De Moivre's Theorem and n-th Roots of Unity

De Moivre's Theorem

De Moivre's Theorem provides a formula for computing powers of complex numbers in polar form.

For an integral index n, the theorem states:

$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$

In Euler's form, this is simply $(e^{i\theta})^n = e^{in\theta}$. The theorem also holds for rational indices, which is the basis for finding roots.

n-th Roots of a Complex Number

To find the n-th roots of a complex number $z = r(\cos\theta + i\sin\theta)$, we solve the equation $w^n = z$. The n distinct roots are given by:

$$w_k = r^{1/n} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right]$$

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

Geometrically, the n-th roots of a complex number lie on the vertices of a regular n-sided polygon inscribed in a circle of radius $r^{1/n}$.

n-th Roots of Unity

The n-th roots of unity are the solutions to the equation $z^n=1$. Since $1 = \cos(0) + i\sin(0)$, the n roots are:

$$z_k = \cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right) = e^{i(2k\pi/n)}$$

for $k=0, 1, 2, ..., n-1$. These roots are often written as $1, \omega, \omega^2, ..., \omega^{n-1}$, where $\omega = e^{i(2\pi/n)}$.

Properties of n-th Roots of Unity:

• They form a Geometric Progression.
• Sum of the roots is 0: $1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$.
• Product of the roots is $(-1)^{n-1}$.

Cube Roots of Unity ($z^3=1$):

This is a special and very important case. The three cube roots of unity are:

• $1$
• $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i(2\pi/3)}$
• $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = e^{i(4\pi/3)}$

Properties of Cube Roots of Unity:

• $1 + \omega + \omega^2 = 0$
• $\omega^3 = 1$

Example 1: Using De Moivre's Theorem

Question: Simplify $(1+i)^8$.

Solution: First, convert $1+i$ to polar form. $r = |1+i| = \sqrt{1^2+1^2} = \sqrt{2}$. $\theta = \arg(1+i) = \pi/4$. So, $1+i = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$. Now, apply De Moivre's Theorem:

$$(1+i)^8 = [\sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))]^8$$

$$= (\sqrt{2})^8 (\cos(8 \cdot \pi/4) + i\sin(8 \cdot \pi/4))$$

$$= 16 (\cos(2\pi) + i\sin(2\pi)) = 16(1 + 0i) = 16$$

Example 2: Finding n-th Roots

Question: Find the cube roots of $8i$.

Solution: First, write $8i$ in polar form. $z = 0+8i$. $r = |8i| = 8$. The argument $\theta = \pi/2$. So, $8i = 8(\cos(\pi/2) + i\sin(\pi/2))$. The cube roots are given by:

$$w_k = 8^{1/3} \left[ \cos\left(\frac{\pi/2 + 2k\pi}{3}\right) + i\sin\left(\frac{\pi/2 + 2k\pi}{3}\right) \right]$$

for $k=0, 1, 2$.

• For k=0: $w_0 = 2(\cos(\pi/6) + i\sin(\pi/6)) = 2(\frac{\sqrt{3}}{2} + i\frac{1}{2}) = \sqrt{3}+i$.
• For k=1: $w_1 = 2(\cos(5\pi/6) + i\sin(5\pi/6)) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3}+i$.
• For k=2: $w_2 = 2(\cos(9\pi/6) + i\sin(9\pi/6)) = 2(\cos(3\pi/2) + i\sin(3\pi/2)) = 2(0 - i) = -2i$.

Example 3: Using Properties of Cube Roots of Unity

Question: If $\omega$ is a complex cube root of unity, find the value of $(1 - \omega + \omega^2)^5$.

Solution: We know the property $1 + \omega + \omega^2 = 0$. From this, we can write $1 + \omega^2 = -\omega$. Substitute this into the expression:

$$(1 - \omega + \omega^2)^5 = ((1+\omega^2) - \omega)^5 = (-\omega - \omega)^5 = (-2\omega)^5$$

$$= (-2)^5 (\omega)^5 = -32 \omega^5$$

Since $\omega^3 = 1$, we can simplify $\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$. So the final value is $-32\omega^2$.