The answer key says $\pi i$. By Cauchy's Integral Formula 5.1 for $f''(w)$, I think the answer is $2 \pi i$:
$$\int_{\square} \frac{e^{z^2}}{z^3} dz = \pi i \frac{d^2}{dz^2} e^{z^2}|_{z=0} $$ = $ \pi i (2) = 2 \pi i$.
I also tried on Wolfram Alpha 1 2 for circles which I believe are homotopic to the square. I think I get the same answer with Residue Thm 9.10 and Prop 9.11:
$$\int_{\square} \frac{e^{z^2}}{z^3} dz = 2 \pi i \frac{1}{2!} \lim_{z \to 0} \frac{d^2}{dz^2} (z-0)^3 \frac{e^{z^2}}{z^3}$$
$$ = \pi i (2)$$
What am I doing wrong?
Best Answer
The correct answer is $2\pi i$. I will suggest another method: expand the exponential in a power series and look at the coefficient of $\frac 1 z$. This shows that the residue at $0$ is $1$ so the integral is $2\pi i$.