I'd like to prove that $f(z)=z^3 \cot(z)$ is analytic in $0$. I know that $\cot(z)$ is not analytic in $\pm n \pi$ (and so in $0$), but this doesn't imply $z^3 \cot(z)$ being not analytic in $0$.
Everything I tried resulted in a very complicated combination of trigonometric and hyperbolic functions. I tried to show that $f(z)$ is continuous in $0$, that is differentiable in $0$, that the Cauchy-Reimann equation applies and that the partial derivatives of the real and imaginary part are continuous in $0$.
How to do that in a doable way?
From "Advanced engineering mathematics 10th edition - Kreyszig", "Problem Set 14.2", exercise 19, page 659. The answer is "yes", that is the Cauchy's integral theorem applies inside the unit circle, that is $f(z)$ is analytic there, and so the line integral around the unit circle is $0$.
Thanks, Luca
From the Bernoulli numbers generating function : $$ \frac{z}{e^z -1} = \sum_{n=0}^\infty \frac{B_n}{n!} \, z^n \text{ valid for } \vert z\vert <2\pi $$ It follows that for $\vert z\vert < \pi$ $$ z\cot z = \sum_{n=0}^\infty \frac{(-1)^{n} 2^{2n}B_{2n}}{(2n)!}\,z^{2n} $$ Then your function $z\mapsto z^3\cot z$ has a convergent Taylor series inside the open disk $D(0,\pi)$ , thus it's analytic inside the unit circle.