How do I find Laurent series of $f(z)=\frac{\operatorname{sh}(z)}{(z^2+\pi^2)^2}$ at $z=\mathrm i\pi$

Question

How do I find Laurent series of $f(z)=\dfrac{\operatorname{sh}(z)}{(z^2+\pi^2)^{2}}$ at $z=\mathrm i\pi$?

Series for $\operatorname{sh}(z)$ is obvious, but I'm really stuck with denominator part for now.

Answer 1

Hint: $1/(z^2+π^2)^2=(1/(2πi)(1/(z-πi)-1/(z+\pi i)))^2$.

Now note that $1/(z+πi)=1/(2πi+(z-πi))=1/(2πi)(1/(1+(z-πi)/(2πi))=1/(2πi)\sum(-1)^n((z-πi)/(2πi))^n$. You can use the Cauchy product for the square.

Once you're done squaring, multiply by the series for $\sinh$.

Answer 2

A bit of algebra should help you get started $$\frac{1}{(z^2+\pi ^2)^2}=\frac{a}{z+\pi i}+\frac{b}{(z+\pi i)^2}+\frac{c}{z-\pi i}+\frac{d}{(z-\pi i)^2}$$ $$\text {equation} (1)\quad1=a(z+\pi i)(z-\pi i)^2+b(z-\pi i)^2+c(z-\pi i)(z+\pi i)^2+d(z+\pi i)^2$$ Differentiate this equation. $$\text{equation} (2) \quad 0=a(z-\pi i)^2+2a(z+\pi i)(z-\pi i)+2b(z-\pi i)+c(z+\pi i)^2+2c(z-\pi i)(z+\pi i)+2d(z+\pi i)$$ Substitute $z=-\pi i$ in equation (1) to obtain $b.$ Substitute $z=-\pi i$ in equation (2) to obtain $a.$ Substitute $z=\pi i$ in equation (1) to obtain $d.$ Substitute $z=\pi i$ in equation (2) to obtain $c.$ You can take it from here.