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Let's take the principal branch $L(Re^{i\varphi})=\ln(R)+i\varphi$ of the logarithm. I got myself confused now over this: Where is the error in the line $$i\pi=L(e^{i\pi})=L(e^{i\pi}\cdot 1)=L(e^{i\pi}\cdot e^{2i\pi})=L(e^{3i\pi})=3i\pi$$

Also: the principal branch isn't even defined for $e^{i\pi}$ but still we can compute it there. What's going on exactly?

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  • $\begingroup$ If you look at the principal branch, then $L_{principal}(e^{3i\pi})=i\pi$. You are not allowed to simply "cancell" $L$ with $e$. :) $\endgroup$ – C. Brendel May 8 at 10:09
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In the last equality: $L(e^{3i\pi})=i\pi$, not $3i\pi$. In the definition of principal branch of the algorithm, we assume that $\varphi\in(-\pi,\pi)$ when we del with the expression $L(Re^{i\varphi})=\ln R+i\varphi$. Note that otherwise the definition would be ambiguous.

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