Laurent series for $\frac{1}{\sin^2(z)}$
by Hidenori
Proposition
Find the terms $c_nz^n$ in the Laurent series for $\frac{1}{\sin^2(z)}$ centered at $z = 0$ for $-4 \leq n \leq 4$.
Solution
Example 8.23 [A first course in complex analysis] has the Laurent series for $\frac{1}{\sin(z)}$. By simply squaring it, we obtain
\[\begin{align*} c_i &= \begin{cases} 1 & (i = -2) \\ 1/3 & (i = 0) \\ 1/15 & (i = 2) \\ 2/189 & (i = 4) \\ 0 & \text{(otherwise)}. \end{cases} \end{align*}\]Subscribe via RSS