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Question Number 29049 by NECx last updated on 03/Feb/18

Prove that       e^(iπ) +1=0

$${Prove}\:{that} \\ $$$$\:\:\:\:\:{e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$

Answered by Penguin last updated on 03/Feb/18

e^(iθ) =cos(θ)+isin(θ)  θ=π  ∴e^(iπ) =−1+i(0)  =−1

$${e}^{{i}\theta} =\mathrm{cos}\left(\theta\right)+{i}\mathrm{sin}\left(\theta\right) \\ $$$$\theta=\pi \\ $$$$\therefore{e}^{{i}\pi} =−\mathrm{1}+{i}\left(\mathrm{0}\right) \\ $$$$=−\mathrm{1} \\ $$

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