Calculate integrals concerning a branch i am trying to calculate this integral:
$$\int_{|z|=5}\frac{dz}{\sqrt{z^2+11}}$$
Using the branch that gives : $\sqrt{36} = -6$
The function has 2 poles at $|z| < 5$, lets call them $\alpha$ and $\beta$,
We still have the problem of the branch so our function is not analytic at 
$\{z | |z| < 5 \wedge z \neq \alpha \wedge z \neq \beta \}$
, so we can't use the residue theorem, 
Tried re parametrization ($z=e^{i\theta} | 0 < \theta < 2\pi$) and maybe finding the primitive function, yet it seems to not be working, 
I thought maybe i should split the curve to 2 parts (twice half circle), and there we can find a branch, so we still get the sum of the residues multiplied by 
$2\pi i$,
Is that correct ? 
 A: Writing explicitly the integral, we want to compute:
$$I = \int_{0}^{2\pi}\text{Sign}(\cos t)\frac{5i e^{it}}{\sqrt{25 e^{2it}+11}}\,dt \tag{1}$$
where $\sqrt{w}$ stands for the square root of $w$ with non-negative imaginary part. The symmetry of the real and imaginary parts of the integrand function lead to:
$$\begin{eqnarray*} I &=& 4i\int_{0}^{\pi/2}\text{Re}\left(\frac{5}{\sqrt{25+11e^{-2i\theta}}}\right)d\theta = 10i\int_{0}^{\pi}\text{Re}\left(\frac{1}{\sqrt{25+11e^{-i\theta}}}\right)d\theta\\&=&5i\int_{-\pi}^{\pi}\text{Re}\left(\frac{1}{\sqrt{25-11 e^{i\theta}}}\right)\,d\theta\tag{2}\end{eqnarray*}$$
Now exploiting the fact that $\frac{1}{\sqrt{25-11z}}$ is an analytic function over $|z|\leq 1$ it follows that:
$$ I = 5i\cdot 2\pi\cdot\frac{1}{\sqrt{25}} = \color{red}{2\pi i}.\tag{3}$$
A: Note that for $|z|>\sqrt{11}$ the integrand is single valued and for that particular branch of the square root it has the Laurent expansion
$$ \frac{1}{\sqrt{z^2+11}}=\frac{1}{z}\cdot\frac{1}{\sqrt{1+\frac{11}{z^2}}} = -\frac{1}{z} + O(\frac{1}{z^3}). $$
So by Cauchy
$$\frac{1}{2\pi \textrm{i}}\int_{|z|=5}\frac{dz}{\sqrt{z^2+11}}=-1.$$
Another way to get this result is to contract the loop $|z|=5$ to the contour that starts at $z=-\sqrt{11}\textrm{i}$, goes up to $\sqrt{11}\textrm{i}$ makes a small counter clockwise turn around it and goes back down. Turning around the branch point picks up a factor $-1$ so the integral becomes
$$2\textrm{i}\int_{-\sqrt{11}}^{\sqrt{11}}\frac{dx}{\sqrt{11-x^2}}=2\textrm{i}\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}}=-2\pi \textrm{i}$$
with the negative branch of the square root in the integrand.
