## De Moivre's Formula

Let $$z$$ be a complex number given in polar form: $$r \operatorname{cis} \theta$$. Let $$n$$ be an integer. De Moivre's Formula states that $z^n = r^n \operatorname{cis} (n\theta).$ This formula simplifies computing powers of a complex number provided one has its polar form.

### Example

Let $$z = 1+i$$. What is $$z^{10}$$?

There is nothing stopping us from computing $$z^{10}$$ directly. However, we will use De Moivre's formula. First, we need to convert $$z$$ to polar form. Note that $$r = |~z~| = \sqrt{2}$$ and $$\arg (z) = \arctan(1) = \frac{\pi}{4}$$. Hence, $$z = \sqrt{2} \operatorname{cis} \frac{\pi}{4}$$, giving $\begin{eqnarray} z^{10} & = & \sqrt{2}^{10}\operatorname{cis} \left(10 \cdot \frac{\pi}{4}\right) \\ & = & 32 \operatorname{cis} \frac{5\pi}{2} \\ & = & 32 \operatorname{cis} \left(2\pi + \frac{\pi}{2} \right) \\ & = & 32 \operatorname{cis} \frac{\pi}{2} \\ & = & 32 i \end{eqnarray}$

In the above, we used the fact that $$\cos (2k\pi + \theta) = \cos \theta$$ and $$\sin (2k\pi + \theta) = \sin \theta$$ for all integers $$k$$.

In general, if the argument is outside the interval $$[0,2\pi)$$, it is customary to reduce it to a value within this interval as we have done above.

## Exercises

1. Compute $$(\sqrt{3}-i)^{12}$$.

2. Compute $$(-1+i)^{8}$$.