# Useful identities from Euler’s formula

Euler’s formula:

$e^{i\theta} =\cos\theta + i\sin\theta$

From the Euler’s formula, we can derive that

$e^{ik\pi} = \begin{cases} 1 & \text{if } k \text{ is even}\\ -1 & \text{if } k \text{ is odd} \end{cases}$

If we write the above equation in reverse, we get

$-1=e^{i(2k+1)\pi}$

$1=e^{i2k\pi}$

These identities are useful for quick manipulations of complex numbers. Below are two examples:

Example 1: Solve $z^n = 1$.

Solution

$z = 1^{1/n} = e^{i\frac{2k\pi}{n}} \qquad k = 0,1,...,n-1$

Example 2: Solve $z^3=-8$.

Solution

$z = (-8)^{1/3} = 2(-1)^{1/3} = 2e^{i\frac{(2k+1)\pi}{3}} \qquad k =0, 1, 2$

• $k=0 \Rightarrow z=1+\sqrt{3}i$
• $k=1 \Rightarrow z=-2$
• $k=2 \Rightarrow z=1-\sqrt{3}i$

Note that we can also write $k=-1, 0, 1$. On the trigonometric circle, $k = 2$ and $k = -1$ yield the same angle.