How to calculate $i^i$ [duplicate]

I've been struggling with this problem, actually I was doing a program in python and did

1j ** 1j(complex numbers) (In python a**b = $a^b$ )

and found out the answer to be a real number with value $0.2079$, How to calculate this value of $i^i$?

Solution 1:

First you need to realize that this is a multi-valued function.

$$i=0+i$$ $$i=\cos\left(\frac{(4k+1)\pi}{2}\right)+i\sin\left(\frac{(4k+1)\pi}{2}\right)$$ $k\in \mathbb Z$, Using Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$

$$i=e^{\large{i(4k+1)\pi}/{2}}$$

Now

$$\begin{align}i^i&=\left(e^{\large {i(4k+1)\pi}/{2}}\right)^i\\ &=e^{\large{i\times i(4k+1)\pi}/{2}}\\ &=e^{\large{-(4k+1)\pi}/{2}}\\ \end{align}$$

Depending on the branch cut (part of $\mathbb{C}$ excluded) and branch of log (what value $\log(z_0)$ has for some $z_0$ not in the branch cut), there is a $k\in\mathbb{Z}$ so that $i^i=e^{-(4k+1)\pi/2}$. Using the most standard branch cut, $i^i=e^{-\pi/2}$

And python didn't seem to care about that and returned

$$i^i=e^{\large{-\pi}/{2}}\approx 0.2078795$$

Solution 2:

$e^{i\pi} = -1$

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

$i^i = e^{-\frac{\pi}{2}}$

I just wanted to point out the derivation!