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I have the expression $$f(z)=\frac{-i}{\sqrt{z^2-a^2}},$$ where $a$ is a purely real number and $z$ is a complex variable. Numerical plotting gives the following.

enter image description here

This leads me to the following observations about the branch structure due to the square-root.

  1. There are branch points at $z=\pm a$
  2. The associated branch cuts go from $\pm a$ to the origin and on the from the origin to $\pm i\infty$.
  3. $\Re[f(z)]$ is discontinuous (flips signs) along this cut.
  4. $\Im[f(z)]<0$ everywhere and achieves the bound $\Im[f(z)]=0$ along the cut.

Looking at the behavior near $z=\pm a$, we also see what looks like kind of like pole singularities; certainly, the function diverges there, but because the real part flips signs there, they aren't really poles. The pole-like behavior is somehow obscured by the branch cut. I certainly can't integrate around these singularities to find residues because I can't cross the cut.

My questions are:

  1. What is this kind of singularity called?
  2. If I have integrals of the form $\int_0^\infty f(z)g(z) dz$ where $g(z)$ is nice and holomorphic everywhere except along the same branch cuts as $f(z)$, are there any standard complex analysis techniques that apply? The residue theorem does not seem to help since I can't integrate around the singularity.
  3. Are there some extensions to, or applications of, the residue or Cauchy integral theorems that can be used?
  4. Is the only contribution to such integrals the integral around branch cut? Is there a pole contribution when I change contours at all?
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  • $\begingroup$ For completeness sake, there is also a pole at infinity. $\endgroup$ – Semiclassical Jul 28 '14 at 18:02
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    $\begingroup$ Your numerical plotting led you up the garden path. You have two branch points, $a$ and $-a$. The potential third branch point is $\infty$, but since $z^2-a^2$ attains the value $\infty$ with multiplicity $2$ at $\infty$, that is not actually a branch point. You can take the (real) interval $[-a,a]$ as the branch cut and have a holomorphic function defined on $\widehat{\mathbb{C}}\setminus [-a,a]$. Or you can take $(-\infty,-a] \cup [a,+\infty)$ as the branch cut and have a holomorphic function on $\mathbb{C}\setminus \{x\in \mathbb{R} : \lvert x\rvert \geqslant a\}$. Or any other curve ... $\endgroup$ – Daniel Fischer Jul 28 '14 at 18:10
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    $\begingroup$ connecting $a$ and $-a$. You can indeed also introduce artificial cuts along the imaginary axis, and have two unrelated branches of $f$ on the left and on the right half-plane (minus $[-a,0)$ resp. $(0,a]$), but those are not natural. $\endgroup$ – Daniel Fischer Jul 28 '14 at 18:12
  • $\begingroup$ Keep in mind that those are the branch cuts selected by the programming system: they are not definitive, and indeed within this context they're not helpful. (To confirm this point, try replacing $\sqrt{z^2-1}\to i \sqrt{1-z^2}$. They shouldn't give the same plot.) $\endgroup$ – Semiclassical Jul 28 '14 at 19:26
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    $\begingroup$ The straightest course is just to do it by hand. (It's annouying that it's hard to get the plotters to do it smartly, but they rely on the branch cut being defined by arg ${z}=\pi$.) $\endgroup$ – Semiclassical Jul 28 '14 at 19:45

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