Powers and Roots of a Complex Number

Learning Objectives

1. Powers of complex numbers using De Moivre’s Theorem 2. Finding roots of complex numbers.

3. Solving equations with complex roots.

Connections between complex numbers and trigonometry streamline many calculations that would otherwise prove to be very tedious. In fact, we will extend the ideas presented in this section to find the complex roots of certain polynomials.

Objective: Powers of complex numbers using De Moivre’s Theorem.

In this section, we introduce a powerful theorem that will allow us to streamline the process of raising complex numbers to powers. For example, to raise z= 2+i 2=2cis 45°to the fourth power in the traditional manner we would begin as follows:

( )

As you can see, this would be quite the cumbersome calculation. However, if we convert it to trigonometric form, then we could apply the product theorem. For example, let

(

^{cos sin}

)

^cis

Now multiplying by another factor of z and applying the product theorem again we have,

( )

There is a pattern; raise the magnitude r to the power and use the power as a multiple of the argument θ. This is true in general and leads to De Moivre’s theorem which states that given any real number n,

Now it’s just a matter of replacing the cosine and sine values and simplifying.

( )

( )

4 16 cos180 sin180

16 1 0

Solution: Begin by finding trigonometric form of the complex number z= − +1 i 3. 1 3 2cis120

z= − +i = °

Next, raise this to the fifth power and apply De Moivre’s theorem.

[ ]

⁵

Now simplify to find the result.

600

( )

The theorem can be applied when the argument is expressed using radians.

Example: Find

(

^{1 i}−

)

⁴.

Solution: Begin by finding trigonometric for the complex number z= −1 i. In this case, we choose to express the argument using radians.

1 2 cis7 z_{= − =}i 4π

Next, raise this to the fourth power and apply De Moivre’s theorem.

4 Now simplify to find the result.

( )

Objective: Finding roots of complex numbers.

Recall from algebra that an nth root of a number is another number that when raised to the nth power gives the number and is denoted a^{1/ n} = ⁿa. Some examples follow, of 25 as well. In fact, there are n distinct nth roots for any complex number. We can see that 3 is a cube root of 27, the other two cube roots are complex numbers. De Moivre’s theorem helps us find all of the nth roots of any complex number. Let’s consider 27 as a complex number,

( )

In this case, k is an integer and we can substitute as follows:

0

When k =3 the angle is coterminal with 360° and the results begin to repeat. These are the angles that give us the 3 distinct cube roots of 27.

( )

Notice that the two complex cube roots of 27 are conjugates. This process can be worked in general, which leads to the nth root theorem. If ^z=r

(

^cosθ+isinθ

)

^=rcisθ is any complex number, then the n distinct complex roots w , where k is an integer, are _k

( )

^1/

Note that the formula is very similar to De Moivre’s theorem, with only an adjustment to the argument.

Example: Find the two square roots of 1+i 3.

Solution: Begin by expressing the complex number in trigonometric form.

( )

1

k= ^{60 360} 210 2

° + °⋅1= °

These are the angles that allow us to find the two square roots.

( )

Sometimes the complex number will be given in trigonometric form.

Example: Find the three cube roots ^{z=8 cos 45}

(

^{° +isin 45}° . Leave your answers in

)

These are the angles that allow us to find the three cube roots.

( )

8 cos15 sin15 2 cis15 8 cos135 sin135 2 cis135

8 cos 255 sin 255 2 cis 255

Answer: w₀ =2 cis15° , w₁=2 cis135° , w₂ =2 cis 255°

Try this! Find the four 4th roots of i in trigonometric form.

Answer: w₀ =cis 22.5° , w₁ =cis112.5° , w₂ =cis 405° , w₃ =cis 292.5°

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Objective: Solving equations with complex roots.

The fundamental theorem of algebra states that for any degree-n polynomial with complex coefficients, there are exactly n roots and the nth root theorem can help find them. For example, consider the polynomial equation

2

The solutions are the two square roots of the imaginary unit, i. Writing the imaginary unit in trigonometric form we have,

( )

These are the angles that allow us to find the two square roots.

( )

Therefore we can say,

Certainly, you could check by multiplying.

2 2 2

The second square root found here is left to the reader to check.

Example: Solve: x⁵− = . Round to 4 decimal places. 1 0 Solution: Since the degree is 5, we expect five solutions.

5

In other words, we are trying to find the five roots of 1. Writing 1 in trigonometric form, we have

Substitute 0,1,2,3,4 in for k.

0

2

These are the angles that allow us to find the five cube roots.

Answer: [ WolframAlpha Solution ]

0

cos 72 sin 72 0.3090 0.9511 cos144 sin144 0.8090 0.5878 cos 216 sin 216 0.8090 0.5878 cos 288 sin 288 0.3090 0.9511

w i i

The above example is of particular importance in higher level mathematics. Any complex number such that

n 1 z =

is called a root of unity or a De Moivre number. Furthermore, these numbers have interesting symmetries on the unit circle. The five roots of 1 found in the previous example are plotted in the complex plane below.

Notice that they come in conjugate pairs and are evenly spaced around the unit circle.

Try this! Solve: x³+ =1 0. Answer: −1, ^{1 3}

2+ 2 , ^{1 3} 2− 2

607

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Key Takeaways ***** ***** *****

• De Moivre’s theorem streamlines the process of raising complex number to a large power.

n ncis z =r nθ

• De Moivre’s theorem leads to the nth root theorem, which allows us to find roots of complex numbers.

• The fundamental theorem of algebra states that there are exactly n roots for any nth degree polynomial with complex coefficients. The nth root theorem gives us a method that we can use to solve polynomials.

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Section Exercises Part A

Simplify and write your answer in rectangular form. Use exact values.

1. ^_^{3 cos 60}

(

^{° +isin 60°}

)

^_²

16.

Find the two square roots of the following numbers. Use exact values and express your answer in standard form.

25. ^{4 cos120}

(

^{° +isin120°}

)

Find the three cube roots of the following numbers. Leave answers in trigonometric form.

Find the three cube roots. Use exact values and give answers in rectangular form.

45. 8i

Part C

Solve. Give exact solutions.

53. x³− =1 0 54. x³− =8 0 55. x³+27= 0 56. x³−27= 0

57. x⁴− =1 0 58. x⁴+ =1 0 59. x³− = i 0 60. x³+ = i 0 Solve. Round to 4 decimal places.

61. x⁵+ =1 0 62. x⁵+ =i 0

63. x⁴+ =i 0 64. x⁴− =i 0

Solve. Round to 4 decimal places. (Hint: Use the quadratic formula to solve for x^2 and then find the square roots of the result.)

65. x⁴ −2x² + =2 0 66. x⁴−4x²+ =5 0

67. x⁴−2x²+ =4 0 68. x⁴−2x+ =3 0

610

Answers to Section Exercises

50. − , 1 ^{1 3}

In document Chapter 8 Chapter 8 - Complex Numbers and Polar Graphs (Page 33-47)