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With the integration I mean one counter-clockwise turn around the origin, i.e.

$\int_{\phi=0}^{2\pi}\ln(1-re^{i\phi})ire^{i\phi}d\phi$

For $r<1$, this is simply a contour integration on a holomorphic part of the logarithm and thus one should obtain 0, but for $r>1$ the integration path winds up to the next branch, so this is no longer a contour integral. According to Wolfram Alpha the indefinite integral is $2\pi$ periodic so the definite integral would vanish, but can I simply add the $2\pi i$ branch difference to obtain

$\begin{cases}0 & r<1 \\ 2\pi i & r>1\end{cases}$

or is the branch-respecting integration more involved? I was considering using $\ln(1-z)=\ln(|1-z|)+i\arg(1-z)$ demanding $\arg$ to remain continuous, but this seems to get rather a mess...

update I used MATLAB for numerical integration (which uses $\ln(1-z)=\ln(|1-z|)+i\text{atan2}(\Im(-z),1-\Re(z))$) and obtained this:

So, the solution should rather be

$\begin{cases}0 & r<1 \\ i(r-1) & r>1\end{cases}$

The question remains though, how to solve this analytically?

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    Maybe start by writing the real and imaginary parts of the log. To do that, write $1-e^{i\phi}$ in polar form.2012-02-20

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The derivative of $z\log(z) - z$ is $\log(z)$, so the derivative of $(1- z) - (1 - z)\log(1 - z)$ is $\log(1 - z)$. If you make the branch cut $(1,\infty)$ as you are doing, your answer is $\big((1- re^{i\theta}) - (1 - re^{i\theta})\log(1 - re^{i\theta})\big)|_0^{2\pi}$. Here $\log$ denotes whatever branch you're using. The $(1- re^{i\theta})$ terms cancel upon subtraction, while the second terms give you (using $\log(1 - r) = \ln(r - 1) + \pi i$)

$ -(1 - r)(\ln(r - 1) + \pi i + 2\pi i) + (1 -r)(\ln(r - 1) + \pi i)$ $= 2\pi i(r - 1)$