Evaluate complex integral: $\int_{\left| z \right| =2} \frac{1}{z^{741} +1}dz$ I want to evaluate the integral $$\int_{\left| z \right| =2} \frac{1}{z^{741} +1}dz.$$ It is clear that all singularities of this function are contained in the region of integration. Therefore, the residue theorem would give us that $$\int_{\left| z \right| =2} \frac{1}{z^{741} +1}dz = 2\pi i  \sum_{k=1}^{741} \text{Res}_{z_k}.$$ I can't calculate the residues however, can someone assist me? 
 A: At any rate, replacing the path of integration by a larger circle $|z|=r$, and then letting the radius $r$ tend to infinity, the triangle inequality for complex line integrals shows that the integral tends to zero. On the other hand, the integral (as a function of $r$) is constant by Cauchy's theorem (because the singularities of the integrand are on the unit circle $|z|=1$), so the integral is zero (for any $r>1$, actually).
A: Actually, using residues is not the best way to deal with this problem, but it can be done. Since I don't want to keep writing $741$, I'll just denote it by $n$. Actually, $n$ can be any natural greater than $1$.
The singularities of $\frac1{z^n+1}$ are the $n^\text{nh}$ roots of $-1$, which are the numbers of the form $\exp\left(\frac{k\pi i}n\right)$, with $k\in\{1,3,5,\ldots,2n-1\}$. The residue of $\frac1{z^n+1}$ at this point is$$\frac1{n\exp\left(\frac{k\pi i}n\right)^{n-1}}=\frac1n\exp\left(-\frac{n-1}n\pi i\right)^k$$and therefore\begin{align}\int_{|z|=2}\frac1{z^n+1}\,\mathrm dz&=2\pi i\sum_{k=1}^n\frac1n\exp\left(-\frac{n-1}n\pi i\right)^k\\&=\frac{2\pi i}n\cdot\frac{\exp\left(-\frac{n-1}n\pi i\right)-\exp\left(-\frac{n-1}n\pi i\right)^{2n+1}}{1-\exp\left(-\frac{n-1}n\pi i\right)}\\&=0.\end{align}
A: Let $\theta_{r,y}=\sin^{-1}\left(\frac{y}{r}\right)\sim\frac{y}{r}$ and $x_{r,y}=\sqrt{r^2-y^2}\sim r$.
Since there are no singularities outside the circle of radius $2$, the integral along the contour
$$
\overbrace{2e^{i\left[\theta_{2,\epsilon},2\pi-\theta_{2,\epsilon}\right]}}^{\substack{\text{counterclockwise}\\\text{along $|z|=2$}}}\cup\overbrace{\left[x_{2,\epsilon}-i\epsilon,x_{R,\epsilon}-i\epsilon\right]\vphantom{e^{\left[2\pi-\theta_{2,\epsilon}\right]}}}^\text{left to right along $[2,R]$}\cup\overbrace{Re^{i\left[2\pi-\theta_{R,\epsilon},\theta_{R,\epsilon}\right]}}^{\substack{\text{clockwise}\\\text{along $|z|=R$}}}\cup\overbrace{[x_{R,\epsilon}+i\epsilon,x_{2,\epsilon}+i\epsilon]\vphantom{e^{\left[2\pi-\theta_{2,\epsilon}\right]}}}^\text{right to left along $[2,R]$}
$$
will be $0$. Since the integrals along $[2,R]$ cancel, this means that the counterclockwise integral along $|z|=2$ is equal to the counterclockwise integral along $|z|=R$, and therefore,
$$
\begin{align}
\left|\,\int_{|z|=2}\frac{\mathrm{d}z}{z^{741}+1}\,\right|
&=\left|\,\lim_{R\to\infty}\int_{|z|=R}\frac{\mathrm{d}z}{z^{741}+1}\,\right|\\
&\le\lim_{R\to\infty}\frac{2\pi R}{R^{741}-1}\\[6pt]
&=0
\end{align}
$$
