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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.

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    By evaluate around that circle you mean find the contour integral? Because this value of z lies outside the contour. – 2017-01-31
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    This isn't an answer, but there's no reason a residue *can't* be 0... – 2017-01-31
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    @Triatticus is right here, the value lies outside the contour thus the contour equals 0. I didn't catch that. For solving with a circle of radius greater than $i\pi$ I managed to figure out too. Let $e^{2z} = g(z)$ and $1+e^{z} = h(z)$. The residue is simply $\frac{g(z_{0})}{h'(z_{0})} = -1$ and finding the contour from there is trivial. Thanks! – 2017-01-31

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

Apply L'hospital on first limit from left.

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