I am not getting complex variable answer $\sqrt{i}^{\sqrt{i}}$

Question

I am solving complex variable and I have solve to one problem that $\sqrt{i}^{\sqrt{i}}$. If I had $i^i$ then it become $e^{i \log i}$ and: $$i=\cos(\pi/2)+i\sin(\pi/2)\implies i=e^{\pi/2}$$ so $\log i=2ni\pi+\log e^{\pi/2}$ it become $i(4n+1)\pi/2$ which shows $e^{i(i(4n+1)\pi/2}= e^{-(4n+1)\pi/2}$ therefore $e^{i \log i}=e^{-(4n+1)\pi/2}$.

I tried stack exchange for the first time and this is my first question please help me out. I tried my best to explain.

question: show that $\sqrt{i}^{\sqrt{i}}$= $e^{-\pi/4\sqrt{2}}(\cos π/4(√2)+i \sin π/4(√2) )$

Answer 1

$$\sqrt{i}^{\sqrt{i}} = e^{i \frac{\pi}{4} e^{i \frac{\pi}{4}}} $$

$$ = e^{i \frac{\pi}{4} \left [ \cos{\left ( \frac{\pi}{4} \right )} + i \sin{\left ( \frac{\pi}{4} \right )} \right ] } $$

$$ = e^{i \frac{\pi}{4} \left ( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right )} $$

$$ = e^{-\frac{\sqrt{2} \pi}{8}} \left [ \cos{\frac{\sqrt{2} \pi}{8}} + i \sin{\frac{\sqrt{2} \pi}{8}} \right ] $$

Answer 2

Surely, you can find $a+ib$ such that for a given $t$ we have $t=\sqrt{a+ib}$. So we don't have any difficulties to solve $t=\sqrt{i}$ for a given $t$. Now take $x=\sqrt{i}^{\sqrt{i}}$ so $$x^2=i^{\sqrt{i}}=i^t$$ where $t=a+ib$ for some $a,b$. Therefore $$x^2=i^{\sqrt{i}}=i^t=i^{a+ib}=i^a\cdot (i^i)^b$$ and you already noted you could evaluate $i^i$, so find $x$.

Answer 3

Try substituting e.g. $\sqrt i = \pm \left( \dfrac{1 + i}{\sqrt 2} \right)$.