TITLE: Arnold Trivium 49 : $ \oint_{|z|=2}\frac{dz}{\sqrt{1+z^{10}}}$.
QUESTION [7 upvotes]: Calculate : $$ \oint_{|z|=2}\frac{dz}{\sqrt{1+z^{10}}}.$$

If you find it too easy, then just post hints.

REPLY [6 votes]: Choosing the squareroot which is positive on the positive real axis we have
$${1\over\sqrt{1+z^{10}}}={1\over z^5}(1+z^{-10})^{-1/2}={1\over z^5}\sum_{k=0}^\infty{-1/2\choose k}z^{-10k}\ ,$$
where the binomial series converges uniformly on $\partial D_2$. Therefore
$$J:=\int_{\partial D_2}{dz\over\sqrt{1+z^{10}}}=\sum_{k=0}^\infty{-1/2\choose k}\int_{\partial D_2}z^{-10k-5}\ dz\ .$$
As $-10k-5\ne-1$ for all $k\geq0$ all summands on the right hand side are zero. It follows that $J=0$.