Calculate the 3 first terms of the Laurent series for $f(z)=\displaystyle\frac{1}{z^2\sinh(z)}$ where $0<|z|<R$ and calculate the highest possible value for $R$.
I've figured out I can do the series expansion of the following using the Cauchy product: $z^2\sinh(z)f(z)=1$, and find out the terms $a_1,a_2,a_3...$ But I think it's too difficult.
The zeros of $sinh (z) $ are $\{k\pi i\}_{k \in \mathbb{Z}}$ so the highest possible value for $R$ is $\pi$.
To find the Laurent series note that $z = 0$ is a pole of the first order for $ g(z) = \frac{1}{sinhz} = \sum_{n =- 1}^{n =+ \infty} a_{n}z^{n}$. Then we have $$\sum_{n =- 1}^{n =+ \infty} a_{n}z^{n} \cdot sinh(z) = 1$$ and so after recalling the Taylor series for $sinh(z)$ we can calculate the $a_{n}$ term by term equating the coefficients. Now it's sufficient to multiply by $\frac{1}{z^{2}}$ to obtain the Laurent series of $f(z)$.
