Laurent Series for $\frac{e^z}{\sin^2(z)}$ centered at the origin.

Question

I want to find the coefficient for $z^{-1}$ in the Laurent series for $$f(z)=\frac{e^z}{\sin^2(z)}$$ centered at $0$, but I am struggling a bit.

Answer 1

First of all, $e^z = 1 + z + O(z^2)$. Next, since $\sin z = z + O(z^3)$, $\sin^2 z = z^2 + O(z^4)$. Hence $${1 \over \sin^2 z} = {1 \over z^2(1 + O(z^2))} = {1 \over z^2}(1 + O(z^2))$$ Thus $${e^z \over \sin^2 z} = {1 \over z^2}(1 + z + O(z^2))(1 + O(z^2))$$ $$ = {1 \over z^2}(1 + z + O(z^2))$$ $$= {1 \over z^2} + {1 \over z} + O(1)$$ The residue is the ${1 \over z}$ coefficient, which is $1$ here.

Answer 2

Note that $1/sin(z)$ has a pole of first order at $z_0 = 0$. Thus $1/sin^2(z)$ has a pole of order $2$ at $z_0 = 0$. Also note that $e^z$ is holomorphic on $C$. Thus $Res(f; 0) = c_1$ where $e^z = \Sigma_{n = 0}^{\infty}c_nz^n$. So you can compute $Res(f; 0) = c_1 = (e^z)'(0) = 1$.