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# 4.8: Powers and Roots of Complex Numbers

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use De Moivre’s Equation to find powers of complex numbers
• Prove De Moivre’s Theorem
• Use De Moivre’s theorem to find all solutions to an equation

## Introduction

Whether it is adding, subtracting, multiplying, dividing or some other mathematical operation that is being done on two or more complex numbers, there will be more than one method- using rectangular form or polar form

## De Moivre’s Theorem

How do we raise a complex number to a power? Let’s start with an example

(-4 - 4i)3 = (-4 - 4i) x (-4 - 4i) x (-4 - 4i) =

In rectangular form, this can get very complex. What about in r cis θ form?

(44i)=42 cis(5π4)\begin{align*}(-4 - 4i) = 4\sqrt{2}\ cis \left ( \frac{5\pi}{4} \right )\end{align*}

So the problem becomes

\begin{align*}4\sqrt{2} \ \mbox{cis} \left ( \frac{5\pi}{4} \right ) \cdot \ 4\sqrt{2} \ \mbox{cis}\left ( \frac{5\pi}{4} \right ) \cdot \ 4\sqrt{2} \ \mbox{cis}\left ( \frac{5\pi}{4} \right )\end{align*}

and using our multiplication rule from the previous section,

\begin{align*}(-4 - 4i)^3 = (4\sqrt{2})^3 \ \mbox{cis}\left ( \frac{15\pi}{4} \right )\end{align*}

Notice, (a + bi)3 = r3 cis 3 θ

In words: Raise the r-value to the same degree as the complex number is raised and then multiply that by cis of the angle multiplied by the number of the degree.

Based on the example above, it appears that we can generalize

(a + bi)n = r cis() = r(cos() + i sin())

If this is correct, then the polar form provides a much faster result for raising a complex number to a power than doing the problem in rectangular form.

Example: Find the value of \begin{align*}(1 + \sqrt{3}i)^4\end{align*}

\begin{align*}r = \sqrt{(1)^2 + (\sqrt{3})^2} = 2\end{align*}

\begin{align*}\mbox{tan} \ \theta_{ref} = \frac{\sqrt{3}}{1},\end{align*}

and θ is in the 1st quadrant, so

\begin{align*}\theta = \frac{\pi}{3}\end{align*}

Using our equation from above:

\begin{align*}z^4 = r^4\ cis\ 4\theta\end{align*}

\begin{align*}z^4 = (2)^4\ cis\ 4\frac{\pi}{3}\end{align*}

Expanding cis form:

\begin{align*}z^4 = 16\left ( \mbox{cos} \left ( \frac{4\pi}{3} \right ) + i \ \mbox{sin}\left ( \frac{4\pi}{3} \right ) \right )\end{align*}

\begin{align*}= 16\left ( (-0.5) - 0.866i \right )\end{align*}

Finally we have

z4 = -8 - 13.856i

Check by complex rectangular multiplication. First we will find z2, then square that result:

\begin{align*}(1 + \sqrt{3}i)^2 = 1 + 2\sqrt{3}i - 3\end{align*}

\begin{align*}= -2 + 2\sqrt{3}i\end{align*}

Now by squaring the square, the fourth power is found or \begin{align*}(-2 + 2\sqrt{3}i)^2\end{align*}

\begin{align*}4 - 8\sqrt{3}i - 12 = -8 - 8\sqrt{3}i\end{align*}

\begin{align*}= -8 - 13.856i\end{align*}

The general rule for raising a complex number to any power is stated by De Moivre’s

Equation:

Let z = r(cos θ + i sin θ) be a complex number in rcisθ form. If n is a positive integer, zn is
zn = rn (cos() + i sin())

Proof:

The proof of De Moivre’s equation uses mathematical induction. Mathematical induction is a method of proving something is true for an infinite number of cases. In this example, we want to show that De Moivre’s equation works for all integer values of n.

The way we do this (and any induction proof) is first we will check that the theorem is true for n = 1. Then we assume it is true for n = k (called the induction step). Finally, we use the case n = k to show that if the theorem is true for n = k then it is also true for n = k + 1. Once we have done this, we have proven the theorem always works for any integer n.

If n = 1, then the theorem is

z1 = r1 (cos(1 \begin{align*}\cdot\end{align*} θ) + i sin(1 \begin{align*}\cdot\end{align*} θ))

Which is the rcis form of z

Now we assume that

zk = rk (cos() + i sin())

Then we check the case of zk + 1.

zk + 1 = zk z

By the induction step,

\begin{align*}z^{k+1} = r^k (\mbox{cos}(k\theta) + i \ \mbox{sin}(k\theta))r(\mbox{cos}(\theta) + i \ \mbox{sin}(\theta))\end{align*}

\begin{align*}= r^{k+1} (\mbox{cos}(k\theta) + i \ \mbox{sin}(k\theta))(\mbox{cos}(\theta) + i \ \mbox{sin}(\theta))\end{align*}

\begin{align*}= r^{k+1} (\mbox{cos}(k\theta)\mbox{cos}(\theta) + i \ \mbox{sin}(k\theta)\mbox{cos}(\theta) + i \ \mbox{cos}(k\theta)\mbox{sin}(\theta) + i^2 \ \mbox{sin}(k\theta)\mbox{sin}(\theta))\end{align*}

Gathering like terms and using i2 = -1

\begin{align*}= r^{k+1} (\mbox{cos}(k\theta)\mbox{cos}(\theta) - \mbox{sin}(k\theta)\mbox{sin}(\theta) + i \ (\mbox{sin}(k\theta)\mbox{cos}(\theta) + \mbox{cos}(k\theta)\mbox{sin}(\theta)))\end{align*}

Finally, we use the angle addition identities, and we get

\begin{align*}= r^{k+1} (\mbox{cos}(k\theta + \theta) + i \ \mbox{sin}(k\theta + \theta))\end{align*}

\begin{align*}= r^{k+1} (\mbox{cos}((k + 1)\theta) + i \ \mbox{sin}((k + 1)\theta))\end{align*}

And now we are done. We have shown for any positive integer n,

\begin{align*}z^n = r^n (\mbox{cos}(n\theta) + i \ \mbox{sin}(n\theta)).\end{align*}

## nth Root Theorem

Often in mathematics whenever an operation is presented, the inverse operation follows. This will be the case for the current topic. The inverse operation of finding a power for a number is to find a root of the same number.

The problem, for example is find the square root of (1 + i).

Previously you have learned that the square root of a number, x, was shown to equal x1/2.

Any root, say the nth root can be written as x1/n

The formula for De Moivre’s theorem also works for fractional powers (though we won’t prove that here). That means that this same formula can be used for finding roots:

\begin{align*}z^{1/n} = (a + bi)^{1/n} = r^{1/n} cis \left ( \frac{\theta}{n} \right )\end{align*}

Example: Find \begin{align*}\sqrt{1 + i}\end{align*}

Fist, rewriting in exponential form: (1 + i)½

And now in polar form:

\begin{align*}\sqrt{1 + i} = \left (\sqrt{2}\ cis \left (\frac{\pi}{4} \right ) \right )^{1/2}\end{align*}

Expanding cis form,

\begin{align*}= \left (\sqrt{2} \left ( \mbox{cos} \left ( \frac{\pi}{4} \right ) + i \ \mbox{sin} \left ( \frac{\pi}{4} \right ) \right ) \right )^{1/2}\end{align*}

Using the formula:

\begin{align*}=(2^{1/2})^{1/2} \left ( \mbox{cos} \left ( \frac{1}{2} \cdot \frac{\pi}{4} \right ) + i \ \mbox{sin} \left ( \frac{1}{2} \cdot \frac{\pi}{4} \right ) \right )\end{align*}

\begin{align*}= 2^{1/4} \left ( \mbox{cos} \left ( \frac{\pi}{8} \right ) + i \ \mbox{sin} \left ( \frac{\pi}{8} \right ) \right )\end{align*}

In decimal form, we get

=1.189( 0.924 + 0.383i)

=1.099 + 0.455i

To check, we will multiply the result by itself in rectangular form:

\begin{align*}(1.099 + 0.455i)\ \cdot \ (1.099 + 0.455\mbox{i}) = 1.099^2 + 1.099(0.455i) + 1.099(0.455i) \ +\end{align*} \begin{align*}(0.455i)^2\end{align*}

\begin{align*}= 1.208 + 0.500i + 0.500i + 0.208i^2\end{align*}

\begin{align*}= 1.208 + i - 0.208 \ \mbox{or}\end{align*}

\begin{align*}= 1 + i\end{align*}

The Fundamental Theorem of Algebra states that the number of solutions for an equation is equal to the degree of an equation. Finding \begin{align*}\sqrt{1 + i}\end{align*}, is the same as solving for x in x2 = 1 + i.

That would mean that there are two values for x, and in the solution above, but we only found one value. What is the other value? How is it found?

If the square root of a real number, n, is b, then both b2 and (-b)2 will equal n. Can this also be true for complex numbers?

In the above example \begin{align*}\sqrt{1 + i}\end{align*} is equivalent to \begin{align*}\sqrt{\sqrt{2}} \ \left ( \mbox{cos}\left [ \frac{\pi}{8} \right ] + i \ \mbox{sin} \left [ \frac{\pi}{8} \right ] \right )\end{align*},

or in approximate decimal form: 1.099 + 0.455i.

The opposite of this number is (-1.099 – 0.455i).

Squaring the opposite,

\begin{align*}(-1.099 - 0.455\mbox{i})^2 = (-1.099)^2 + (-1.099)(-0.455i) + (-1.099)(-0.455i) \ +\end{align*} \begin{align*}(-0.455i)^2\end{align*}

\begin{align*}= 1.208 + 0.500i + 0.500i + 0.208i^2\end{align*}

\begin{align*}= 1.208 + i - 0.208\end{align*}

\begin{align*}= 1 + i\end{align*}

Graphing both of these complex solutions on the same polar axis reveals an important feature of the roots of complex numbers:

What this reveals is that both solutions are the same distance from the origin, and but they are rotated by π radians when graphed. This suggests is that the roots of complex numbers, when graphed, are spaced apart in such a manner, that the difference between their angles is equal. In this case, both solutions are separated by π radians. Finding the cube root of a complex number would have the three solutions be \begin{align*}\frac{2\pi}{3}\end{align*} radians or 120º apart when graphed. Finding the fourth root of a complex number would have the four solutions be \begin{align*}\frac{2\pi}{4} = \frac{\pi}{2}\end{align*} radians or 90º apart.

The general rule would be to find the first root by the method seen in the above example. Leave the answer in cis form. To find the additional roots use rotations of \begin{align*}\frac{2\pi}{n}\end{align*} radians or \begin{align*}\frac{360^\circ}{n}\end{align*} to find the angle difference between the roots. Add this angle “n - 1” times to the first angle, until a complete circle has been formed.

The procedures, when combined, are known as De Moivre’s Theorm.

## Solve Equations using De Moivre’s Theorem

Example: Find the value of \begin{align*}x : x^3 = \left (1 - \sqrt{3}i \right )\end{align*}

First we put \begin{align*}1 - \sqrt{3}i\end{align*} in polar form.

Use \begin{align*}x = 1, \ y = -\sqrt{3}\end{align*} to obtain \begin{align*}r = 2, \ \theta = \frac{5\pi}{3}\end{align*}

let \begin{align*}z = \left ( 1 - \sqrt{3}i \right )\end{align*} in rectangular form

\begin{align*}z = 2\ \mbox{cis}\ \left ( \frac{5\pi}{3} \right )\end{align*} in polar form

\begin{align*}x = \left (1 - \sqrt{3}i \right )^{1/3}\end{align*}

\begin{align*}x = \left [2cis \left ( \frac{5\pi}{3} \right ) \right ]^{1/3}\end{align*}

Use De Moivre’s Equation to find the first solution:

\begin{align*}x_1 = 2^{1/3} cis \left ( \frac{5\pi /3}{3} \right )\end{align*} or \begin{align*}2^{1/3} cis \left ( \frac{5\pi}{9} \right )\end{align*}

Leave answer in cis form to find the remaining solutions:

n = 3 which means that the 3 solutions are \begin{align*}\frac{2\pi}{3}\end{align*} radians apart or

\begin{align*}x_2 = 2^{1/3} cis \left ( \frac{5\pi}{9} + \frac{2\pi}{3} \right )\end{align*} and \begin{align*}x_3 = 2^{1/3} cis \left ( \frac{5\pi}{9} + \frac{2\pi}{3} + \frac{2\pi}{3} \right )\end{align*}

Note: It is not necessary to add \begin{align*}\frac{2\pi}{3}\end{align*} again. Adding \begin{align*}\frac{2\pi}{3}\end{align*} three times equals 2π. That would result in rotating around a full circle and to start where it all began- that is the first solution.

The three solutions are:

\begin{align*}x_1 = 2^{1/3} cis \left (\frac{5\pi}{9} \right )\end{align*}

\begin{align*}x_2 = 2^{1/3} cis \left (\frac{11\pi}{9} \right )\end{align*}

\begin{align*}x_3 = 2^{1/3} cis \left (\frac{17\pi}{9} \right )\end{align*}

Each of these solutions, when graphed will be \begin{align*}\frac{2\pi}{3}\end{align*} apart.

Check any one of these solutions, for example the second solution, to see if the results are confirmed. To check the first and third solution will be given as an exercise at the back of the chapter.

\begin{align*}x_2 = 2^{1/3} cis \left (\frac{11\pi}{9} \right )\end{align*}

\begin{align*}= 1.260 \left [ \mbox{cos} \left (\frac{11\pi}{9} \right ) + i \ \mbox{sin} \left (\frac{11\pi}{9} \right ) \right ]\end{align*}

\begin{align*}= 1.260[-0.766 - 0.643i]\end{align*}

\begin{align*}= -0.965 - 0.810i\end{align*}

Does (-0.965 – 0.810i)3 or (-0.965 – 0.810i) (-0.965 – 0.810i) (-0.965 – 0.810i)

\begin{align*}= \left (1-\sqrt{3}i\right )?\end{align*}

Using polar multiplication would in effect be the reverse of De Moivre’s Theorem. Using rectangular or algebraic multiplication would yield an independent check!

(-0.965 – 0.810i) (-0.965 – 0.810i) = 0.931 + 0.782i + 0.782i + 0.656i2

or

(0.932 – 0.656) + 2(0.782)i

or

0.276 + 1.564i

Now multiply again

(0.276 + 1.564i)(-0.965 - 0.810i) = -0.266 – 0.224i – 1.509i – 1.266i2 or

(-0.266 +1.266) – (0.224 + 1.509)i or

1 – 1.733i or approximately \begin{align*}1 - \sqrt{3}i\end{align*}

Example: What are the two square roots of i? Let \begin{align*}z = \sqrt{0 + i}\end{align*}

\begin{align*}r = 1, \ \theta = \pi/2\end{align*} or \begin{align*}z = \left [1 \times \ cis \frac{\pi}{2} \right ]^{1/2}\end{align*} Utilizing De Moivre’s Theorem:

\begin{align*}z_1 = \left [1 \times \ cis \frac{\pi}{4} \right ]\end{align*} or \begin{align*}z_2 = \left [1 \times \ cis \frac{5\pi}{4} \right ]\end{align*}

\begin{align*}z_1 = 1\left ( \mbox{cos}\frac{\pi}{4} + i \ \mbox{sin}\frac{\pi}{4} \right )\end{align*} or \begin{align*}z_2 = 1 \left (\mbox{cos} \frac{5\pi}{4} + i \ \mbox{sin} \frac{5\pi}{4} \right )\end{align*}

\begin{align*}z_1 = 0.707 + 0.707i\end{align*} or \begin{align*}z_2 = -0.707 - 0.707i\end{align*}

Check for z1 solution: (0.707 + 0.707i)2 = i?

0.500 + 0.500i + 0.500i + 0.500i2 = 0.500 + i + 0.500(-1) or i

Example: What are the four fourth roots of 1?

Let z = 1 or z = 1 + 0i Then the problem becomes find z1/4 = (1 + 0i)1/4

Since \begin{align*}r = 1 \ \theta = 0, \ z^{1/4} = [1 \times cis \ 0]^{1/4}\end{align*} with \begin{align*}z_1 = 1^{1/4} \left (\mbox{cos}\ \frac{0}{4} + i \ \mbox{sin}\ \frac{0}{4} \right )\end{align*} or \begin{align*}1(1 + 0)\end{align*} or \begin{align*}1\end{align*}

That root is not a surprise. Now use De Moivre’s to find the other roots:

\begin{align*}z_2 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{\pi}{2} \right ) \right ]\end{align*} Since there are 4 roots, dividing 2π by 4 yields 0.5π

or 0 + i or just i \begin{align*}z_3 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{2\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{2\pi}{2} \right ) \right ]\end{align*} which yields z3 = -1

Finally \begin{align*}z_4 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{3\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{3\pi}{2} \right ) \right ]\end{align*} or \begin{align*}z_4 = -i\end{align*}

The four fourth roots of 1 are 1, i, -1 and -i

Example: What are the three cube roots of 1?

Let z = 1 or z = 1 + 0i Since r = 1 θ = 0 Then the problem becomes find z1/3 = [1 × cis 0]1/3

\begin{align*}z_1 = 1^{1/3}[\mbox{cos}(0) + i \ \mbox{sin}(0)]\end{align*} or \begin{align*}1\end{align*}

\begin{align*}z_2 = 1^{1/3} \left [\mbox{cos}\left (0 + \frac{2\pi}{3}\right ) + i \ \mbox{sin}\left (0 + \frac{2\pi}{3} \right ) \right ]\end{align*} or \begin{align*}-0.500 + 0.866i\end{align*}

\begin{align*}z_3 = 1^{1/3} \left [\mbox{cos}\left (0 + \frac{4\pi}{3}\right ) + i \ \mbox{sin}\left (0 + \frac{4\pi}{3} \right ) \right ]\end{align*} or \begin{align*}-0.500 + 0.866i\end{align*}

Check z2 : (-0.500 + 0.866i)3 = 1?

(-0.500 + 0.866i)(-0.500 + 0.866i) = 0.25 - 0.866i + .750i2 or -0.500 - 0.866i

(-0.500 - 0.866i)(-0.500 + 0.866i) = 0.25 – 0.866i + 0.866i -0.750 i2 or 1.

## Applications, Technological Tools

After utilizing De Moivre’s Theorem, answers can be checked using a graphing calculator. Convert the given values into a + bi form and then put the calculator into “pol” mode. Perform the operation.

## Lesson Summary

There is only one practical method for finding the power or roots of a complex number. Before raising a complex number to a power or finding a root or roots of a complex number, the complex number must be in polar or cis form, and then De Moivre’s Theorem can be utilized.

## Points to Consider

Having a number in polar form can yield results that the rectangular form could NOT! Recall the Correspondence Principle. Sometimes a method learned at an earlier level in mathematics will work up to a point. When that method will no longer work, newer, sometimes more complicated, or seemingly more complicated methods are introduced, often to the protestations of students. Often these new methods are helpful for solving a wider range of problems.

## Review Questions

1. Perform indicated operation on these complex numbers:
1. Divide: \begin{align*}\frac{2 + 3i}{1 - i}\end{align*}
2. Multiply: \begin{align*}(-6 - i)(-6 + i)\end{align*}
3. Multiply: \begin{align*}\left (\frac{\sqrt{3}}{2} - \frac{1}{2} i \right )^2\end{align*}
4. Find the product using polar form: \begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*}
5. Multiply: \begin{align*}2(\mbox{cos} \ 40^\circ + i \ \mbox{sin} \ 40^\circ) \times 4(\mbox{cos} \ 20^\circ + i \ \mbox{sin} \ 20^\circ)\end{align*}
6. Multiply: \begin{align*}2 \left (\mbox{cos} \ \frac{\pi}{8} + i \ \mbox{sin} \ \frac{\pi}{8} \right ) \times 2 \left (\mbox{cos} \ \frac{\pi}{10} + i \ \mbox{sin} \ \frac{\pi}{10} \right )\end{align*}
7. Divide: \begin{align*}2(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ) \div 6(\mbox{cos} \ 200^\circ + i \ \mbox{sin} \ 200^\circ)\end{align*}
8. Divide: \begin{align*}3\ \mbox{cis}(130^\circ) \div 4\ \mbox{cis}(270^\circ)\end{align*}
2. Use De Moivre’s Theorem:
1. \begin{align*}[3(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ)]^3\end{align*}
2. \begin{align*}\left [\sqrt{2} \left (\mbox{cos}\ \frac{5\pi}{16} + i \ \mbox{sin} \ \frac{5\pi}{16} \right ) \right ]^4\end{align*}
3. \begin{align*}\left (\sqrt{3} - i \right )^6\end{align*}
4. The 3 complex cube roots of 1 + i
5. The 4 complex fourth roots of - 16i
6. The five complex fifth roots of i

## Date Created:

Feb 23, 2012

Jun 08, 2015
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