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Suppose I have the following function of a complex variable $$f(z)=\log(z)(z^2+1)^{1/2}.$$ Wolfram Alpha tells me the branch cuts of $f(z)$ are $z\leq 0$ (presumably for the logarithmic term), and $\text{Re}(z)=0$,$|z|\geq 1$ (presumably for the square root), i.e. they are restricted to subsets of either the real or imaginary axes. If I want to use the residue calculus with $f(z)$, my contour had better avoid (i.e. not cross) these branch cuts.

I think I'm right in saying that I can choose my branches to suite my needs. Hence, if it turns out that the Wolfram choice of branch cuts permits an awkward contour, can I make a different choice of branch (by way of defining my branch cuts differently) to make life easier?

For example, instead of the above choice, could I choose my branch cuts to be $z\geq 0$ for the logarithm, and $z\leq -1$ and $z\geq 1$ for the square root? This choice leaves the entire upper and lower half planes "free of problems", whereas the initial choice defines branches along both positive and imaginary axes.

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You can certainly put your branch cuts where you want to. The logarithmic term forces a branch cut from $0$ to infinity in whatever direction you desire – it doesn't even have to be a straight line. And for the other factor, a branch cut between $i$ and $-i$ is required. You can take the shorter way, or you can do as Wolfram Alpha suggests and let it pass through infinity.

Of course, this all depends on what you want to do. The application often dictates some special choice of branch cuts. And sometimes, merely using branch cuts isn't sufficient, and you have to use more of the associated Riemann surface – as when you use the Pochhammer contour to study the beta function. There is no choice of branch cuts that will avoid that contour!

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  • $\begingroup$ @Harold thanks for your reply. Just to clarify - you say "And for the other factor, a branch cut between i and −i is required." Does this mean I can't take the alternative branch cut I suggested for the square root term? $\endgroup$ – Pixel Jan 15 '14 at 19:05
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    $\begingroup$ A branch cut involving $\pm i$ is essential, since $(z^2+1)^{1/2}$ has branch points there (where $z^2+1=0$). I did mean either of the alternatives you mention. The Wolfram Alpha choice is the former. Think of the Riemann sphere: $i$ to $\infty$ to $-i$ can be though of as one path, going up the imaginary axis, passing through $\infty$, and coming back up the negative imaginary axis. On the Riemann sphere, there is only one infinity, as it is the same in any direction. $\endgroup$ – Harald Hanche-Olsen Jan 15 '14 at 19:09
  • $\begingroup$ @Harold just one more query: If the branch cut involving $\pm i$ is essential, then am I right in thinking by "You can certainly put your branch cuts where you want to," that the branch cuts can be in any direction possible, so long as overall the complex plane is "cut" ? For example, they could be horizontal to $-\infty$ and $+\infty$ or even both $-\infty$ or both $+\infty$ (from the branch points), etc. $\endgroup$ – Pixel Jan 15 '14 at 19:21
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    $\begingroup$ Basically, yes. What you need to ensure is that, if you follow any closed curve in the cut plane, the function value will return to its original value when you return to the starting point. Moving once around a square root branching point will multiply by $-1$. So you must disallow that, but it's okay to move once about two such points, since the sign changes will cancel out. Logarithmic branch points are more difficult to deal with. $\endgroup$ – Harald Hanche-Olsen Jan 15 '14 at 19:46
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    $\begingroup$ Sure, I don't see why not. But the devil is in the details as always. $\endgroup$ – Harald Hanche-Olsen Jan 15 '14 at 21:37

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