I was working on a problem stating as
Find all possible values $z\in\mathbb{C}$ such that $\log\dfrac{i\pi}{3}=z$.
I think I worked it out pretty quick but then I was a little bit confused about the principal branch.
Here is my attempt:
Firstly we have $$\log\dfrac{i\pi}{3}=\log i+\log\dfrac{\pi}{3}.$$
Then, the principal branch of $\log i$ is \begin{align*} \text{Log}(i)&=\log|i|+i\arg(i)\\ &=0+i\dfrac{\pi}{2}. \end{align*}
Therefore, we can conclude that $$z=\log\dfrac{\pi}{3}+i\dfrac{\pi}{2}.$$
However, the words "find all possible values" in the question alerts me to double check my answer, then I found a problem with the principal branch.
I used the principal branch to solve the problem, but should it be $$\log(i)=0+i(\dfrac{\pi}{2}+2k\pi),$$ so that $$z=\log\dfrac{\pi}{3}+i(\dfrac{\pi}{2}+2k\pi)?$$
So the question here is that, when we deal with the logarithm of a complex number, we only use principal value or we use the infinitely many value?
Perhaps I totally misunderstood the principal branch.
Thank you for any corrections and inspirations.
First of all you cannot use $\log (ab)=\log\, a +\log\, b$ for complex numbers. There are infinitely many logarithms of $\frac {i \pi} 3$ and they are $\log (\frac {\pi} 3)+i(\frac {\pi} 2+2n\pi)$ where $n$ is an integer. The principal logarithm is $\log (\frac {\pi} 3)+i\frac {\pi} 2$. For the principal logarithm the imaginary part $\theta$ of the logarithm must satisfy $-\pi <\theta \leq \pi$.
You are right in saying that there is only one value for the principal branch of logarithm.