Evaluate $\frac{e^{2z}}{1+e^{z}}$ at $ z=i\pi$ by residue

Question

I'm having a little trouble figuring out how to do this question. I have the residue theorem which states $$\text{R}(i\pi) = \lim_{z\to i\pi}\big((z-z_0)f(z) \big)$$ but I'm having trouble working it out from there, as the residue I keep getting is 0. Any help is appreciated!

Edit: We are supposed to evaluate it around a unit circle of radius $\dfrac{3}{2}$, center at the origin.

Answer 1

$R(i\pi)=\lim_{z\to i\pi} \frac {(z-i\pi)(e^{2z})}{1+e^z} = (\lim_{z \to i\pi} \frac {z-i\pi}{1+e^z})(\lim_{z\to i\pi} e^{2z}).$

Apply L'hospital on first limit from left.

Then you'll get $R(i\pi)=(-1)(1)=-1.$