De Moivre’s theorem

Let $z = r(\cos\theta + i\sin\theta)$. We want to show that $z^n = r^n(\cos(n\theta) + i\sin(n\theta))$.

Base Case ($n = 1$):

$z^1 = r^1(\cos(1\theta) + i\sin(1\theta)) = r(\cos\theta + i\sin\theta)$

which is true.

Inductive Hypothesis:

Assume the statement is true for $n = k$:

$z^k = r^k(\cos(k\theta) + i\sin(k\theta)) \quad (6.2)$

Inductive Step:

We need to prove that the result is also true for $n = k + 1$:

$z^{k+1} = z^k \cdot z = r^k(\cos(k\theta) + i\sin(k\theta)) \cdot r(\cos\theta + i\sin\theta)$

$= r^{k+1}[(\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta) + i(\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta)]$

Using the trigonometric sum formulas:

$z^{k+1} = r^{k+1}[\cos(k\theta + \theta) + i\sin(k\theta + \theta)]$

$z^{k+1} = r^{k+1}[\cos((k+1)\theta) + i\sin((k+1)\theta)]$

Thus, if the statement is true for $n = k$, it is also true for $n = k + 1$. Since it is true for $n = 1$, by mathematical induction, it is true for all positive integers $n$.

If $n$ is a negative integer, let $n = -m$, where $m$ is a positive integer:

$z^n = z^{-m} = \frac{1}{z^m} = \frac{1}{r^m(\cos(m\theta) + i\sin(m\theta))} = \frac{1}{r^m}\frac{(\cos(m\theta) - i\sin(m\theta))}{(\cos(m\theta) + i\sin(m\theta))(\cos(m\theta) - i\sin(m\theta))}$

$= r^{-m}(\cos(-m\theta) + i\sin(-m\theta))$

Consider a positive integer $n \ge 2$ and a complex number $z_0 \ne 0$. The equation $z^n - z_0 = 0$ (6.4) defines the nth roots of $z_0$. Any solution $z$ of equation (6.4) is called an nth root of $z_0$.

Let $z_0 = r(\cos\theta + i\sin\theta)$ be a complex number with $r > 0$ and $0 \le \theta < 2\pi$. The number $z_0$ has $n$ distinct nth roots given by:

$z_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right)\right] \quad (6.5)$

for $k = 0, 1, 2, ..., n - 1$.

Using the polar representation of the complex number $z$ with the extended argument:

$z^n = \rho^n(\cos(n\alpha) + i\sin(n\alpha))$

Thus, equation (6.4) becomes:

$\rho^n(\cos(n\alpha) + i\sin(n\alpha)) = r(\cos\theta + i\sin\theta) \quad (6.6)$

This implies that $\rho^n = r$ and $n\alpha = \theta + 2\pi k$, where $k \in \mathbb{Z}$.

$\rho = \sqrt[n]{r} \quad \text{and} \quad \alpha = \frac{\theta + 2\pi k}{n}$

Therefore, the roots are:

$z_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right)\right], \quad \text{where } k \in \mathbb{Z}.$

Example 1

Find the complex cube roots of $8(\cos 60^\circ + i\sin 60^\circ)$ and express your answers in polar form.

Solution:

Let $z = 8(\cos 60^\circ + i\sin 60^\circ)$. Thus, $r = 8$ and $\theta = 60^\circ$.

The $n$th roots of a complex number $z$ are given by:

$z_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta + 360^\circ k}{n}\right) + i\sin\left(\frac{\theta + 360^\circ k}{n}\right)\right], \quad \text{where } k \in \mathbb{Z}.$

For cube roots, $n = 3$ and $k = 0, 1, 2$. Thus:

$z_k = \sqrt[3]{8}\left[\cos\left(\frac{60^\circ + 360^\circ k}{3}\right) + i\sin\left(\frac{60^\circ + 360^\circ k}{3}\right)\right]$

When $k = 0$:

$z_0 = 2(\cos 20^\circ + i\sin 20^\circ)$

When $k = 1$:

$z_1 = 2(\cos 140^\circ + i\sin 140^\circ)$

When $k = 2$:

$z_2 = 2(\cos 260^\circ + i\sin 260^\circ)$

Therefore, the complex cube roots are $2(\cos 20^\circ + i\sin 20^\circ)$, $2(\cos 140^\circ + i\sin 140^\circ)$, and $2(\cos 260^\circ + i\sin 260^\circ)$.

Example 2

Find the roots of the equation $z^5 = 1 + i$. Express your answers in polar form.

Solution:

Let $w = 1 + i$. Then $z^5 = w$.

Converting $w$ to polar form:

$r = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$

Thus, $w = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})$.

The $n$th roots are given by:

$z_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right)\right]$

For the fifth roots, $n = 5$ and $k = 0, 1, 2, 3, 4$. Thus:

$z_k = \sqrt[10]{2}\left[\cos\left(\frac{\frac{\pi}{4} + 2\pi k}{5}\right) + i\sin\left(\frac{\frac{\pi}{4} + 2\pi k}{5}\right)\right]$

Simplifying the argument:

$z_k = \sqrt[10]{2}\left[\cos\left(\frac{\pi + 8\pi k}{20}\right) + i\sin\left(\frac{\pi + 8\pi k}{20}\right)\right]$

Substituting $k = 0, 1, 2, 3, 4$ gives the five roots:

$z_0 = \sqrt[10]{2}\left(\cos\frac{\pi}{20} + i\sin\frac{\pi}{20}\right)$

$z_1 = \sqrt[10]{2}\left(\cos\frac{9\pi}{20} + i\sin\frac{9\pi}{20}\right)$

$z_2 = \sqrt[10]{2}\left(\cos\frac{17\pi}{20} + i\sin\frac{17\pi}{20}\right)$

$z_3 = \sqrt[10]{2}\left(\cos\frac{25\pi}{20} + i\sin\frac{25\pi}{20}\right) = \sqrt[10]{2}\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right)$

$z_4 = \sqrt[10]{2}\left(\cos\frac{33\pi}{20} + i\sin\frac{33\pi}{20}\right)$

Example 3

Find the cube roots of the complex number $z = -8i$. Express your answers in the form $a + bi$, where $a$ and $b$ are real numbers.

Solution:

Let $w = -8i$. Converting $w$ to polar form:

$r = |-8i| = 8$

$\theta = \frac{3\pi}{2} \quad \text{or} \quad 270^\circ.$

Thus, $w = 8(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2})$.

For the cube roots, $n = 3$ and $k = 0, 1, 2$. Thus:

$z_k = \sqrt[3]{8}\left[\cos\left(\frac{\frac{3\pi}{2} + 2\pi k}{3}\right) + i\sin\left(\frac{\frac{3\pi}{2} + 2\pi k}{3}\right)\right]$

$z_k = 2\left[\cos\left(\frac{3\pi + 4\pi k}{6}\right) + i\sin\left(\frac{3\pi + 4\pi k}{6}\right)\right]$

When $k = 0$: $z_0 = 2(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) = 2(0 + i) = 2i$

When $k = 1$: $z_1 = 2(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}) = 2(-\frac{\sqrt{3}}{2} - \frac{1}{2}i) = -\sqrt{3} - i$

When $k = 2$: $z_2 = 2(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}) = 2(\frac{\sqrt{3}}{2} - \frac{1}{2}i) = \sqrt{3} - i$

Therefore, the roots are $2i$, $-\sqrt{3} - i$, and $\sqrt{3} - i$.

De Moivre's theorem is used in finding solutions of equations, simplifying, expressing, and proving trigonometric identities.

Example 1

Solve the equation $z^4 - z^3 + z - 1 = 0$.

Solution:

$z^4 - z^3 + z - 1 = 0$

$z^3(z - 1) + 1(z - 1) = 0$

$(z^3 + 1)(z - 1) = 0$

Either $z - 1 = 0$ or $z^3 + 1 = 0$.

Thus, $z = 1$ or $z^3 = -1$.

Let's find the cube roots of $-1$.

$-1 = \cos(\pi + 2\pi k) + i\sin(\pi + 2\pi k) \quad \text{where } k \text{ is an integer}$

$z = (-1)^{\frac{1}{3}} = \cos\left(\frac{\pi + 2\pi k}{3}\right) + i\sin\left(\frac{\pi + 2\pi k}{3}\right)$

When $k = 0$, $z = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$

When $k = 1$, $z = \cos\pi + i\sin\pi = -1$

When $k = 2$, $z = \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

Therefore, the roots are $1$, $-1$, $\frac{1}{2} + i\frac{\sqrt{3}}{2}$, and $\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

Example 2

Use De Moivre's theorem to find the value of $(1 + i)^8$ in the form $a + ib$, where $a$ and $b$ are real numbers.

Solution:

Let $z = 1 + i$.

$r = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$

Thus, $z = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})$.

$z^8 = (\sqrt{2})^8\left(\cos\frac{8\pi}{4} + i\sin\frac{8\pi}{4}\right)$

$= 16(\cos 2\pi + i\sin 2\pi)$

$= 16(1 + 0i)$

$= 16$

Therefore, $(1 + i)^8 = 16$.

Example 3

If $(1 + i\sqrt{3})^n = x + iy$, where $x$ and $y$ are real numbers and $n$ is a positive integer, show that $x^2 + y^2 = 4^n$.

Solution:

Let $1 + i\sqrt{3} = r(\cos\theta + i\sin\theta)$.

$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$

$\theta = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$

Thus, $1 + i\sqrt{3} = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$.

$(1 + i\sqrt{3})^n = 2^n\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right)$

Since $(1 + i\sqrt{3})^n = x + iy$, then

$x = 2^n\cos\frac{n\pi}{3} \quad \text{and} \quad y = 2^n\sin\frac{n\pi}{3}$

$x^2 + y^2 = \left(2^n\cos\frac{n\pi}{3}\right)^2 + \left(2^n\sin\frac{n\pi}{3}\right)^2$

$= 4^n\left(\cos^2\frac{n\pi}{3} + \sin^2\frac{n\pi}{3}\right)$

$= 4^n(1)$

$x^2 + y^2 = 4^n$

Example 4

Use De Moivre's theorem to simplify the expression $\frac{(\cos 3\theta + i\sin 3\theta)^8(\cos 4\theta + i\sin 4\theta)^5}{(\cos 2\theta + i\sin 2\theta)^4(\cos \theta + i\sin \theta)^3}$.

Solution:

$\frac{(\cos 3\theta + i\sin 3\theta)^8(\cos 4\theta + i\sin 4\theta)^5}{(\cos 2\theta + i\sin 2\theta)^4(\cos \theta + i\sin \theta)^3}$

$= \frac{(\cos 24\theta + i\sin 24\theta)(\cos 20\theta + i\sin 20\theta)}{(\cos 8\theta + i\sin 8\theta)(\cos 3\theta + i\sin 3\theta)}$

$= \frac{\cos 44\theta + i\sin 44\theta}{\cos 11\theta + i\sin 11\theta}$

$= \cos(44\theta - 11\theta) + i\sin(44\theta - 11\theta)$

$= \cos 33\theta + i\sin 33\theta$