I have not had much exposure to more advanced complex analysis exercises, although this one might be easy, and the solution just completely escapes me. I am determining the set of complex numbers for which the identity
\begin{equation}\log\left(1-\frac{1}{z}\right) = a(1-z)-b(z)-i\pi \end{equation} holds, where
\begin{align*} \log\left(1-\frac{1}{z}\right), &\quad \quad \arg\left(1-\frac{1}{z}\right) \in [-\pi, \pi) \\ a(1-z) = \log\left(1-z\right), &\quad \quad \arg\left(1-z\right) \in [0, 2\pi)\\ b(z) = \log(z), &\quad \quad \arg(z) \in [-\pi, \pi). \end{align*}
I was tempted to observe simply the complex parts of $\log\left(1-\frac{1}{z}\right)$ such that we have
\begin{align*} \log\left(1-\frac{1}{z}\right) &= \ln\left|\frac{z-1}{z}\right|+i\arg\left(\frac{z-1}{z}\right) \\ &= \ln|1-z| - \ln|z| + i\arg\left(\frac{z-1}{z}\right) \end{align*} and compare the arguments of each side, though I'm not sure how to proceed from here. My other question, albeit off-topic, is if the power series for the left side in the first equation would hold true for the right side (in this case, we would have a Laurent series). That is, does it make sense that we would have $$ a(1-z)-b(z)-i\pi = -\sum_{n \geq 1}\frac{1}{nz^n}$$ or this some absurdity? I am missing some fundamentals here. Your support is very much appreciated!