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Question Number 114857 by Rio Michael last updated on 21/Sep/20

we know that         e^(πi)  = −1   ⇒ ln (e^(πi) ) = ln(−1)    πi = ln (−1).   How good is this prove?

$$\mathrm{we}\:\mathrm{know}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:{e}^{\pi{i}} \:=\:−\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{ln}\:\left({e}^{\pi{i}} \right)\:=\:\mathrm{ln}\left(−\mathrm{1}\right) \\ $$$$\:\:\pi{i}\:=\:\mathrm{ln}\:\left(−\mathrm{1}\right).\: \\ $$$$\mathrm{How}\:\mathrm{good}\:\mathrm{is}\:\mathrm{this}\:\mathrm{prove}? \\ $$

Commented by malwan last updated on 21/Sep/20

is that mean π=ln(^(√(−1)) (√(−1)) ) ?

$${is}\:{that}\:{mean}\:\pi={ln}\left(\:^{\sqrt{−\mathrm{1}}} \sqrt{−\mathrm{1}}\:\right)\:? \\ $$

Commented by mr W last updated on 21/Sep/20

you can even say  π=ln ((−1))^(1/i)

$${you}\:{can}\:{even}\:{say} \\ $$$$\pi=\mathrm{ln}\:\sqrt[{{i}}]{−\mathrm{1}} \\ $$

Commented by Rio Michael last updated on 21/Sep/20

really sir?

$$\mathrm{really}\:\mathrm{sir}? \\ $$

Commented by MJS_new last updated on 21/Sep/20

(−1)^(1/i) =(−1)^(−i)   −1=e^(iπ)  ⇒ (−1)^(−i) =e^π

$$\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{i}}} =\left(−\mathrm{1}\right)^{−\mathrm{i}} \\ $$$$−\mathrm{1}=\mathrm{e}^{\mathrm{i}\pi} \:\Rightarrow\:\left(−\mathrm{1}\right)^{−\mathrm{i}} =\mathrm{e}^{\pi} \\ $$

Commented by Rio Michael last updated on 21/Sep/20

thanks prof

$$\mathrm{thanks}\:\mathrm{prof} \\ $$

Commented by Dwaipayan Shikari last updated on 21/Sep/20

log(−1)=log(e^(πi) )  log(e^(2kπi+πi) )=log(−1)⇒log(−1)=πi(2k+1)  i^(1/i) =e^(π/2)   i^i =e^(−(π/2))

$${log}\left(−\mathrm{1}\right)={log}\left({e}^{\pi{i}} \right) \\ $$$${log}\left({e}^{\mathrm{2}{k}\pi{i}+\pi{i}} \right)={log}\left(−\mathrm{1}\right)\Rightarrow{log}\left(−\mathrm{1}\right)=\pi{i}\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$${i}^{\frac{\mathrm{1}}{{i}}} ={e}^{\frac{\pi}{\mathrm{2}}} \\ $$$${i}^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \\ $$

Commented by MJS_new last updated on 21/Sep/20

yes

$$\mathrm{yes} \\ $$

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