Finding distinct values for $\log(-e)$

Question

I want to find three distinct values for $\log(-e)$. I found the first by using the $\log$ rules to get $\log({e^{i\pi}}) + \log(e)$ which is equivalent to $i\pi + 1$. Now, since I want to find another distinct solution, I would have to add $2\pi$ to our number's phase in order to produce a full revolution in the unit circle. Would I need to add $2\pi$ to the phase in the polar form of $-e$ or would I just add $2\pi$ to our original solution?

Answer 1

It's the imaginary part of the logarithm that can be adjusted by any multiple of $2\pi$, so the possible logarithms of $-e$ are $(2k+1)\pi i + 1$ for $k\in\mathbb Z$.

Answer 2

Note that $\log(z)=\log(|z|)+i\arg(z)$, where $\arg(z)=\text{Arg}(z)+2n\pi$ and where $-\pi<\text{Arg}(z)\le \pi$ is the principal argument of $z$.

Hence, if $z=-e$, we find $|z|=e$, $\text{Arg}(z)=\pi$, and $\arg(z)=(2n+1)\pi$. Putting it all together, we can write

$$\log(-e)=i(2n+1)\pi$$