Veva Garcia edited untitled.tex  almost 8 years ago

Commit id: b86cc7fabcae8520fd37b4e0b00d99df0427b1d2

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We need to find values of $z$ where it is multiplied by itself $n$ times and is equal to $re^{i\theta}.$ \\So, let $z=se^{i\theta}.$\\ Thus, we have $s^n=r.$\\ Hence, $s=\sqrt[n]{r}$\\  To solve for $\theta$, we have $n\theta= \theta +2\pi k$, for some $k\in\mathbb{Z}$ \\  Then $\theta=\frac{\theta + 2 \pi k}{n}$\\  Therefore we have $z=\sqrt{[n]{r}e^i}$ $z=\sqrt[n]{r}e^i$  \end{proof}  \end{problem}