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Question Number 8838 by tawakalitu last updated on 31/Oct/16

Show that :  e^(iπ + 1)  = 0

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

Answered by FilupSmith last updated on 31/Oct/16

do you mean e^(iπ) +1=0  e^(ix) =cos(x)+isin(x)  x=π  e^(iπ) =(−1)+i(0)  e^(iπ) =−1  ∴e^(iπ) +1=(−1)+1=0

$$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:{e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$$${e}^{{ix}} =\mathrm{cos}\left({x}\right)+{i}\mathrm{sin}\left({x}\right) \\ $$$${x}=\pi \\ $$$${e}^{{i}\pi} =\left(−\mathrm{1}\right)+{i}\left(\mathrm{0}\right) \\ $$$${e}^{{i}\pi} =−\mathrm{1} \\ $$$$\therefore{e}^{{i}\pi} +\mathrm{1}=\left(−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$

Commented by tawakalitu last updated on 31/Oct/16

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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