) ⋅ overline(w)$] Suppose #math.equation(block: false, alt: "t , w")[$t , " " w$] and #math.equation(block: false, alt: "z")[$z$] are complex numbers. If #math.equation(block: false, alt: "w equals t plus z")[$w = t + z$], is it necessarily true that #math.equation(block: false, alt: "vertical bar w vertical bar equals vertical bar t vertical bar plus vertical bar z vertical bar")[$| w | = | t | + | z |$]? Provide examples to support your conclusion. No. Let #math.equation(block: false, alt: "t equals i")[$t = i$] and #math.equation(block: false, alt: "z equals minus i")[$z = − i$]. Then #math.equation(block: false, alt: "w equals t plus z equals i minus i equals 0 ,")[$w = t + z = i − i = 0 ,$] so #math.equation(block: false, alt: "vertical bar w vertical bar equals 0 ,")[$| w | = 0 ,$] but #math.equation(block: false, alt: "vertical bar t vertical bar plus vertical bar z vertical bar equals vertical bar i vertical bar plus vertical bar minus i vertical bar equals 1 plus 1 equals 2.")[$| t | + | z | = | i | + | − i | = 1 + 1 = 2.$] Prove the triangle inequality for complex numbers: #math.equation(block: true, alt: "vertical bar w plus z vertical bar less than or equal to vertical bar w vertical bar plus vertical bar z vertical bar")[$| w + z | ≤ | w | + | z |$] In Problems 67–70, + Given one solution of a quadratic equation with rational coefficients, find the other solution. + Write a quadratic equation that has those solutions. #math.equation(block: true, alt: "2 plus the square root of 5")[$2 + sqrt(5)$] + #math.equation(block: false, alt: "2 minus the square root of 5")[$2 − sqrt(5)$] + #math.equation(block: false, alt: "x squared minus 4 x minus 1 equals 0")[$x^(2) − 4 x − 1 = 0$] #math.equation(block: true, alt: "3 minus the square root of 2")[$3 − sqrt(2)$] #math.equation(block: true, alt: "4 minus 3 i")[$4 − 3 i$] + #math.equation(block: false, alt: "4 plus 3 i")[$4 + 3 i$] + #math.equation(block: false, alt: "x squared minus 8 x plus 25 equals 0")[$x^(2) − 8 x + 25 = 0$] #math.equation(block: true, alt: "5 plus i")[$5 + i$] For Problems 71–74, find a fourth degree polynomial with real coefficients that has the given complex numbers as two of its zeros. #math.equation(block: true, alt: "1 minus 3 i , 2 minus i")[$1 − 3 i , " " 2 − i$] #math.equation(block: true, alt: "x to the power 4 minus 6 x cubed plus 23 x squared minus 50 x plus 50 equals 0")[$x^(4) − 6 x^(3) + 23 x^(2) − 50 x + 50 = 0$] #math.equation(block: true, alt: "5 minus 4 i , minus i")[$5 − 4 i , " " − i$] #math.equation(block: true, alt: "the fraction 1 over 2 minus the fraction the square root of 3 over 2 i , 3 plus 2 i")[$display(frac(1, 2)) − display(frac(sqrt(3), 2)) i , " " 3 + 2 i$] #math.equation(block: true, alt: "x to the power 4 minus 7 x cubed plus 20 x squared minus 19 x plus 13 equals 0")[$x^(4) − 7 x^(3) + 20 x^(2) − 19 x + 13 = 0$] #math.equation(block: true, alt: "minus the fraction the square root of 2 over 2 plus the fraction the square root of 2 over 2 i , 4 minus i")[$− display(frac(sqrt(2), 2)) + display(frac(sqrt(2), 2)) i , " " 4 − i$]