Prove that ${\sqrt i} ^ {\sqrt i} = e^{-\alpha}\operatorname{cis} \alpha$ , where $\alpha = \frac {π}{4 \sqrt 2}$

Question

Prove that ${\sqrt i} ^ {\sqrt i} = e^{-\alpha}\operatorname{cis} \alpha$ , where $\alpha = \frac {π}{4 \sqrt 2}$

My Attempt:

I know that $Z^{\alpha}= e^{\alpha\log Z} = e^{\alpha (\ln |Z| + \operatorname{arg}Z)}$ So by using it I get\begin{align}{\sqrt i} ^ {\sqrt i}&= e^{\sqrt i\log \sqrt i}\\&=e^{\sqrt i \left(\ln |\sqrt i| +\operatorname{arg} \sqrt i\right)}\\&=e^{\sqrt i\left(\ln 1 + \operatorname{arg} \sqrt{\cos\frac{π}{2} + i\sin\frac{π}{2}}\right)}\\&= e^{\sqrt i\left(0 + \operatorname{arg} \sqrt{\cos\frac{π}{2} + i\sin\frac{π}{2}}\right)}\\&= e^{\sqrt i \left(\operatorname{arg} \left(\cos\frac{π}{4} + i\sin\frac{π}{4}\right)\right)}\\&= e^{\sqrt i\left(\operatorname{arg}\left(e^{i\frac{π}{4}}\right)\right)}.\end{align}I can't proceed it further. Please help me.

Answer 1

Hint : Use Euler's identity, $e^{ix} = \cos (x) + i\sin(x)$. Write $\sqrt i $ in terms of exponential function.

Answer 2

The number $i$ has two square roots, which are $\pm e^{\pi i/4}$. Let us consider just $e^{\pi i/4}$. A logarithm of $e^{\pi i/4}$ is, of course, $\frac{\pi i}4$ and, more generally, the set of all logarithms of $e^{\pi i/4}$ is $\left\{\frac{\pi i}4+2\pi in\,\middle|\,n\in\Bbb Z\right\}$. Then, one of the possible values of $\sqrt i^{\sqrt i}$ is$$\exp\left(e^{\pi i/4}\frac{\pi i}4\right)=\exp\left(\frac{\pi}{4\sqrt2}(-1+i)\right)=e^{-\alpha}(\cos\alpha+i\sin\alpha).$$

Answer 3

$$\sqrt i = e^\frac{\pi i}{4}=\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}=\sqrt2/2+i\sqrt2/2$$ $$\sqrt i^{\sqrt i} = e^\frac{\pi i \sqrt i}{4}=e^\frac{\pi i (\sqrt2/2+i\sqrt2/2)}{4}=e^\frac{\pi i (\sqrt2/2)}{4}e^{-\frac{\pi (\sqrt2/2)}{4}}$$