Question Number 103253 by Study last updated on 13/Jul/20

(i)^(1/i) =?????

Answered by Dwaipayan Shikari last updated on 13/Jul/20

i^(1/i) =i^(−i) ((1/i)=(i/i^2 )=−i) e^(πi) =−1 e^((πi )/2) =i e^(((πi)/2).(−i)) =i^(−i) e^(π/2) =i^(−i) ⇒4.81077... For regularaization e^((π/2)+2πk) (k∈Z)

Commented byDwaipayan Shikari last updated on 13/Jul/20

approximately 4.810477380965351655473035666703833126390170874664534940020... (using the principal branch of the logarithm for complex exponentiation)

Commented byStudy last updated on 13/Jul/20

(e^((π/2)i) )^(−i) =(i)^(−i ) ok

Commented byStudy last updated on 13/Jul/20

(1/i)=i^(−1) ok

Commented byDwaipayan Shikari last updated on 13/Jul/20

(1/i)=(i/i^2 )=−i

Commented byDwaipayan Shikari last updated on 13/Jul/20

it means e^((π/2)i.(−i)) =(i)^(−i)

Answered by abdomsup last updated on 13/Jul/20

=i^(1/i) =i^(−i) =(e^((iπ)/2) )^(−i) =e^(π/2)

Answered by OlafThorendsen last updated on 13/Jul/20

z = (i)^(1/i) = i^(1/i) lnz = (1/i)lni = (1/i)lne^(i(π/2)) = (1/i)i(π/2) lnz = (π/2) ⇒ z = e^(π/2)