Is there any function $f(z)$ such that $f(z)^i = z$?

Question

Basically, is there a formula for $\sqrt[i]{z}$.

I was thinking about complex numbers and trying to come up with formulas like $log_i(x)$ and $x^i$. I then thought about this:

$$\sqrt[i]{z} = f(z)$$

So I started trying. We know that $x^i = \cos(\ln(x))+i\sin(\ln(x))$, from Euler's formula. I then tried to reverse it:

$$\ln(x) \to e^x$$ $$\cos(x) \to \arccos(x)$$ $$...e^{\arccos(x)}$$

But, obviously, it will give you the wrong answer for $i\sin(\ln x)$.

I'm also unsure if $\arccos$ is even defined for values > 1.

Answer 1

Hint:

$$\sqrt[i]{z}=z^{1/i}=z^{-i}$$

And then apply Euler's formula.

Answer 2

It's a multi-valued function (see here for an explanation of branch cuts). Specifying a branch and taking an appropriate subset of $\mathbb{C}$ let's you define it on that subset.

To see how to define it, notice that $\frac{1}{i}=-i$ and so $z^{\frac{1}{i}}=z^{-i}=\frac{1}{z^i}$ and see if you can derive it from there via Euler's formula.

Answer 3

Note that $1/i=-i$. So, we may in fact define $$ f(z)= z^{-i} $$ Keep in mind, however, that $z^w$ is a multi-valued function for any non-real complex number $w$.