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?


migrated from mathoverflow.net Oct 1 '17 at 4:30

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  • $\begingroup$ If it looks like homework ... and it quacks like homework ... $\endgroup$ – Michael Renardy Oct 1 '17 at 1:54
  • 2
    $\begingroup$ @Jamai-Con: I used to pose this problem on my complex analysis final exams. One can also argue on the Riemann sphere: then the only singularity is at infinity, and the residue there can be seen to vanish. $\endgroup$ – GH from MO Oct 1 '17 at 2:14
  • $\begingroup$ Consider the map $z \to e^{2\pi i /741} z$. $\endgroup$ – anomaly Oct 1 '17 at 7:13
  • $\begingroup$ An elementary version of GH from MO's explanation: Apply the substitution $z = 1/w$ and notice that (1) this maps the CCW-oriented circle $|z| = 2$ to CW-oriented circle $|w| = 1/2$ and that (2) $dz = -dw/w^2$. So we have $$ \int_{|z| = 2} \frac{1}{z^{741} + 1} \, dz = \int_{|w| = 1/2} \frac{w^{739}}{w^{741} + 1} \, dw = 0. $$ $\endgroup$ – Sangchul Lee Oct 1 '17 at 7:23

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).


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}


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, $$ \begin{align} \left|\,\int_{|z|=2}\frac{\mathrm{d}z}{z^{741}+1}\,\right| &=\left|\,\lim_{R\to\infty}\int_{|z|=R}\frac{\mathrm{d}z}{z^{741}+1}\,\right|\\ &\le\lim_{R\to\infty}\frac{2\pi R}{R^{741}-1}\\[6pt] &=0 \end{align} $$


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