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?

Commented bymalwan last updated on 21/Sep/20

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

Commented bymr W last updated on 21/Sep/20

you can even say  π=ln ((−1))^(1/i)

Commented byRio Michael last updated on 21/Sep/20

really sir?

Commented byMJS_new last updated on 21/Sep/20

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

Commented byRio Michael last updated on 21/Sep/20

thanks prof

Commented byDwaipayan 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))

Commented byMJS_new last updated on 21/Sep/20

yes