Let $f(x)=\frac{z}{z^{31}-1}$ and $\Gamma = \{|z| = 2\}$. I'm asked to compute $\int_\Gamma f(z) \, dz$.
By the Cauchy Residue Theorem, we know that $\int_\Gamma f(z)\,dz = \sum_{a\in R} \operatorname{Res}(f,a)$, for $R$ the set of poles of $f$. We could just compute each residue, but is there another way that can make this proceeding shorter?
I thought of representing $f$ as a Laurent series like $g(z) = z^{32}-z^{63}+z^{97}-\cdots$ and since $g$ is holomorphic, by the Cauchy Theorem, the integral is $0.$ Is this alright? Or is there another way of doing it?
Best Answer
It is true that the integral is $0$ but not because of your reasoning: The power series only converges for $|z|<1$ and $f$ is not holomorphic in $|z|<2$.
One way of calculating the integral without computing the residues would be to note that $\int_\Gamma f(z) \, dz = \int_{\Gamma_R}f(z)\,dz$ where $\Gamma_R=\{|z|=R\}$ (with the same orientation), $R>1$. Then let $R\to\infty$.
Edit: Note that $|{f(z)}|=o(|z|^{-1})$ as $|z|\to\infty$, hence $$|\int_{\Gamma_R}f(z)\,dz|\leq \operatorname{length}(\Gamma_R)\max_{z\in\Gamma_R}|f(z)|=2\pi Ro(R^{-1})\to0$$ as $R\to\infty$. Since $\int_{\Gamma_R}f(z)\,dz$ is independent of $R$ (if $R>1$) this implies $\int_{\Gamma_R}f(z)\,dz=0$