Arnold Trivium 49 : $ \oint_{|z|=2}\frac{dz}{\sqrt{1+z^{10}}}$. Calculate : $$ \oint_{|z|=2}\frac{dz}{\sqrt{1+z^{10}}}.$$

If you find it too easy, then just post hints.
 A: 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 $J=0$.
A: First of all to make all things well-defined, we have to fix the branch of the square root. It will be chosen so that for real $z>0$ we have $\sqrt{1+z^{10}}\in\mathbb{R}_{>0}$.
Next let us analyze the singularities of the integrand in the complex $z$-plane. They are given by $10$ square root branch points on the unit circle, of the form $z_k=\exp\frac{i\pi(1+2k)}{10}$, $k=0,\ldots,9$. We can choose the branch cuts as shown on the figure below.
Now let us make the change of variables $z=1/s$. The contour of integration will change the orientation which will compensate the minus sign in $dz=-ds/s^2$, and the integral itself becomes
$$I=\oint_{|z|=2}\frac{dz}{\sqrt{1+z^{10}}}=\oint_{|z|=2}\frac{dz}{z^5\sqrt{1+z^{-10}}}
=\oint_{|s|=\frac12}\frac{s^3ds}{\sqrt{1+s^{10}}},$$
where all contours are oriented counterclockwise. Note that in the complex $s$-plane all branch cuts are outside the integration contour. Further, the point $s=0$ (which corresponds to $z=\infty$) is not a pole. Therefore we can shrink the contour in the $s$-plane and conclude that $I=0$. In terms of variable $z$, that would correspond to expanding the contour to infinity.

A: Consider the function
$$
\begin{align}
f(z)
&=\log(1025)+\sum_{k=0}^9\int_2^z\frac{\mathrm{d}w}{w-\xi^{2k+1}}\\
&=\log(1025)+\int_2^z\frac{10w^9\,\mathrm{d}w}{1+w^{10}}\tag{1}
\end{align}
$$
where $\xi=e^{\pi i/10}$ is a primitive $20^\text{th}$ root of $1$.
The residue at each singularity of the integrand is $1$. Thus, the integral over a path which circles all $10$ singularities, is $20\pi i$. This means that $g=e^{-f/2}$ is well-defined over any path that encompasses all of the singularities since $e^{-10\pi i}=1$. Therefore, define $g$ by integrating $(1)$ over any path that does not pass inside the unit circle.
From $(1)$,
$$
f'(z)=\frac{10z^9}{1+z^{10}}\tag{2}
$$
Thus, $f(z)$ is locally $\log\left(1+z^{10}\right)$. Therefore, since $g(2)=\frac1{\sqrt{1025}}$, we have
$$
g(z)=\frac1{\sqrt{1+z^{10}}}\tag{3}
$$
is well defined and analytic on $\mathbb{C}$ outside the unit circle.
Now, that we have that $\frac1{\sqrt{1+z^{10}}}$ is analytic outside the unit circle, it should be simple to compute
$$
\oint_{|z|=2}\frac{\mathrm{d}z}{\sqrt{1+z^{10}}}\tag{4}
$$
Hint: consider increasing the radius of the circle of integration to $\infty$.
Hint: The difference of the large counter-clockwise circular path and the small counter-clockwise circular path is the path below, which encloses the colored C-shaped region. Note that the subtracted path, the small circle, is in the opposite direction from the small counter-clockwise circular path.
$\hspace{4.9cm}$
The integral along the paths connecting the circular paths cancel each other out since they are along the same points in opposite directions.
Thus, the integral along the large counter-clockwise circular path equals the integral along the small counter-clockwise circular path since the function is analytic in the colored region.
The absolute value of $\frac1{\sqrt{1+z^{10}}}$ over the outer path is less than $\frac1{\sqrt{r^{10}-1}}\sim\frac1{r^5}$ and the path is $2\pi r$ long, so the integral over the outer path is $O\left(\frac1{r^4}\right)\to0$.
