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Question Number 19301 by saa last updated on 09/Aug/17
e^(iπ) +1=0
$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$
Answered by ajfour last updated on 09/Aug/17
e^(iπ) =−1 .
$$\mathrm{e}^{\mathrm{i}\pi} =−\mathrm{1}\:. \\ $$

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