Lesson Objectives
• Learn how to find powers of complex numbers
• Learn how to find roots of complex numbers

## How to find Powers and Roots of Complex Numbers Using De Moivre's Theorem

In this lesson, we will learn how to find powers and roots of complex numbers. In the last lesson, we learned how to multiply two complex numbers using the product theorem for complex numbers. Since raising a complex number in polar form to a positive integer power is a repeated application of the product theorem, we can consider the following: $$[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]^{2}$$ $$=[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)] \cdot [r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]$$ $$=r \cdot r[\text{cos}\hspace{.1em}(θ + θ) + i \hspace{.1em}\text{sin}\hspace{.1em}(θ + θ)]$$ $$=r^{2}(\text{cos}\hspace{.1em}2θ + i \hspace{.1em}\text{sin}\hspace{.1em}2θ)$$ Similarly: $$[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]^{3}$$ $$=r^{3}(\text{cos}\hspace{.1em}3θ + i \hspace{.1em}\text{sin}\hspace{.1em}3θ)$$ $$[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]^{4}$$ $$=r^{4}(\text{cos}\hspace{.1em}4θ + i \hspace{.1em}\text{sin}\hspace{.1em}4θ)$$ These results lead us to De Moivre's Theorem.

### De Moivre's Theorem

For any complex number in polar form: $$r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ And any real number n, we can state the following: $$[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]^{n}$$ $$=r^{n}(\text{cos}\hspace{.1em}nθ + i \hspace{.1em}\text{sin}\hspace{.1em}nθ)$$ In compact form: $$[r \hspace{.1em}\text{cis}\hspace{.1em}θ]^{n}=r^{n}(\text{cis}\hspace{.1em}nθ)$$ Let's look at a few examples.
Example #1: Simplify, write the answer in polar form. $$\left(-\frac{5\sqrt{3}}{2}+ \frac{5}{2}i\right)^{2}$$ For this problem, we are given our complex number in rectangular form. Let's convert over to polar form in order to use De Moivre's theorem.
Let's start by finding r, the absolute value of the complex number. $$r=\sqrt{\left(-\frac{5\sqrt{3}}{2}\right)^2 + \left(\frac{5}{2}\right)^{2}}$$ $$=\sqrt{\frac{75}{4}+ \frac{25}{4}}$$ $$=\sqrt{\frac{100}{4}}$$ $$=\sqrt{25}$$ $$=5$$ The r value will be 5. Let's find our argument for the complex number. $$\text{tan}\hspace{.1em}θ=\frac{5}{2}\cdot -\frac{2}{5\sqrt{3}}$$ $$=-\frac{1}{\sqrt{3}}$$ $$=-\frac{\sqrt{3}}{3}$$ Let's use the inverse tangent function to find our reference angle: $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=\frac{π}{6}$$ Note: If you want to work with degrees: $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ Our complex number lies in quadrant II since the real part is negative and the imaginary part is positive. To find an angle in quadrant II with our given reference angle: $$θ=π - \frac{π}{6}$$ $$=\frac{6π}{6}- \frac{π}{6}$$ $$=\frac{5π}{6}$$ Let's change our complex number in the problem to polar form: $$\left(-\frac{5\sqrt{3}}{2}+ \frac{5}{2}i\right)^{2}$$ $$=\left[5 \left(\text{cos}\hspace{.1em}\frac{5π}{6}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{5π}{6}\right)\right]^{2}$$ Using De Moivre's Theorem: $$[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]^{n}$$ $$=r^{n}(\text{cos}\hspace{.1em}nθ + i \hspace{.1em}\text{sin}\hspace{.1em}nθ)$$ $$\left[5 \left(\text{cos}\hspace{.1em}\frac{5π}{6}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{5π}{6}\right)\right]^{2}$$ $$=5^{2}\left(\text{cos}\hspace{.1em}2 \cdot \frac{5π}{6}+ i \hspace{.1em}\text{sin}\hspace{.1em}2 \cdot \frac{5π}{6}\right)$$ $$=25 \left(\text{cos}\hspace{.1em}\frac{10π}{6}+ i \hspace{.1em}\text{sin}\hspace{.1em}\cdot \frac{10π}{6}\right)$$ $$=25 \left(\text{cos}\hspace{.1em}\frac{5π}{3}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{5π}{3}\right)$$ Example #2: Simplify, write the answer in rectangular form. $$\left[2\left(\text{cos}\hspace{.1em}\frac{7π}{4}+ i \hspace{.1em}\text{sin}\frac{7π}{4}\right)\right]^{5}$$ Since our complex number is already in polar form, we can immediately use De Moivre's Theorem. $$[r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)]^{n}$$ $$=r^{n}(\text{cos}\hspace{.1em}nθ + i \hspace{.1em}\text{sin}\hspace{.1em}nθ)$$ $$\left[2 \left(\text{cos}\hspace{.1em}\frac{7π}{4}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{7π}{4}\right)\right]^{5}$$ $$=2^{5}\left(\text{cos}\hspace{.1em}5 \cdot \frac{7π}{4}+ i \hspace{.1em}\text{sin}\hspace{.1em}5 \cdot \frac{7π}{4}\right)$$ $$=32\left(\text{cos}\hspace{.1em}\frac{35π}{4}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{35π}{4}\right)$$ Find the coterminal angle (not necessary but makes our calculations easier): $$\frac{35π}{4}- 8π$$ $$\frac{35π}{4}- \frac{32π}{4}=\frac{3π}{4}$$ $$32\left(\text{cos}\hspace{.1em}\frac{35π}{4}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{35π}{4}\right)$$ $$=32\left(\text{cos}\hspace{.1em}\frac{3π}{4}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{3π}{4}\right)$$ Let's now write this in rectangular form: $$\text{cos}\frac{3π}{4}=-\frac{\sqrt{2}}{2}$$ $$\text{sin}\frac{3π}{4}=\frac{\sqrt{2}}{2}$$ $$32\left(\text{cos}\hspace{.1em}\frac{3π}{4}+ i \hspace{.1em}\text{sin}\hspace{.1em}\frac{3π}{4}\right)$$ $$=32\left(-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}\right)$$ $$=-\frac{\sqrt{2}}{2}\cdot 32 + i\frac{\sqrt{2}}{2}\cdot 32$$ $$=-16\sqrt{2}+ 16i\sqrt{2}$$

### Fundamental Theorem of Algebra Revisited

Previously in our course, we talked about the fundamental theorem of algebra. The fundamental theorem of algebra tells us that a polynomial equation of degree n has n solutions in the complex number system. $$3x^3-4x^2+6x-20=0$$ Since this polynomial is of degree 3, we will have 3 complex solutions: $$3x^3-4x^2+6x-20=0$$ $$(x - 2)(3x^2 + 2x + 10)=0$$ Zero Product Property: $$x -2=0$$ $$x=2$$ $$3x^2 + 2x + 10=0$$ $$x=\frac{-2 \pm \sqrt{2^2 - 4(3)(10)}}{2(3)}$$ $$x=\frac{-2 \pm \sqrt{-116}}{6}$$ $$x=\frac{-2 \pm 2i\sqrt{29}}{6}$$ $$x=-\frac{1 \pm i\sqrt{29}}{3}$$ Our equations has exactly 3 complex solutions. $$x=2, -\frac{1 \pm i\sqrt{29}}{3}$$ Every nonzero complex number will have exactly n distinct complex nth roots. In other words, a given nonzero complex number will have exactly 2 square roots, exactly 3 cube roots, exactly 4 fourth roots, and so on and so forth... We can use De Moivre's theorem to find all nth roots of a complex number.

### Roots of Complex Numbers (nth Root)

For a positive integer n, the complex number a + b$i$ is an nth root of the complex number x + y$i$ if: $$x + yi=(a + bi)^{n}$$ To find a formula for an nth root of a complex number, let's work through a simple example. $$3(\text{cos}\hspace{.1em}45° + i \hspace{.1em}\text{sin}\hspace{.1em}45°)$$ Let's suppose we were asked to find the three complex cube roots.
Let's think about the following: $$3(\text{cos}\hspace{.1em}45° + i \hspace{.1em}\text{sin}\hspace{.1em}45°)=[r(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)]^{3}$$ Using De Moivre's Theorem, our right side becomes: $$3(\text{cos}\hspace{.1em}45° + i \hspace{.1em}\text{sin}\hspace{.1em}45°)=r^{3}(\text{cos}\hspace{.1em}3 \hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}3 \hspace{.1em}α)$$ To solve this equation, we will set: $$r^3=3$$ $$\text{cos}\hspace{.1em}45°=\text{cos}\hspace{.1em}3α$$ $$\text{sin}\hspace{.1em}45°=\text{cos}\hspace{.1em}3α$$ Our first equation can be solved by simply taking the cube root of each side: $$r^3=3$$ $$\sqrt[3]{r^3}=\sqrt[3]{3}$$ $$r=\sqrt[3]{3}$$ The last two equations require some consideration. For these equations to be satisfied 3$α$ must be an angle that is coterminal with 45°. $$\text{cos}\hspace{.1em}45°=\text{cos}\hspace{.1em}3α$$ $$\text{sin}\hspace{.1em}45°=\text{sin}\hspace{.1em}3α$$ Let k be any integer: $$3α=45° + 360° \cdot k$$ Divide both sides by 3 to isolate $α$: $$α=\frac{45° + 360° \cdot k}{3}$$ $$α=15° + 120° \cdot k$$ Let's plug in the integer values of 0, 1, 2, 3, 4, and 5 for k: $$α=15° + 120° \cdot 0=15°$$ $$α=15° + 120° \cdot 1=135°$$ $$α=15° + 120° \cdot 2=255°$$ $$α=15° + 120° \cdot 3=375°$$ Note: 375° - 360° = 15°
This means sin 375° = sin 15° and cos 375° = cos 15.
$$α=15° + 120° \cdot 4=495°$$ Note: 495° - 360° = 135°
This means sin 495° = sin 135° and cos 495° = cos 135.
$$α=15° + 120° \cdot 5=615°$$ Note: 615° - 360° = 255°
This means sin 615° = sin 255° and cos 615° = cos 255.
We can see that the last three solutions are repeats, continuing with larger values of k would continue this pattern and show only solutions that have already been found. This is as expected since, for every nonzero complex number, we will have exactly n distinct complex nth roots. So here our complex number has 3 distinct cube roots as expected.
When k = 0, the root is:$$\sqrt[3]{3}(\text{cos}\hspace{.1em}15° + i\hspace{.1em}\text{sin}\hspace{.1em}15°)$$ When k = 1, the root is:$$\sqrt[3]{3}(\text{cos}\hspace{.1em}135° + i\hspace{.1em}\text{sin}\hspace{.1em}135°)$$ When k = 2, the root is:$$\sqrt[3]{3}(\text{cos}\hspace{.1em}255° + i\hspace{.1em}\text{sin}\hspace{.1em}255°)$$ This leads us to the nth root theorem.

### nth Root Theorem

If n is any positive integer, r is a positive real number, and θ is in degrees, then the nonzero complex number: $$r(\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ will have exactly n distinct nth roots given by the following: $$\sqrt[n]{r}(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)$$ Where: $$α=\frac{θ}{n}+ \frac{360° \cdot k}{n}$$ $$k=0, 1, 2, 3,...,n - 1$$ If θ is given in radians, we can adjust our formula: $$α=\frac{θ}{n}+ \frac{2πk}{n}$$ $$k=0, 1, 2, 3,...,n - 1$$ Let's look at a few examples.
Example #3: Find all nth roots. Write the answer in polar form. $$-3\sqrt{3}- 3i$$ $$n=4$$ Since n is 4, this means we want to find the four 4th roots of our complex number.
Let's begin by putting our number in polar form: $$r=\sqrt{(-3\sqrt{3})^2 + (-3)^2}$$ $$r=\sqrt{27 + 9}$$ $$r=\sqrt{36}$$ $$r=6$$ $$\text{tan}\hspace{.1em}θ=\frac{-3}{-3\sqrt{3}}=\frac{\sqrt{3}}{3}$$ Our reference angle: $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ Since our complex number has a negative real part and a negative imaginary part, it will lie in quadrant III. $$θ=180° + 30°=210°$$ $$-3\sqrt{3}- 3i$$ $$=6(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ Let's use the nth root theorem to find our four 4th roots: $$\sqrt[4]{6}(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)$$ Where: $$α=\frac{210°}{4}+ \frac{360° \cdot k}{4}$$ $$α=52.5° + 90° \cdot k$$ $$k=0, 1, 2, 3$$ $$6(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ Our four 4th roots are: $$\sqrt[4]{6}\left(\text{cos}\hspace{.1em}52.5° + i \hspace{.1em}\text{sin}\hspace{.1em}52.5° \right)$$ $$\sqrt[4]{6}\left(\text{cos}\hspace{.1em}142.5° + i \hspace{.1em}\text{sin}\hspace{.1em}142.5° \right)$$ $$\sqrt[4]{6}\left(\text{cos}\hspace{.1em}232.5° + i \hspace{.1em}\text{sin}\hspace{.1em}232.5° \right)$$ $$\sqrt[4]{6}\left(\text{cos}\hspace{.1em}322.5° + i \hspace{.1em}\text{sin}\hspace{.1em}322.5° \right)$$ Example #4: Find all nth roots. Write the answer in polar form. Graph the roots on the Argand diagram. $$-8 + 8i\sqrt{3}, n=4$$ First, let's write our complex number in polar form. $$r=\sqrt{(-8)^2 + (8\sqrt{3})^2}$$ $$r=\sqrt{64 + 192}$$ $$r=\sqrt{256}$$ $$r=16$$ $$\text{tan}\hspace{.1em}θ=\frac{8\sqrt{3}}{-8}=-\sqrt{3}$$ Our reference angle: $$\text{tan}^{-1}(\sqrt{3})=60°$$ Since our complex number has a negative real part and a positive imaginary part, it will lie in quadrant II. $$θ=180° - 60°=120°$$ $$-8 + 8i\sqrt{3}$$ $$=16(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120°)$$ Let's use the nth root theorem to find our four 4th roots: $$\sqrt[4]{16}(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)$$ $$=2(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)$$ Where: $$α=\frac{120°}{4}+ \frac{360° \cdot k}{4}$$ $$α=30° + 90° \cdot k$$ $$k=0, 1, 2, 3$$ $$16(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120°)$$ Our four 4th roots are: $$2\left(\text{cos}\hspace{.1em}30° + i \hspace{.1em}\text{sin}\hspace{.1em}30° \right)$$ $$2\left(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120° \right)$$ $$2\left(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210° \right)$$ $$2\left(\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300° \right)$$ To graph our roots, each complex number can be represented as a vector in our complex plane with a magnitude of 2 and a direction angle given by the argument. This means our graphs of the roots lie on a circle with the center at the origin and a radius of 2. Notice that the roots are equally spaced 90° apart: Note: We may also label the y-axis as the imaginary axis and the x-axis as the real axis when working with the Argand diagram. If we think about a complex number as x + y$i$, we may use the traditional x and y.
In some cases, we may be asked to find the complex roots to an equation using the nth root theorem. Let's look at an example.
Example #5: Find all solutions of the given equation and represent the solutions graphically. $$x^3 - 27=0$$ To begin, let's write our equation as: $$x^3=27$$ Since we have a 3rd-degree polynomial, we know there will be exactly 3 complex solutions. To find them, let's write 27 in polar form: $$27=27 + 0i$$ $$r=\sqrt{(27)^2 + 0^{2}}$$ $$r=27$$ $$\text{tan}\hspace{.1em}θ=\frac{0}{27}=0$$ $$\text{tan}^{-1}(0)=0°$$ $$27 + 0i=27(\text{cos}\hspace{.1em}0° + i \hspace{.1em}\text{sin}\hspace{.1em}0°)$$ Let's use the nth root theorem to find our three cube roots: $$\sqrt[3]{27}(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)$$ $$=3(\text{cos}\hspace{.1em}α + i \hspace{.1em}\text{sin}\hspace{.1em}α)$$ Where: $$α=\frac{0°}{3}+ \frac{360° \cdot k}{3}$$ $$α=0° + 120° \cdot k$$ $$α=120° \cdot k$$ $$k=0, 1, 2$$ $$27(\text{cos}\hspace{.1em}0° + i \hspace{.1em}\text{sin}\hspace{.1em}0°)$$ Our three cube roots are: $$3\left(\text{cos}\hspace{.1em}0° + i \hspace{.1em}\text{sin}\hspace{.1em}0° \right)$$ $$3\left(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120° \right)$$ $$3\left(\text{cos}\hspace{.1em}240° + i \hspace{.1em}\text{sin}\hspace{.1em}240° \right)$$ Graphically, we can show our solutions on the Argand diagram, where the tips of the arrows representing the three roots lie on a circle with a radius of 3 and are equally spaced around it every 120°. In rectangular form: $$3\left(\text{cos}\hspace{.1em}0° + i \hspace{.1em}\text{sin}\hspace{.1em}0° \right)$$ $$=3$$ $$3\left(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120° \right)$$ $$=-\frac{3}{2}+ \frac{3\sqrt{3}}{2}i$$ $$3\left(\text{cos}\hspace{.1em}240° + i \hspace{.1em}\text{sin}\hspace{.1em}240° \right)$$ $$=-\frac{3}{2}- \frac{3\sqrt{3}}{2}i$$ These are the solutions for our equation: $$x^3 - 27=0$$ $$x=3, -\frac{3}{2}\pm \frac{3\sqrt{3}}{2}i$$

#### Skills Check:

Example #1

Simplify, write the answer in rectangular form. $$[\sqrt{2}(\text{cos}\hspace{.1em}45° + i \hspace{.1em}\text{sin}\hspace{.1em}45°)]^{5}$$

A
$$-\frac{243}{2}- \frac{243\sqrt{3}}{2}i$$
B
$$-4 - 4i$$
C
$$4 + 4i$$
D
$$-\sqrt{2}+ 5i$$
E
$$-\frac{243}{2}+ \frac{243\sqrt{3}}{2}i$$

Example #2

Simplify, write the answer in rectangular form. $$\left[6\left(\text{cos}\hspace{.1em}\frac{2π}{3}+ i \hspace{.1em}\text{sin}\frac{2π}{3}\right)\right]^{3}$$

A
$$-64i$$
B
$$216$$
C
$$-4\sqrt{2}- 4i\sqrt{2}$$
D
$$128i\sqrt{2}$$
E
$$2 - 3i$$

Example #3

Determine which answer is a fifth root for the given complex number. $$\frac{3\sqrt{2}}{2}+ \frac{3\sqrt{2}}{2}i$$

A
$$\sqrt[5]{3}\hspace{.1em}\text{cis}\hspace{.1em}9°$$
B
$$\sqrt[5]{3}\hspace{.1em}\text{cis}\hspace{.1em}24°$$
C
$$\sqrt[5]{3}\hspace{.1em}\text{cis}\hspace{.1em}157°$$
D
$$\sqrt[5]{3}\hspace{.1em}\text{cis}\hspace{.1em}229°$$
E
$$\sqrt[5]{3}\hspace{.1em}\text{cis}\hspace{.1em}300°$$