$\int_{|z| = 2} \frac{1}{f(z)(1+f(z))^2} dz$ where $f(z) = z^{1/2}$ with branch such that $\Re f(z) \geq 0$

Question

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.

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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.

Answer 1

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 $F(e^{i\pi}) - F(e^{-i\pi}) = 2i.$