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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.

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    $\begingroup$ The first answer is $\log\left((1-i)^4\right)$ while the second is $4\log(1-i)$. $\endgroup$
    – robjohn
    Feb 10, 2019 at 22:55

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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.

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The discrepancy you've found here is analogous to what you get if you believe that

$$(-1)^{1/3}=(-1)^{2/6}=((-1)^2)^{1/6}=1^{1/6}$$

Note that there are three complex possibilities for $(-1)^{1/3}$ but six possibilities for $1^{1/6}$. That is, the "obvious" equality $z^{ab}=(z^a)^b$ is not true. Likewise the "obvious" simplification $\log(z^n)=n\log z$ can get you into trouble. I would say that you did the calculation correctly, and the book got it wrong.

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