Find the principle value and all other values of $i^{2/\pi}$ I'm a little confused about how to go about this? Any help would be appreciated, thanks.
 A: Complex exponentiation is a delicate process. The typical way of defining complex exponentiation $a^b$, where both $a,b \in \mathbb{C}$, is to say
$$a^b := e^{b\log a}.$$
But the logarithm function isn't well-defined. It has infinitely many branches. When $b$ is an integer, it doesn't matter what branch we choose, so $a^b$ is well-defined. When $b$ is rational, there are finitely many choices. When $b$ is irrational, $a^b$ is extremely ambiguous and might represent infinitely many numbers based on the choice of the logarithm.
Recall that $\log z = \ln r + i\theta + 2\pi i k$, where $z = re^{i\theta}$, $k$ is some integer, and I use $\ln (\cdot)$ to represent the real logarithm. The ambiguity lies in the choice of $k$. The so-called principal branch is when we set $k=0$.
So here, 
$$i^{2/\pi} = e^{(2/\pi) \log i} = e^ {(2/\pi)(i\pi/2 + 2\pi i k)} = e^ {i}\color{#0000A0}{e^{4 i k}},$$
where the principal value arises when $k=0$ and the other (infinitely many) values come from $k \neq 0$, and which I denoted in blue.
A: The computation is straightforward. The values of $z^w$ are all the possible values of $e^{w\log z}$. So here, you want the possible values of $e^{\frac{2}{\pi}\log i}$.
Since you can write $i=e^{\frac{\pi}{2}i}$, the values of $\log i$ are $(\frac{\pi}{2}+ 2\pi k)i$ for integral $k$. This means you have the values
$$e^{\frac{2}{\pi}(\frac{\pi}{2}+ 2\pi k)i}
=e^{(1+4k)i}=\boxed{\cos(1+4k) +i\sin(1+4k)}
$$ for integral $k$.
The principal value, obtained by using the principal value in the logarithm (so $k=0$), is $$\cos 1 + i\sin 1.$$
A: using $a^b=e^{b\log a}$ and $\log z=\ln |z|+i(arg(z)+2\pi k) \forall k\in \mathbb Z $ we have that
$i^{\frac2{\pi}} = e^{\frac2{\pi} \log i} = e^ {\frac2{\pi}(\ln 1+ i(\frac{\pi}{2} + 2\pi k))} = e^ {\frac2{\pi} i(\frac{\pi}{2} + 2\pi k)}=e^{(1+4k)i}=\cos (1+4k)+i\sin (1+4k) \forall k\in \mathbb Z$
Principal value is obtained using $k=0$ i.e it is $e^i=\cos 1+i\sin 1$
