6.5: De Moivre's and the nth Root Theorem

Learning Objectives

• Find powers of complex numbers in polar form.
• Find roots of complex numbers in polar form.

Finding Powers of Complex Numbers in Polar Form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer \(n\), \(z^n\) is found by raising the modulus to the \(n^{th}\) power and multiplying the argument by \(n\). It is the standard method used in modern mathematics.

DE MOIVRE’S THEOREM

If \(z=r(\cos \theta+i \sin \theta)\) is a complex number, then

\[\begin{align} z^n &= r^n[\cos(n\theta)+i \sin(n\theta) ] \\ z^n &= r^n\space cis(n\theta) \end{align}\]

where \(n\) is a positive integer.

Example \(\PageIndex{1}\): Evaluating an Expression Using De Moivre’s Theorem

Evaluate the expression \({(1+i)}^5\) using De Moivre’s Theorem.

Solution

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write \((1+i)\) in polar form. Let us find \(r\).

\[\begin{align*} r &= \sqrt{x^2+y^2} \\ r &= \sqrt{{(1)}^2+{(1)}^2} \\ r &= \sqrt{2} \end{align*}\]

Then we find \(\theta\). Using the formula \(\tan \theta=\dfrac{y}{x}\) gives

\[\begin{align*} \tan \theta &= \dfrac{1}{1} \\ \tan \theta &= 1 \\ \theta &= \dfrac{\pi}{4} \end{align*}\]

Use De Moivre’s Theorem to evaluate the expression.

\[\begin{align*} {(a+bi)}^n &= r^n[\cos(n\theta)+i \sin(n\theta)] \\ {(1+i)}^5 &= {(\sqrt{2})}^5\left[ \cos\left(5⋅\dfrac{\pi}{4}\right)+i \sin\left(5⋅\dfrac{\pi}{4}\right) \right] \\ {(1+i)}^5 &= 4\sqrt{2}\left[ \cos\left(\dfrac{5\pi}{4}\right)+i \sin\left(\dfrac{5\pi}{4}\right) \right] \\ {(1+i)}^5 &= 4\sqrt{2}\left[ −\dfrac{\sqrt{2}}{2}+i\left(−\dfrac{\sqrt{2}}{2}\right) \right] \\ {(1+i)}^5 &= −4−4i \end{align*}\]

Finding Roots of Complex Numbers in Polar Form

To find the \(n^{th}\) root of a complex number in polar form, we use the \(n^{th}\) Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding \(n^{th}\) roots of complex numbers in polar form.

THE \(N^{TH}\) ROOT THEOREM

To find the \(n^{th}\) root of a complex number in polar form, use the formula given as

\[z^{\tfrac{1}{n}}=r^{\tfrac{1}{n}}\left[ \cos\left(\dfrac{\theta}{n}+\dfrac{2k\pi}{n}\right)+i \sin\left(\dfrac{\theta}{n}+\dfrac{2k\pi}{n}\right) \right]\]

where \(k=0, 1, 2, 3, . . . , n−1\). We add \(\dfrac{2k\pi}{n}\) to \(\dfrac{\theta}{n}\) in order to obtain the periodic roots.

Example \(\PageIndex{2}\): the Root of a Complex Number

Evaluate the cube roots of \(z=8\left(\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right)\).

Solution

We have

\[\begin{align*} z^{\frac{1}{3}} &= 8^{\frac{1}{3}}\left[ \cos\left(\frac{\frac{2\pi}{3}}{3}+\frac{2k\pi}{3}\right)+i \sin\left(\frac{\frac{2\pi}{3}}{3}+\frac{2k\pi}{3}\right) \right] \\ z^{\frac{1}{3}} &= 2\left[ \cos\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right)+i \sin\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right) \right] \end{align*}\]

There will be three roots: \(k=0, 1, 2\). When \(k=0\), we have

\(z^{\frac{1}{3}}=2\left(\cos\left(\dfrac{2\pi}{9}\right)+i \sin\left(\dfrac{2\pi}{9}\right)\right)\)

When \(k=1\), we have

\[\begin{align*} z^{\frac{1}{3}} &=2\left[ \cos\left(\dfrac{2\pi}{9}+\dfrac{6\pi}{9}\right)+i \sin\left(\dfrac{2\pi}{9}+\dfrac{6\pi}{9}\right) \right] \;\;\;\;\;\;\;\;\; \text{Add }\dfrac{2(1)\pi}{3} \text{ to each angle.} \\ z^{\frac{1}{3}} &= 2\left(\cos\left(\dfrac{8\pi}{9}\right)+i \sin\left(\dfrac{8\pi}{9}\right)\right) \end{align*}\]

When \(k=2\), we have

\[\begin{align*} z^{\frac{1}{3}} &= 2\left[ \cos\left(\dfrac{2\pi}{9}+\dfrac{12\pi}{9}\right)+i \sin\left(\dfrac{2\pi}{9}+\dfrac{12\pi}{9}\right) \right] \;\;\;\;\;\;\; \text{Add }\dfrac{2(2)\pi}{3} \text{ to each angle.} \\ z^{\frac{1}{3}} &= 2\left(\cos\left(\dfrac{14\pi}{9}\right)+i \sin\left(\dfrac{14\pi}{9}\right)\right) \end{align*}\]

Remember to find the common denominator to simplify fractions in situations like this one. For \(k=1\), the angle simplification is

\[\begin{align*} \dfrac{\dfrac{2\pi}{3}}{3}+\dfrac{2(1)\pi}{3} &= \dfrac{2\pi}{3}(\dfrac{1}{3})+\dfrac{2(1)\pi}{3}\left(\dfrac{3}{3}\right) \\ &=\dfrac{2\pi}{9}+\dfrac{6\pi}{9} \\ &=\dfrac{8\pi}{9} \end{align*}\]

Exercise \(\PageIndex{1}\)

Find the four fourth roots of \(16(\cos(120°)+i \sin(120°))\).

Answer

\(z_0=2(\cos(30°)+i \sin(30°))\)

\(z_1=2(\cos(120°)+i \sin(120°))\)

\(z_2=2(\cos(210°)+i \sin(210°))\)

\(z_3=2(\cos(300°)+i \sin(300°))\)

Extra Practice

For the following exercises, find the powers of each complex number in polar form.

1. Find \(z^{4}\) when \(z=2\) cis \(\left(70^{\circ}\right)\)
2. Find \(z^{2}\) when \(z=5\) cis \(\left(\frac{3 \pi}{4}\right)\)

For the following exercises, evaluate each root.

1. Evaluate the cube root of \(z\) when \(z=64\) cis \(\left(210^{\circ}\right)\).
2. Evaluate the square root of \(z\) when \(z=25\) cis \(\left(\frac{3 \pi}{2}\right)\).

Key Concepts

• To find the power of a complex number \(z^n\), raise \(r\) to the power \(n\), and multiply \(\theta\) by \(n\). See Example \(\PageIndex{1}\).
• Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See Example \(\PageIndex{2}\).

Contributors and Attributions

• Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.