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I have the following problem:

Consider the function $$f(z)=\ln(z+1)-\ln(z-1)$$ This has branch points at $z=1$ and $z=-1$, and we place the branch cut joining the two branch points along the real line. Consider the contour segments about the branch points:

A: Anticlockwise arc of infinitesimal radius, encircling $z=1$ from just below the cut to just above.

B: Anticlockwise arc of infinitesimal radius encircling $z=-1$ from just above the cut to just below.

Calculate the integral $\int f(z) dz$ over the contours A and B.

My attempt:

I parameterize segment A as such: $$ A \equiv \{z(\theta)=1+\epsilon e^{i\theta} \,\vert\, \theta\in(-\pi,\pi)\} $$

Now computing the integral (choosing the principal branch): $$\begin{align} \int_Af(z)dz &= \lim_{\epsilon\to 0}\int_{-\pi}^\pi f(z(\theta))\,z'(\theta)\,d\theta \\ &= \lim_{\epsilon\to 0}\int_{-\pi}^\pi \ln \bigg\vert\frac{\epsilon e^{i\theta}+2}{\epsilon e^{i\theta}}\bigg\vert \, i\epsilon e^{i\theta}\,d\theta \\ &= \lim_{\epsilon\to 0}\int_{-\pi}^\pi \ln \bigg\vert 1+\frac{2}{\epsilon e^{i\theta}}\bigg\vert \, i\epsilon e^{i\theta}\,d\theta \end{align}$$

At which point I'm stuck on how to evaluate the integral. I want to check if I have parameterized A correctly, and that it is just a matter of finding a way to plough through the integration; Or whether there is a smarter way of parameterizing/setting up the integral.

I only require assistance on A, as B will be very similar once I get an idea of how to compute the integral. Thanks all!

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  • $\begingroup$ $$\oint_A (\log(z+1)-\log(z-1))\,dz=\lim_{\epsilon \to 0}\int_{-\pi}^\pi (\log(2+\epsilon e^{i\phi})-\log(\epsilon e^{i\phi}))\,\epsilon e^{i\phi}\,d\phi=0$$ $\endgroup$ – Mark Viola Nov 28 '16 at 14:23
  • $\begingroup$ @Dr.MV The integral around a branch point can be zero? $\endgroup$ – Troy Nov 28 '16 at 14:27
  • $\begingroup$ Yes. In the limit as $\epsilon \to 0$. $\endgroup$ – Mark Viola Nov 28 '16 at 14:51

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