Effective Branch cut

Asked May 27 '15 at 13:02

Active May 27 '15 at 13:02

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I have problems understanding how to get to the "effective Brach cut" in the top answer of this post: Dog Bone Contour Integral ?

The answer says that one has a Branch cut for $\frac{1}{\sqrt{z-1}}$ at $(-\infty,1]$ and for $\frac{1}{\sqrt{z+1}}$ at $(-\infty,-1]$ and then one gets the "effective" Branch cut for $f(z) = \frac{1}{\sqrt{(z-1)(z+1)}}$ at $[-1,+1]$.

I don't understand this. Why is the function $f(z)$ suddenly defined on $(-\infty,-1)$? What cancels out ? Thanks in advance for the help!

asked May 27 '15 at 13:02

benbln

This MSE link may prove relevant, as may this MSE link II. – Marko Riedel May 27 '15 at 19:39
Thank you!This was helpful! – benbln Jun 04 '15 at 17:38

Effective Branch cut

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