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Then,  \begin{align*}  z^{n}&=re^{i\theta}\\  &=re^{i\theta}e^{(2\pi m)i}\text{For some $m\in\mathbb{N}$} $m\in\mathbb{N}$}\\  &=re^{i(\theta+2\pi m)}\\  \end{align*}  Thus,