In Complex Variables and Applications (Churchill, Brown), Chapter 3 Exercise 13.5, we are asked to show that:
The set of values of $\log(i^\frac{1}{2})$ is $$ (n + \frac{1}{4})\pi i \quad (n = 0, \pm 1, \pm 2, \dots) $$
I know the definitions of $\log$ and $\exp$. Particularly, $\log(z) = \ln\lvert z\rvert + i \arg z$ where $\log$ is a multi-valued function.
There are several points that I am unsure of:
$i^\frac{1}{2} = e^{i \frac{\pi}{4}}, e^{i \frac{5\pi}{4}}$ is multi-valued. What does it mean to take the $\log$ of a multi-valued variable?
In my attempt to solve it, I tried to reduce to the case of taking $\log$ of a single value: By following the definitions of $\log$, we can obtain (i) $\log(e^{i \frac{\pi}{4}}) = i \pi (2n+\frac{1}{4})$ for $n \in \mathbb{Z}$. Similarly, we have (ii) $\log(e^{i \frac{5\pi}{4}}) = i \pi (2n + \frac{5}{4})$ fo $n \in \mathbb{Z}$. So, combining the solutions for (i) and (ii), the solutions for $\log(i^\frac{1}{2})$ is $(n + \frac{1}{4})\pi i$ for $n \in \mathbb{Z}$.
Is this correct?