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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?

weknowthateπi=1ln(eπi)=ln(1)πi=ln(1).Howgoodisthisprove?

Commented by malwan last updated on 21/Sep/20

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

isthatmeanπ=ln(11)?

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

you can even say  π=ln ((−1))^(1/i)

youcanevensayπ=ln1i

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

really sir?

reallysir?

Commented by MJS_new last updated on 21/Sep/20

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

(1)1i=(1)i1=eiπ(1)i=eπ

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

thanks prof

thanksprof

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(1)=log(eπi)log(e2kπi+πi)=log(1)log(1)=πi(2k+1)i1i=eπ2ii=eπ2

Commented by MJS_new last updated on 21/Sep/20

yes

yes

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