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.

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