Consider $\log\bigl((1-i)^4\bigr)$.
If I first expand the inside, I get $$\log((1-i)^2\cdot(1-i)^2)=\log((1-2i-1)\cdot(1-2i-1))=\log(4i^2)=\log(-4)=\ln|-4|+i\operatorname{Arg}(-4)+2ki\pi=\ln(4)-i\pi+2ki\pi$$
Here $\operatorname{Arg}$ is the argument of the principal branch, and its value $\in (-\pi,\pi]$,and $k \in \mathbb{Z}$.
But if I start with exponential, then $$\log((1-i)^4)=\log(e^{4\log(1-i)})=\log(e^{4(ln|1-i|+i \operatorname{Arg}(1-i)+2ki\pi)})=\log(e^{4(\ln(\sqrt2)-i\frac{\pi}{4}+2ki\pi)})=4(\ln(\sqrt2)-i\frac{\pi}{4}+2ki\pi)=\ln(4)-i\pi+8ki\pi$$
In fact, the second one's answer is the final answer given by Stephen Fisher's book. But I felt the first one is more correct than the second one. Which one is correct?
Thanks a lot for any insights. Much appreciated.
Best Answer
I suggest that you don't try to deal with logarithms as if there were a logarithm function $\log\colon\mathbb{C}\setminus\{0\}\longrightarrow\mathbb C$. If your goal is to compute all logarithms of $(1-i)^4$, then compute one of them and then add all integral multiples of $2\pi i$ to it. Now, since $(1-i)^4=-4$ and since a logarithm of $-4$ is $\log4-\pi i$, the set of all logarithms of $(1-i)^4$ is$$\{\log(4)-\pi i+2k\pi i\,|\,k\in\mathbb{Z}\}.$$In paticular, your first answer is correct.