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DeMoivre's Theorem and nth Roots

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De Moivre's Theorem and nth Roots

You know how to multiply two complex numbers together and you’ve seen the advantages of using trigonometric polar form, especially when multiplying more than two complex numbers at the same time.  Because raising a number to a whole number power is repeated multiplication, you also know how to raise a complex number to a whole number power. 

What is a geometric interpretation of squaring a complex number? 

Watch This

http://www.youtube.com/watch?v=Sf9gEzcVZkU James Sousa: De Moivre’s Theorem: Powers of Complex Numbers in Trig Form

Guidance

Recall that if z_1=r_1 \cdot \text{cis} \ \theta_1  and  z_2=r_2 \cdot \text{cis} \ \theta_2 with r_2 \neq 0 , then z_1 \cdot z_2=r_1 \cdot r_2 \cdot \text{cis} \ (\theta_1+\theta_2) .

If z_1=z_2=z=r \ \text{cis} \ \theta  then you can determine z^2  and z^3 :

z^2 &= r \cdot r \cdot \text{cis} \ (\theta+\theta)=r^2 \ \text{cis} \ (2 \cdot \theta)\\z^3 &=  r^3 \ \text{cis} \ (3 \cdot \theta)

De Moivre’s Theorem simply generalizes this pattern to the power of any positive integer.

z^n=r^n \cdot \text{cis} \ (n \cdot \theta)

In addition to raising a complex number to a power, you can also take square roots, cube roots and  n^{th} roots of complex numbers.  Suppose you have complex number z=r \ \text{cis} \ \theta  and you want to take the  n^{th} root of z .  In other words, you want to find a number v=s \cdot \text{cis} \ \beta  such that v^n=z . Do some substitution and manipulation:

v^n &= z\\(s \cdot \text{cis} \ \beta)^n &= r \cdot \text{cis} \ \theta\\s^n \cdot \text{cis} \ (n \cdot \beta) &= r \cdot \text{cis} \ \theta

You can see at this point that to find s  you need to take the  n^{th} root of r .  The trickier part is to find the angles, because n \cdot \beta  could be any angle coterminal with \theta .  This means that there are n  different  n^{th} roots of z .

n \cdot \beta &= \theta+2 \pi k\\\beta &= \frac{\theta + 2 \pi k}{n}

The number k  can be all of the counting numbers including zeros up to n-1 .  So if you are taking the  4^{th} root, then k=0,1,2,3

Thus the  n^{th} root of a complex number requires n  different calculations, one for each root:

v=\sqrt[n]{r} \cdot \text{cis} \ \left(\frac{\theta+2 \pi k}{n}\right) \ for \ \{ k \ \epsilon \ I | 0 \le k \le n-1 \}

Example A

Find the cube root of the number 8. 

Solution:  Most students know that 2^3=8  and so know that 2 is the cube root of 8.  However, they don’t realize that there are two other cube roots that they are missing.  Remember to write out k=0,1,2  and use the unit circle whenever possible to help you to find all three cube roots.

8 &= 8 \ \text{cis} \ 0=(s \cdot \text{cis} \ \beta)^3\\z_1 &= 2 \cdot \text{cis} \ \left(\frac{0+2 \pi \cdot 0}{3}\right)=2 \ \text{cis} \ 0=2 (\cos 0+i \cdot \sin 0)=2 (1+0)=2\\z_2 &= 2 \cdot \text{cis} \ \left(\frac{0+2 \pi \cdot 1}{3} \right)=2 \ \text{cis} \left(\frac{2 \pi}{3}\right) \\ & =2 \left(\cos \left(\frac{2 \pi}{3}\right)+i \cdot \sin \left(\frac{2 \pi}{3}\right) \right)=2 \left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i \right)=-1+i \sqrt{3}\\z_3 &= 2 \cdot \text{cis} \left(\frac{0+2 \pi \cdot 2}{3}\right)=2 \ \text{cis} \left(\frac{4 \pi}{3}\right) \\ & =2 \left(\cos \left(\frac{4 \pi}{3}\right)+i \cdot \sin \left(\frac{4 \pi}{3}\right) \right)=2 \left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)=-1-i \sqrt{3}

The cube roots of 8 are 2, -1+i \sqrt{3}, -1-i \sqrt{3} .

Example B

Plot the roots of 8 graphically and discuss any patterns you notice. 

Solution:


The three points are equally spaced around a circle of radius 2.  Only one of the points, 2+0i , is made up of only real numbers.  The other two points have both a real and an imaginary component which is why they are off of the  x axis. 

As you become more comfortable with roots, you can just determine the number of points that need to be evenly spaced around a certain radius circle and find the first point.  The rest is just logic.

Example C

What are the fourth roots of 16 \ \text{cis}\ 48^\circ ?

Solution:  There will be 4 points, each 90^\circ  apart with the first point at 2 \ \text{cis} \ (12^\circ)

2 \ \text{cis} \ (12^\circ), \ 2 \ \text{cis} \ (102^\circ), \ 2 \ \text{cis} \ (192^\circ), \ 2 \ \text{cis} \ (282^\circ)

Concept Problem Revisited

Squaring a complex number produces a new complex number.  The angle gets doubled and the magnitude gets squared, so geometrically you see a rotation.

Vocabulary

The  n^{th} roots of unity refer to the  n^{th} roots of the number 1. 

The real axis is the  x axis and the imaginary axis is the  y axis.  Together they make the complex coordinate plane. 

Guided Practice

1.  Check the three cube roots of 8 to make sure they are truly cube roots. 

2.  Solve for z  by finding the nth root of the complex number.

z^3=64-64 \sqrt{3}i

3. Use De Moivre’s Theorem to evaluate the following power.

\left(\sqrt{2}-\sqrt{2}i \right)^6

Answers:

1. 

z_1^3 &= 2^3=8\\z^3_2 &= \left(-1+i \sqrt{3}\right)^3\\&= \left(-1+i \sqrt{3}\right) \cdot \left(-1+i \sqrt{3} \right) \cdot (-1+i \sqrt{3})\\&= \left(1-2i \sqrt{3}-3\right) \cdot \left(-1+i \sqrt{3}\right)\\&= \left(-2-2i \sqrt{3} \right) \cdot \left(-1+i \sqrt{3}\right)\\&= 2-2i \sqrt{3}+2i \sqrt{3}+6\\&= 8

Note how many steps and opportunities there are for making a mistake when multiplying multiple terms in rectangular form.  When you check z_3 , use trigonometric polar form.

z_3^3 &= 2^3 \ \text{cis} \ \left(3 \cdot \frac{4 \pi}{3} \right)\\&= 8 (\cos 4 \pi+i \cdot \sin 4 \pi)\\&= 8(1+0)\\&= 8

2.  First write the complex number in \text{cis}  form.  Remember to identify k=0,1,2 . This means the roots will appear every \frac{360^\circ}{3}=120^\circ .

z^3 &= 64-64 \sqrt{3}i=128 \cdot \text{cis} \ 300^\circ\\z_1 &= 128^{\frac{1}{3}} \cdot \text{cis} \ \left(\frac{300}{3}\circ\right)=128^{\frac{1}{3}} \cdot \text{cis} (100^\circ)\\z_2 &= 128^{\frac{1}{3}} \cdot \text{cis} \ (220^\circ)\\z_3 &= 128^{\frac{1}{3}} \cdot \text{cis} \ (340^\circ)

3.   First write the number in trigonometric polar form, then apply De Moivre’s Theorem and simplify.

\left(\sqrt{2}-\sqrt{2}i\right)^6 &= (2 \ \text{cis} \ 315^\circ)^6\\&= 2^6 \cdot \text{cis} \ (6 \cdot 315^\circ)\\&= 64 \cdot \text{cis} \ (1890^\circ)\\&= 64 \cdot \text{cis} \ (1890^\circ)\\&= 64 \cdot \text{cis} \ (90^\circ)\\&= 64(\cos 90^\circ+i \cdot \sin 90^\circ)\\&= 64 (0+i)\\&= 64i  

Practice

Use De Moivre’s Theorem to evaluate each expression.  Write your answers in rectangular form.

1.  (1+i)^5

2.  \left(1-\sqrt{3}i\right)^3

3.  (1+2i)^6

4.  \left(\sqrt{3}-i\right)^5

5.  \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^4

6.  Find the cube roots of 3+4i .

7.  Find the  5^{th} roots of 32i .

8.  Find the  5^{th} roots of 1+\sqrt{5}i .

9.  Find the  6^{th} roots of - 64 and plot them on the complex plane.

10.  Use your answers to #9 to help you solve x^6+64=0 .

For each equation: a) state the number of roots, b) calculate the roots, and c) represent the roots graphically.

11.  x^3=1

12.  x^8=1

13.  x^{12}=1

14.  x^4=16

15.  x^3=27

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