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Question Number 70138 by Scientist0000001 last updated on 01/Oct/19

$$provethat{e}^{i\theta }=e^{i(\theta +2k\mathrm{\Pi })}giventhatk=0,\pm 1,\pm 2...$$

Answered by mind is power last updated on 01/Oct/19

$$e^{ix}=cos\left(x\right)+isin\left(x\right)$$$$\Rightarrow e^{i(\theta +2k\pi )}=cos(\theta +2k\pi )+isin(2k\pi +\theta )$$$$=cos\left(\theta \right)+isin\left(\theta \right)weusedperiodicityofSinandvoswitchis2\pi$$$$cos\left(\theta \right)+isin\left(\theta \right)=e^{i\theta }$$

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