I have to calculate the residuo of $ f(z) $ in $ i\pi $ $$ f(z) = \frac{e^z +1}{sin^2(iz)} $$
I understood it's a 1st order pole, but honestly I'm having trouble calculating this limit:
$$ \lim_{z->i\pi} \frac{e^z +1}{sin^2(iz)} (z-i\pi) $$
Can somebody please help me?
$\lim_{z\to i\pi} \frac{(e^z + 1)(z-i\pi)}{\sin^2(iz)} = \lim_{z\to i\pi} \frac{e^z(z-i\pi) + e^z +1}{i\sin(2iz)}= \lim_{z\to i\pi} \frac{e^z(z-i\pi) + 2e^z}{-2\cos(2iz)} = 1$ by two applications of L'Hopital.
Change variable $z=i\pi+t$. Now $$e^z=-e^t$$ $$\sin(iz)=-i \sinh (t)$$ $$ \frac{e^z +1}{\sin^2(iz)} (z-i\pi)=\frac{1-e^t}{-\sinh ^2(t)}t=\frac{e^t-1}{\sinh ^2(t)}t$$ Now, using Taylor series around $t=0$ $$e^t=1+t+O\left(t^2\right)$$ $$\sinh(t)=t+O\left(t^2\right)$$