i-1-i-

i-1-i-

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

\sqrt[i]{i} = ? ? ? ? ?

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

i^{\frac{1}{i}} = i^{- i} (\frac{1}{i} = \frac{i}{i^{2}} = - i)

e^{π i} = - 1

e^{\frac{π i}{2}} = i

e^{\frac{π i}{2} . (- i)} = i^{- i}

e^{\frac{π}{2}} = i^{- i} \Rightarrow 4.81077 \dots

For regularaization

e^{\frac{π}{2} + 2 π k} (k \in Z)

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

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

Commented by Study last updated on 13/Jul/20

{(e^{\frac{π}{2} i})}^{- i} = {(i)}^{- i} o k

Commented by Study last updated on 13/Jul/20

\frac{1}{i} = i^{- 1} o k

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

\frac{1}{i} = \frac{i}{i^{2}} = - i

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

i t m e a n s

e^{\frac{π}{2} i . (- i)} = {(i)}^{- i}

Answered by abdomsup last updated on 13/Jul/20

= i^{\frac{1}{i}} = i^{- i} = {(e^{\frac{i π}{2}})}^{- i} = e^{\frac{π}{2}}

Answered by OlafThorendsen last updated on 13/Jul/20

z = \sqrt[i]{i} = i^{\frac{1}{i}}

\ln z = \frac{1}{i} \ln i = \frac{1}{i} \ln e^{i \frac{π}{2}} = \frac{1}{i} i \frac{π}{2}

\ln z = \frac{π}{2} \Rightarrow z = e^{\frac{π}{2}}