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As the title states, the definite integral in question is

$$\int_{|z| = 2} \frac{1}{f(z)(1+f(z))^2} dz,$$

where $f(z) = z^{1/2}$ with branch cut such that $\Re f(z) \geq 0$, i.e., the cut is the negative real axis.

Edit: As pointed out in a comment, this isn't an integral that should be done with a keyhole contour (edited out of my original question).

My question thus is, how does one evaluate this contour integral given that the contour is not a closed loop due to the presence of a branch cut? I am studying out of Ahlfors and integrating on a Riemann surface hasn't been introduced yet.

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    $\begingroup$ I'm a little rusty with contour integration so maybe wait for (hopefully) verification from others on this, but: It seems to me that if the problem calls for integration along $|z|=2$ then, no, you don't do the keyhole contour, you stick with the $|z|=2$ circle. Since the square root introduces a branch, you have to do more than one loop around the circle to close the contour. I think two should do it, but double check that. $\endgroup$
    – bob.sacamento
    Aug 18, 2014 at 23:13
  • $\begingroup$ @bob.sacamento Yes, you're right. I've been studying using contour integrals to evaluate definite integrals and got a tad mixed up while tired. Doesn't that approach amount to integrating on a Riemann surface? I'm studying out of Ahlfors and I haven't seen any such method explained so far. (I edited my question.) $\endgroup$
    – user169845
    Aug 19, 2014 at 0:06

1 Answer 1

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This was so simple it's embarrassing.

Let

$$F(z) = \frac{-2}{1 + z^{1/2}}.$$

Then, $F'(z) = f(z)$. Because of the branch cut, the integral is over the curve starting at $e^{-i\pi}$ and going counterclockwise to $e^{i\pi}$, so the integral is just $F(e^{i\pi}) - F(e^{-i\pi}) = 2i.$

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