Quick way of solving the contour integral $\oint \frac{1}{1+z^5} dz$

Do you know what the residue at the infinity is?

There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.

The relation is $$\sum_{k=0}^{n} \operatorname{Res}_{a_k}(f) = -\operatorname{Res}_{\infty}(f),$$ where $\{ a_{k} : n(\Gamma,a_k)\ne 0\}_{k=0}^{n}$ are the poles (inside the path $\Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(\gamma,z)$ is the index of the complex number $z$ with respect to the path $\gamma$. I'm supposing that the curve only winds one time around the origin.

HINT:

All of the poles lie on the unit circle. Hence, for any values of $R_1>1$ and $R_2>1$, we have

$$\oint_{|z|=R_1}\frac1{1+z^5}\,dz=\oint_{|z|=R_2}\frac1{1+z^5}\,dz$$

Now, what happens for $R_1=3$ and $R_2\to\infty$? The answer is zero. This approach is tantamount to invoking the residue at infinity.

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If one insists on summing the residues, then one may proceed in general as follows. First note that if $z^n+1=0$, then $z=z_k=e^{i(2k-1)\pi/n}$ for $k=1,2,\dots,n$.

Next, we calculate the residues of $\frac{1}{z^n+1}$ at $z_k$ using L'Hospital's Rule as follows:

$$\begin{align} \text{Res}\left(\frac1{z^n+1},z=z_k\right)&=\lim_{z\to z_k}\left(\frac{z-z_k}{z^n+1}\right)\\\\ &=\frac{1}{nz_k^{n-1}}\\\\ &=\frac1n e^{-i(2k-1)\pi(n-1)/n}\\\\ &=-\frac1n e^{-i\pi/n}\left(e^{i2\pi/n}\right)^k \end{align}$$

Summing all of the residues reveals for $n>1$

$$\begin{align} \sum_{k=1}^n \text{Res}\left(\frac1{1+z^n},z=z_k\right)&=-\frac1ne^{-i\pi/n}\sum_{k=1}^n\left(e^{i2\pi/n}\right)^k\\\\ &=-\frac1ne^{=i\pi/n} \left(\frac{e^{i2\pi/n}-\left(e^{i2\pi/n}\right)^{n+1}}{1-e^{i2\pi/n}}\right)\\\\ &=0 \end{align}$$

Hence, for $n>1$ the residues of $\frac1{1+z^n}$ add to zero. And we are done!

Here is a subtle approach that completely gets around having to determine the residues.

Cut the plane along the positive real axis. Do your integration around a contour that goes like this:

Part 1: Counterclockwise from $z=3$ on the upper side of the cut to $x=3$ on the lower side of the cut.

Part 2: Follow the lower side of the cut out to $z=R$, where we will take the limit as $R\to\infty$.

Part 3: Clockwise around the circle $|z|=R$ ending back on the upper side of the cut.

Part 4: Back to $z=3$ along the upper side of the cut.

This contour encloses no singularities, therefore the integral around it is zero. But also:

Parts 2 and 4: The real integral converges between $x=3$ and $x=\infty$ by comparison with $1/x^5$ whose antiderivative $(-1/(4x^4))+C$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.

Part 3: The integrand is $O(1/R^5)$ and the length of the contour is $2\pi R$, so by the triangle inequality the Part 3 integral is $O(1/R^4)\to 0$.

So the Part 1 integral which is your original integral must also be zero.

In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $1/z$ as $z\to\infty$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.