Assume that $f(z) \in \{\sqrt{2z^2 + 1}\}$ $,f(0) = 1$
We have a cut: $\gamma = \{|z| = \frac{1}{\sqrt2}, Re(z) \geqslant 0 \}$
$\oint\limits_{|z|=1} \frac{z dz}{(z+2)(f(z) + 3)}$
I found singularities: $z_1 = 2$ and $z_{2,3} = -2$. But they are not in our area $|z| < 1$. According to residue theorem, it's mean that integral is equal $0$.
But in book they have completely different answer: $\pi i( \frac{17}{12} - \sqrt2 )$
What have I missed?
Too long for a comment...
I would suggest:
1) Choose the contour of integration as described above, this means two half circles with radius $1$ arguments $[\pm\pi/2\pm\epsilon]$ and four straight lines $[\pm i/\sqrt{2}\pm\epsilon,\pm i\pm\epsilon]$. The litte circles around the branching point will vanish...
2)Because no residues are inside our contour the only contribution is from the branch cut integrals around the slits.
3)The remaining integrals can be done by flip the cut so that we just have one which reaches from $(-i/\sqrt{2},i\sqrt{2})$ (You have to be really careful about the arguments at this point)
4) The remaining integral can be done by reverting the direction of the contour and pick up the residues in the whole complex plane
5) At this point it will be necessary to pick up residues at infinity, this can be done by changing $f(z)\rightarrow \frac{-1}{z^2}f(1/z)$ and looking at $0$