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== 7.3#h(0.6em)Complex Numbers


    
      
=== Introduction


      You know that not all quadratic equations have real solutions.


      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), and the equation


      #math.equation(block: true, alt: "x squared minus 2 x plus 2 equals 0")[$x^(2) − 2 x + 2 = 0$]
      has no real solutions.


      #figure(figph[parabola without x-intercepts], alt: "parabola without x-intercepts", caption: none)


      We can still use completing the square or the quadratic formula to solve the equation.


      #examplebox("Example 1")[][
Solve the equation #math.equation(block: false, alt: "x squared minus 2 x plus 2 equals 0")[$" " x^(2) − 2 x + 2 = 0 " "$] by using the quadratic formula.


        
        #solutionbox[
We substitute #math.equation(block: false, alt: "a equals 1")[$a = 1$], #math.equation(block: false, alt: "b equals minus 2")[$b = − 2$], and #math.equation(block: false, alt: "c equals 2")[$c = 2$] into the quadratic formula to get


          #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 open parenthesis 1 close parenthesis equals the fraction 2 plus or minus the square root of minus 4 over 2")[$x = frac(− ( − 2 ) ± sqrt(( − 2 )^(2) − 4 ( 1 ) ( 2 )), 2 ( 1 )) = frac(2 ± sqrt(− 4), 2)$]
          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 equals 0")[$x^(2) − 2 x + 2 = 0$] has no real solutions.
]
]


      
        Solve the equation #math.equation(block: false, alt: "x squared minus 6 x plus 13 equals 0")[$x^(2) − 6 x + 13 = 0$] by using the quadratic formula.


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different solutions with a comma.


        #solutionbox[
#math.equation(block: true, alt: "x equals the fraction 6 plus or minus the square root of minus 16 over 2")[$x = display(frac(6 ± sqrt(− 16), 2))$]
]


      
      
        Solve the equation #math.equation(block: false, alt: "x squared minus 6 x plus 13 equals 0")[$x^(2) − 6 x + 13 = 0$] by using the quadratic formula.


        #solutionbox[
#math.equation(block: true, alt: "x equals the fraction 6 plus or minus the square root of minus 16 over 2")[$x = display(frac(6 ± sqrt(− 16), 2))$]
]


      
    
    
      
=== 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 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, #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 #strong[imaginary unit] #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")[$i^(2) = − 1 " " " " " " " or " " " " " " " i = sqrt(− 1)$]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The letter #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.
]


      
        What is the imaginary unit?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: false, alt: "the square root of minus 1")[$sqrt(− 1)$] is the imaginary unit, denoted by #math.equation(block: false, alt: "i")[$i$].
]


      
      
        What is the imaginary unit?


        + #math.equation(block: false, alt: "minus 1")[$− 1$]
+ #math.equation(block: false, alt: "open parenthesis minus 1 close parenthesis squared")[$( − 1 )^(2)$]
+ #math.equation(block: false, alt: "i squared")[$i^(2)$]
+ #math.equation(block: false, alt: "the square root of minus 1")[$sqrt(− 1)$]


      
      The square root of any negative number can be written as the product of a real number and #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 #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 #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 times the square root of a equals i the square root of a")[$sqrt(− a) = sqrt(− 1) ⋅ sqrt(a) = i sqrt(a)$]
]


      Here are some xamples of imaginary numbers.


      #math.equation(block: true, alt: "3 i , the fraction 7 over 8 i , minus 38 i , and i the square root of 5")[$3 i ", " " " " " " " frac(7, 8) i ", " " " " " " " − 38 i ", " " " " " " " " and " " " " " " " i sqrt(5)$]
      #examplebox("Example 2")[][
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 2 the square root of minus 1 the square root of 3; equals 2 i the square root of 3")[$2 sqrt(− 3) & = 2 sqrt(− 1) sqrt(3) \  & = 2 i sqrt(3)$]
]
]


      
        Write each radical as an imaginary number. Use #math.equation(block: false, alt: "i")[$i$] for the imaginary unit, and enter "sqrt(2)" for #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$].


        + #math.equation(block: false, alt: "the square root of minus 18 equals")[$sqrt(− 18) =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "minus 6 the square root of minus 5")[$− 6 sqrt(− 5)$]\_\_\_\_\_


        #solutionbox[
+ #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")[$− 6 i sqrt(5)$]
]


      
      
        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")[$− 6 sqrt(− 5)$]


        #solutionbox[
+ #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")[$− 6 i sqrt(5)$]
]


      
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Every negative real number has two imaginary square roots, #math.equation(block: false, alt: "i the square root of a")[$i sqrt(a)$]  and #math.equation(block: false, alt: "minus i the square root of a")[$− i sqrt(a)$], because


        #math.equation(block: true, alt: "open parenthesis i the square root of a close parenthesis squared equals i squared open parenthesis the square root of a close parenthesis squared equals minus a")[$attach(( i sqrt(a) ), t: 2) = i^(2) ( sqrt(a) )^(2) = − a$]
        and


        #math.equation(block: true, alt: "open parenthesis minus i the square root of a close parenthesis squared equals open parenthesis minus i close parenthesis squared open parenthesis the square root of a close parenthesis squared equals minus a")[$attach(( − i sqrt(a) ), t: 2) = ( − i )^(2) ( sqrt(a) )^(2) = − a$]
        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")[$x = frac(− ( − 2 ) ± sqrt(( − 2 )^(2) − 4 ( 1 ) ( 5 )), 2) = frac(2 ± sqrt(− 16), 2)$]
      If we now replace #math.equation(block: false, alt: "the square root of minus 16")[$sqrt(− 16)$] with #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")[$x = frac(2 ± 4 i, 2) = 1 ± 2 i$]
      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 numbers 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: 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.
]


      Here are some examples of complex numbers.


      #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")[$3 − 5 i ", " " " " " " " 2 + sqrt(7) i ", " " " " " " " frac(4 − i, 3) ", " " " " " " " 6 i ", " " " " " " " " and " − 9$]
      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 the imaginary part equal to zero). A complex number whose real part equals zero is called a #strong[pure imaginary] number.


      #examplebox("Example 3")[][
Write the solutions to Example,
		#math.equation(block: false, alt: "the fraction 2 plus or minus the square root of minus 4 over 2")[$display(frac(2 ± sqrt(− 4), 2))$], as complex numbers.


        
        #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: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different solutions with a comma.


        #solutionbox[
#math.equation(block: true, alt: "x equals the fraction minus 1 over 2 plus or minus the fraction 3 over 2 i")[$x = display(frac(− 1, 2)) ± display(frac(3, 2)) i$]
]


      
      
        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.


        #solutionbox[
#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


      All the properties of real numbers listed in Algebra Skills Refresher  The Real Number System  are also true of complex numbers. We can carry out arithmetic operations with complex numbers.


      
        Which statement about complex numbers is false?


        \_\_\_\_\_


        #solutionbox[
"Complex numbers cannot be combined with real numbers" is a false statement.
]


      
      
        Which statement about complex numbers is false?


        + A complex number has a real part and an imaginary part.
+ We can perform all four arithmetic operations on complex numbers.
+ Every quadratic equation has solutions in the complex numbers.
+ Complex numbers cannot be combined with real 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")[$( 4 + 5 i ) + ( 2 − 3 i ) & = ( 4 + 2 ) + ( 5 − 3 ) i \  & = 6 + 2 i$]
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Sums and Differences of Complex Numbers]


        #math.equation(block: true, alt: "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")[$( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i$]
        #math.equation(block: true, alt: "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")[$( a + b i ) − ( c + d i ) = ( a − c ) + ( b − d ) i$]
]


      #examplebox("Example 4")[][
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 equals")[$( − 3 + 2 i ) − ( − 3 − 2 i ) =$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "4 i")[$4 i$]
]


      
      
        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")[$( − 3 + 2 i ) − ( − 3 − 2 i )$]


        #solutionbox[
#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 times 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 FOIL method, as if the numbers were binomials. For example,


      #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")[$( 6 + 15 ) + ( − 10 i + 9 i ) = 21 − i$]
      #examplebox("Example 5")[][
Multiply #math.equation(block: false, alt: "open parenthesis 7 minus 4 i close parenthesis open parenthesis minus 2 minus i close parenthesis")[$( 7 − 4 i ) ( − 2 − i )$].


        
        #solutionbox[
#math.equation(block: true, alt: "open parenthesis 7 minus 4 i close parenthesis open parenthesis minus 2 minus i close parenthesis, equals minus 14 minus 7 i plus 8 i plus 4 i squared, Replace i squared by minus 1.; equals minus 14 minus 7 i plus 8 i minus 4, Combine real parts and imaginary; equals minus 18 plus i, imaginary parts.")[$( 7 − 4 i ) ( − 2 − i ) & = − 14 − 7 i + 8 i + 4 i^(2) &  & "Replace " bold(italic(i))^(2) " by " − 1. \  & = − 14 − 7 i + 8 i − 4 &  & "Combine real parts and imaginary" \  & = − 18 + i &  & "imaginary parts."$]
]
]


      
        Multiply #math.equation(block: false, alt: "open parenthesis minus 3 plus 2 i close parenthesis open parenthesis minus 3 minus 2 i close parenthesis equals")[$( − 3 + 2 i ) ( − 3 − 2 i ) =$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "13")[$13$]
]


      
      
        Multiply #math.equation(block: false, alt: "open parenthesis minus 3 plus 2 i close parenthesis open parenthesis minus 3 minus 2 i close parenthesis")[$( − 3 + 2 i ) ( − 3 − 2 i )$]


        #solutionbox[
#math.equation(block: true, alt: "13")[$13$]
]


      
      You can verify that in general the following rule holds.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Product of Complex Numbers]


        #math.equation(block: true, alt: "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")[$( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i$]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
One property of real numbers that is not true of complex numbers is #math.equation(block: false, alt: "the square root of a b equals the square root of a times 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 minus 2 times the square root of minus 2 equals the square root of minus 1 times 2 times the square root of minus 1 times 2; 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 in this case


        #math.equation(block: true, alt: "the square root of a b not equal to the square root of a times 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.
]


      
        How is multiplying complex numbers similar to multiplying binomials?


        \_\_\_\_\_


      
      
        How is multiplying complex numbers similar to multiplying binomials?


      
    
    
      
=== Quotients of Complex Numbers


      To find the quotient of two complex numbers, we use the technique of rationalizing the denominator. (See Algebra Skills Refresher Working with Radicals.)


      For example, consider the quotient


      #math.equation(block: true, alt: "the fraction 3 plus 4 i over 2 i")[$frac(3 + 4 i, 2 i)$]
      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 close parenthesis")[$i = sqrt(− 1) \)$], we multiply the numerator and denominator of the quotient by #math.equation(block: false, alt: "i")[$bold(italic(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 to the numerator.; equals the fraction 3 i minus 4 over minus 2, Recall that i squared equals minus 1.")[$frac(( 3 + 4 i ) ⋅ bold(italic(i)), 2 i ⋅ bold(italic(i))) & = frac(3 i + 4 i^(2), 2 i^(2)) &  & "Apply the distributive law to the numerator." \  & = frac(3 i − 4, − 2) &  & "Recall that " bold(italic(i))^(2) = − 1.$]
      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 to get


      #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 equals 2 plus the fraction minus 3 over 2 i")[$frac(3 i, − 2) − frac(4, − 2) = frac(− 3, 2) i + 2 = 2 + frac(− 3, 2) i$]
      #examplebox("Example 6")[][
Divide #math.equation(block: false, alt: "the fraction 10 minus 15 i over 5 i")[$" " " " display(frac(10 − 15 i, 5 i))$]


        
        #solutionbox[
We multiply numerator and denominator by #math.equation(block: false, alt: "i")[$bold(italic(i))$].


          #math.equation(block: true, alt: "the fraction 10 minus 15 i over 5 i, equals the fraction open parenthesis 10 minus 15 i close parenthesis times i over 5 i times i; equals the fraction 10 i minus 15 i squared over 5 i squared, Replace i squared by minus 1.; equals the fraction 10 i plus 15 over minus 5; equals the fraction 10 i over minus 5 plus the fraction 15 over minus 5, Divide minus 5 into each term of numerator.; equals minus 2 i minus 3")[$frac(10 − 15 i, 5 i) & = frac(( 10 − 15 i ) ⋅ bold(italic(i)), 5 i ⋅ bold(italic(i))) &  &  \  & = frac(10 i − 15 i^(2), 5 i^(2)) &  & "Replace " bold(italic(i))^(2) " by " − 1. \  & = frac(10 i + 15, − 5) &  &  \  & = frac(10 i, − 5) + frac(15, − 5) &  & "Divide " − 5 " into each term of numerator." \  & = − 2 i − 3$]
          The quotient is #math.equation(block: false, alt: "minus 3 minus 2 i")[$− 3 − 2 i$].
]
]


      
        Divide #math.equation(block: false, alt: "the fraction 8 plus 9 i over 3 i equals")[$display(frac(8 + 9 i, 3 i) =)$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "3 minus the fraction 8 over 3 i")[$3 − display(frac(8, 3)) i$]
]


      
      
        Divide #math.equation(block: false, alt: "the fraction 8 plus 9 i over 3 i")[$display(frac(8 + 9 i, 3 i))$]


        #solutionbox[
#math.equation(block: true, alt: "3 minus the fraction 8 over 3 i")[$3 − display(frac(8, 3)) i$]
]


      
      If #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$]  is any nonzero complex number, then the number #math.equation(block: false, alt: "z bar equals a minus b i")[$limits(z)^(―) = a − b i$]  is called the #strong[complex conjugate] of #math.equation(block: false, alt: "z")[$z$]. 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")[$z limits(z)^(―) = ( a + b i ) ( a − b i ) = a^(2) − b^(2) i^(2) = a^(2) − b^(2) ( − 1 ) = a^(2) + b^(2)$]
      We use this fact to find the quotient of complex numbers. If the divisor has both a real and an imaginary part, we multiply numerator and denominator by the conjugate of the denominator.


      
        What is the product of #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] with its complex conjugate?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "a squared plus b squared")[$a^(2) + b^(2)$]
]


      
      
        What is the product of #math.equation(block: false, alt: "z equals a plus b i")[$z = a + b i$] with its complex conjugate?


        + #math.equation(block: false, alt: "a minus b i")[$a − b i$]
+ #math.equation(block: false, alt: "a squared plus b squared")[$a^(2) + b^(2)$]
+ #math.equation(block: false, alt: "a squared minus b squared")[$a^(2) − b^(2)$]
+ #math.equation(block: false, alt: "a i plus b")[$a i + b$]


      
      #examplebox("Example 7")[][
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")[$4 + 2 bold(italic(i))$], the 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")[$frac(2 + 3 i, 4 − 2 i) & = frac(( 2 + 3 i ) ( 4 + 2 bold(italic(i)) ), ( 4 − 2 i ) ( 4 + 2 bold(italic(i)) )) &  & "Expand numerator and denominator." \  & = frac(8 + 4 i + 12 i + 6 i^(2), 16 + 8 i − 8 i − 4 i^(2)) &  & "Replace " bold(italic(i))^(2) " by " − 1. \  & = frac(8 + 16 i − 6, 16 − ( − 4 )) &  & "Combine like terms." \  & = frac(2 + 16 i, 20) &  & "Divide 20 into each term of numerator." \  & = frac(2, 20) + frac(16 i, 20) \  & = frac(1, 10) + frac(4, 5) i$]
]
]


      
        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")[$a + b i$].


        Answer: \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "1 minus 3 i")[$1 − 3 i$]
]


      
      
        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")[$a + b i$].


        #solutionbox[
#math.equation(block: true, alt: "1 minus 3 i")[$1 − 3 i$]
]


      
      
        Explain how to divide one complex number by another.


        \_\_\_\_\_


      
      
        Explain how to divide one complex number by another.


      
    
    
      
=== Zeros of Polynomials


      Because we can add, subtract, and multiply any two complex numbers, we can use a complex number as an input for a polynomial function. Thus, we can extend the domain of any polynomial to include all complex numbers.


      #examplebox("Example 8")[][
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")[$x = 1 + i$], then simplify.


        
        #solutionbox[
We substitute #math.equation(block: false, alt: "x equals 1 plus i")[$x = 1 + bold(italic(i))$] 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; equals 0")[$f ( 1 + bold(italic(i)) ) & = ( 1 + bold(italic(i)) )^(2) − 2 ( 1 + bold(italic(i)) ) + 2 \  & = 1^(2) + 2 i + i^(2) − 2 − 2 i + 2 \  & = 1 + 2 i + ( − 1 ) − 2 − 2 i + 2 \  & = 0$]
          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 solution of #math.equation(block: false, alt: "x squared minus 2 x plus 2 equals 0")[$x^(2) − 2 x + 2 = 0$].
]
]


      
        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 equals")[$f ( 3 + 2 i ) =$]\_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "f open parenthesis 3 plus 2 i close parenthesis equals open parenthesis 3 plus 2 i close parenthesis squared minus 6 open parenthesis 3 plus 2 i close parenthesis plus 13 equals 0")[$f ( 3 + 2 i ) = ( 3 + 2 i )^(2) − 6 ( 3 + 2 i ) + 13 = 0$]
]


      
      
        For #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")[$f ( 3 + 2 i )$]


        #solutionbox[
#math.equation(block: true, alt: "f open parenthesis 3 plus 2 i close parenthesis equals open parenthesis 3 plus 2 i close parenthesis squared minus 6 open parenthesis 3 plus 2 i close parenthesis plus 13 equals 0")[$f ( 3 + 2 i ) = ( 3 + 2 i )^(2) − 6 ( 3 + 2 i ) + 13 = 0$]
]


      
      In Quadratic Functions, we learned that irrational solutions of quadratic equations occur in conjugate pairs,


      #math.equation(block: true, alt: "x equals the fraction minus b over 2 a plus the fraction the square root of b squared minus 4 a c over 2 a and x equals the fraction minus b over 2 a minus the fraction the square root of b squared minus 4 a c over 2 a")[$x = frac(− b, 2 a) + frac(sqrt(b^(2) − 4 a c), 2 a) " " " " " and " " " " " x = frac(− b, 2 a) − frac(sqrt(b^(2) − 4 a c), 2 a)$]
      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,


      #math.equation(block: true, alt: "z equals the fraction minus b over 2 a plus the fraction i the square root of vertical bar D vertical bar over 2 a and z bar equals the fraction minus b over 2 a minus the fraction i the square root of vertical bar D vertical bar over 2 a")[$z = frac(− b, 2 a) + frac(i sqrt(| D |), 2 a) " " " " " and " " " " " limits(z)^(―) = frac(− b, 2 a) − frac(i sqrt(| D |), 2 a)$]
      Thus, if we know that #math.equation(block: false, alt: "z")[$z$] is a complex solution of a quadratic equation, we know that #math.equation(block: false, alt: "z bar")[$limits(z)^(―)$] is the other solution. The quadratic equation with solutions #math.equation(block: false, alt: "z")[$z$] and #math.equation(block: false, alt: "z bar")[$limits(z)^(―)$] is


      #math.equation(block: true, alt: "open parenthesis x minus z close parenthesis open parenthesis x minus z bar close parenthesis, equals 0; x squared minus open parenthesis z plus z bar close parenthesis plus z z bar, equals 0")[$( x − z ) ( x − limits(z)^(―) ) & = 0 \ x^(2) − ( z + limits(z)^(―) ) + z limits(z)^(―) & = 0$]
      #examplebox("Example 9")[][
+ Let #math.equation(block: false, alt: "z equals 7 minus 5 i")[$z = 7 − 5 i$]. Compute #math.equation(block: false, alt: "z z bar")[$z limits(z)^(―)$].
+ Find a quadratic equation with one solution 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")[$limits(z)^(―) = 7 + 5 i$], so
			#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")[$z limits(z)^(―) & = ( 7 − 5 i ) ( 7 + 5 i ) \  & = 49 − 25 i^(2) \  & = 49 + 25 \  & = 74$]
+ The other solution of the equation is #math.equation(block: false, alt: "z bar equals 7 plus 5 i")[$limits(z)^(―) = 7 + 5 i$], and the equation is #math.equation(block: false, alt: "open parenthesis x minus z close parenthesis open parenthesis x minus z bar close parenthesis equals 0")[$( x − z ) ( x − limits(z)^(―) ) = 0$]. We expand the product to find
				#math.equation(block: true, alt: "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; equals x squared minus open parenthesis 7 minus 5 i plus 7 plus 5 i close parenthesis x plus 74; equals x squared minus 14 x plus 74")[$( x − z ) ( x − limits(z)^(―) ) & = x^(2) − ( z + limits(z)^(―) ) x + z limits(z)^(―) \  & = x^(2) − ( 7 − 5 i + 7 + 5 i ) x + 74 \  & = x^(2) − 14 x + 74$]
		The equation is #math.equation(block: false, alt: "x squared minus 14 x plus 74 equals 0")[$x^(2) − 14 x + 74 = 0$].
]
]


      
        + Let #math.equation(block: false, alt: "z equals minus 3 plus 4 i")[$z = − 3 + 4 i$]. Compute #math.equation(block: false, alt: "z z bar equals")[$z limits(z)^(―) =$]\_\_\_\_\_
+ Find a quadratic equation with one solution being #math.equation(block: false, alt: "z equals minus 3 plus 4 i")[$z = − 3 + 4 i$]. #linebreak() Answer: \_\_\_\_\_#math.equation(block: false, alt: "equals 0")[$= 0$]


        Use #math.equation(block: false, alt: "x")[$x$] as the variable. The coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] should be 1.


        #solutionbox[
+ #math.equation(block: false, alt: "25")[$25$]
+ #math.equation(block: false, alt: "x squared plus 6 x plus 25 equals 0")[$x^(2) + 6 x + 25 = 0$]
]


      
      
        + Let #math.equation(block: false, alt: "z equals minus 3 plus 4 i")[$z = − 3 + 4 i$]. Compute #math.equation(block: false, alt: "z z bar")[$z limits(z)^(―)$]
+ Find a quadratic equation with one solution being #math.equation(block: false, alt: "z equals minus 3 plus 4 i")[$z = − 3 + 4 i$].


        #solutionbox[
+ #math.equation(block: false, alt: "25")[$25$]
+ #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.
]


      As a result, the factor theorem tells that every polynomial of degree #math.equation(block: false, alt: "n")[$n$] can be factored as the product of #math.equation(block: false, alt: "n")[$n$] 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.


      #figure(figph[quartic with no x-intercepts], alt: "quartic with no x-intercepts", caption: none)


      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: false, alt: "1 plus i")[$1 + i$], #math.equation(block: false, alt: "minus 1 plus i")[$− 1 + i$], #math.equation(block: false, alt: "minus 1 minus i")[$− 1 − i$] , and #math.equation(block: false, alt: "1 minus i")[$1 − i$]. 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 ,")[$x^(4) = attach(( x^(2) ), t: 2) = ( 2 i )^(2) = − 4 ,$]
      so #math.equation(block: false, alt: "x to the power 4 plus 4 equals open parenthesis minus 4 close parenthesis plus 4 equals 0")[$x^(4) + 4 = ( − 4 ) + 4 = 0$].


      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")[$x^(4) + 4 = [ x − ( 1 + i ) ] [ x − ( − 1 + i ) ] [ x − ( − 1 − i ) ] [ x − ( 1 − i ) ]$]
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
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")[$− 1 ± i$]. In fact, for every polynomial with real coefficients, the nonreal zeros always occur in complex conjugate pairs.
]


      
        If #math.equation(block: false, alt: "z equals 3 plus 5 i")[$z = 3 + 5 i$] is a solution to a polynomial equation, then so is:


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "3 minus 5 i")[$3 − 5 i$]
]


      
      
        If #math.equation(block: false, alt: "z equals 3 plus 5 i")[$z = 3 + 5 i$] is a solution to a polynomial equation, then so is:


        + #math.equation(block: false, alt: "5 plus 3 i")[$5 + 3 i$]
+ #math.equation(block: false, alt: "9 plus 25")[$9 + 25$]
+ #math.equation(block: false, alt: "3 minus 5 i")[$3 − 5 i$]
+ #math.equation(block: false, alt: "5 minus 3 i")[$5 − 3 i$]


      
      #examplebox("Example 10")[][
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: "f open parenthesis x close parenthesis equals x to the power 4 plus 15 x squared minus 16")[$f ( x ) = x^(4) + 15 x^(2) − 16$]. #linebreak() \_\_\_\_\_  Separate different solutions with a comma.
+ Write the polynomial in factored form. #linebreak() #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis")[$f \( x \) = \($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis open parenthesis")[$\) \($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis open parenthesis")[$\) \($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis open parenthesis")[$\) \($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis")[$\)$]


        #solutionbox[
+ #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")[$( x − 1 ) ( x + 1 ) ( x − 4 i ) ( x + 4 i )$]
]


      
      
        + Find the zeros of the polynomial #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power 4 plus 15 x squared minus 16")[$f ( x ) = x^(4) + 15 x^(2) − 16$].
+ Write the polynomial in factored form.


        #solutionbox[
+ #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")[$( x − 1 ) ( x + 1 ) ( x − 4 i ) ( x + 4 i )$]
]


      
      
        What can you say about the graph of a quadratic function whose zeros are complex numbers?


        \_\_\_\_\_


      
      
        What can you say about the graph of a quadratic function whose zeros are complex numbers?


      
    
    
      
=== Graphing Complex Numbers


      Real numbers can be plotted on a number line, but to graph a complex number we use a plane, called the complex plane. In the #strong[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].


      #figure(figph[plots of complex conjugates], alt: "plots of complex conjugates", caption: none)


      To plot a complex number a + bi, 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. 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 at left.


      #examplebox("Example 11")[][
Plot the numbers #math.equation(block: false, alt: "z")[$z$], #math.equation(block: false, alt: "z bar")[$limits(z)^(―)$], #math.equation(block: false, alt: "minus z")[$− z$], and #math.equation(block: false, alt: "minus z bar")[$− limits(z)^(―)$] as points on the complex plane, for #math.equation(block: false, alt: "z equals 2 minus 2 i")[$z = 2 − 2 i$].


        
        #solutionbox[
- To plot #math.equation(block: false, alt: "z equals 2 minus 2 i")[$z = 2 − 2 i$], we move from the origin #math.equation(block: false, alt: "2")[$2$] units to the right and #math.equation(block: false, alt: "2")[$2$] units down.
- To plot #math.equation(block: false, alt: "z bar equals 2 plus 2 i")[$limits(z)^(―) = 2 + 2 i$], we move from the origin #math.equation(block: false, alt: "2")[$2$]  units to the right and #math.equation(block: false, alt: "2")[$2$] units up.
- To plot #math.equation(block: false, alt: "minus z equals minus 2 plus 2 i")[$− z = − 2 + 2 i$], we move from the origin #math.equation(block: false, alt: "2")[$2$] units to the left and #math.equation(block: false, alt: "2")[$2$] units up.
- To plot #math.equation(block: false, alt: "minus z bar equals minus 2 minus 2 i")[$− limits(z)^(―) = − 2 − 2 i$], we move from the origin #math.equation(block: false, alt: "2")[$2$] units to the left and #math.equation(block: false, alt: "2")[$2$] units down.


          All four points are plotted at right.


          #figure(figph[figure of comlex numbers arranged symmetrically with respect to the axes], alt: "figure of comlex numbers arranged symmetrically with respect to the axes", caption: none)
]
]


      
        Plot the following numbers as points on the complex plane.


        - #math.equation(block: false, alt: "z equals 1 plus i")[$z = 1 + i$]
- #math.equation(block: false, alt: "i z equals i plus i squared")[$i z = i + i^(2)$]
- #math.equation(block: false, alt: "i squared z equals i squared plus i cubed")[$i^(2) z = i^(2) + i^(3)$]
- #math.equation(block: false, alt: "i cubed z equals i cubed plus i to the power 4")[$i^(3) z = i^(3) + i^(4)$]


        #solutionbox[
A plot is below.
]


        #figure(figph[points in complex plane], alt: "points in complex plane", caption: none)


      
      
        Plot the following numbers as points on the complex plane.


        - #math.equation(block: false, alt: "z equals 1 plus i")[$z = 1 + i$]
- #math.equation(block: false, alt: "i z equals i plus i squared")[$i z = i + i^(2)$]
- #math.equation(block: false, alt: "i squared z equals i squared plus i cubed")[$i^(2) z = i^(2) + i^(3)$]
- #math.equation(block: false, alt: "i cubed z equals i cubed plus i to the power 4")[$i^(3) z = i^(3) + i^(4)$]


        #solutionbox[
#figure(figph[points in complex plane], alt: "points in complex plane", caption: none)
]


      
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
If we draw an arrow from the origin to the point #math.equation(block: false, alt: "a plus b i")[$a + b i$] in the complex plane, we can see that multiplication by #math.equation(block: false, alt: "i")[$i$] corresponds to rotating a point around the origin by #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] in the counterclockwise direction. For example, the figure at right shows the graphs of #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")[$i z = 2 i − 3$].


        #figure(figph[figure of comlex number multiplied by i causing a rotation of 90 degrees], alt: "figure of comlex number multiplied by i causing a rotation of 90 degrees", caption: none)
]


    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Imaginary unit
- Imaginary number
- Complex number
- Imaginary axis
- Complex conjugate
- Real part
- Imaginary part
- Complex plane
- Real axis


      
      
        
==== CONCEPTS


        + The square root of a negative number is an imaginary number.
+ A complex number is the sum of a real number and an imaginary number.
+ We can perform the four arithmetic operations on complex numbers.
+ The product of a nonzero complex number and its conjugate is always a positive real number.
+ #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.
+ We can graph complex numbers in the complex plane.
+ 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.


      
      
        
==== 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


        Practice each skill in the Homework problems listed.


        + Write and simplify complex numbers: \#1–10
+ Perform arithmetic operations on complex numbers: \#11–36
+ Evaluate polynomials at complex numbers, expand polynomials: \#37–48
+ Find a polynomial with given zeros: \#53–56, 59–62
+ Graph complex numbers: \#63–70


      
    
    
      
=== Homework 7.3


      For Problems 1–6, 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: true, alt: "the square root of minus 25 minus 4")[$sqrt(− 25) − 4$]


        
          #math.equation(block: true, alt: "minus 4 plus 5 i")[$− 4 + 5 i$]


        
      
      
        #math.equation(block: true, alt: "the square root of minus 9 plus 3")[$sqrt(− 9) + 3$]


      
      
        #math.equation(block: true, alt: "the fraction minus 8 plus the square root of minus 4 over 2")[$display(frac(− 8 + sqrt(− 4), 2))$]


        
          #math.equation(block: true, alt: "minus 4 plus i")[$− 4 + i$]


        
      
      
        #math.equation(block: true, alt: "the fraction 6 minus the square root of minus 36 over 2")[$display(frac(6 − sqrt(− 36), 2))$]


      
      
        #math.equation(block: true, alt: "the fraction minus 5 minus the square root of minus 2 over 6")[$display(frac(− 5 − sqrt(− 2), 6))$]


        
          #math.equation(block: true, 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: true, alt: "the fraction 7 plus the square root of minus 3 over 4")[$display(frac(7 + sqrt(− 3), 4))$]


      
      For Problems 7–10, find the zeros of the quadratic polynomial. Write each 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 11–14, 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 15–24, 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 plus 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 17 plus 34 i")[$− 17 + 34 i$]


        
      
      
        #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 25–36, 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: "7 plus 4 i")[$7 + 4 i$]


        
      
      
        #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")[$display(frac(4 − 6 i, − 3 − 2 i))$]


      
      For Problems 37–42, 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 43–48, expand each 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 what values of #math.equation(block: false, alt: "x")[$x$] will #math.equation(block: false, alt: "the square root of x minus 5")[$sqrt(x − 5)$] be real? Imaginary?


        
          #math.equation(block: true, alt: "x greater than or equal to 5")[$x ≥ 5$]; #math.equation(block: true, alt: "x less than 5")[$" " x < 5$]


        
      
      
        For what values of #math.equation(block: false, alt: "x")[$x$] will #math.equation(block: false, alt: "the square root of x plus 3")[$sqrt(x + 3)$] be real? Imaginary?


      
      
        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")[$i^(− 6)$]


      
      In Problems 53–56,


      + 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")[$x^(2) − 4 x − 1$]


        
      
      
        #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")[$x^(2) − 8 x + 25$]


        
      
      
        #math.equation(block: true, alt: "5 plus i")[$5 + i$]


      
      Every polynomial factors into a product of a constant and linear factors of the form #math.equation(block: false, alt: "open parenthesis x minus a close parenthesis")[$( x − a )$], where #math.equation(block: false, alt: "a")[$a$] can be either real or complex. In Problems 57–58, how many linear factors are in the factored form of the given polynomial?


      
        + #math.equation(block: false, alt: "x to the power 4 minus 2 x cubed plus 4 x squared plus 8 x minus 6")[$x^(4) − 2 x^(3) + 4 x^(2) + 8 x − 6$]
+ #math.equation(block: false, alt: "2 x to the power 5 minus x cubed plus 6 x minus 4")[$2 x^(5) − x^(3) + 6 x − 4$]


        
          + #math.equation(block: false, alt: "4")[$4$]
+ #math.equation(block: false, alt: "5")[$5$]


        
      
      
        + #math.equation(block: false, alt: "x to the power 6 minus 6 x")[$x^(6) − 6 x$]
+ #math.equation(block: false, alt: "x cubed plus 3 x squared minus 2 x plus 1")[$x^(3) + 3 x^(2) − 2 x + 1$]


      
      For Problems 59–62, 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 plus 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")[$x^(4) − 6 x^(3) + 23 x^(2) − 50 x + 50$]


        
      
      
        #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")[$x^(4) − 7 x^(3) + 20 x^(2) − 19 x + 13$]


        
      
      
        #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")[$− display(frac(sqrt(2), 2)) + display(frac(sqrt(2), 2)) i , " " 4 − i$]


      
      For Problems 63–66, plot each 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)


          The complex conjugates are reflections of each other across the real axis.


        
      
      
        #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)


          The complex conjugates are reflections of each other across the real axis.


        
      
      
        #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 59–62, 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")[$i^(4)$]


        
          #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")[$i ( 1 + 2 i )$]


        
          #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")[$i ( 3 − 4 i )$]


      
      Problems 71–72 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: "m equals the fraction b over a")[$m = display(frac(b, a))$]
+ #math.equation(block: false, alt: "m equals the fraction a over minus b")[$m = display(frac(a, − b))$]
+ #math.equation(block: false, alt: "minus 1")[$− 1$]; The angle is #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 #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: "i z")[$i z$]?
+ 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: "i z")[$i z$]?
+ What can you conclude about the angle between the two segments from parts (a) and (b)?