I am sure the integral is zero, but I am not sure of the specific reason. Is it because the function has no pole? (if so, how do I demonstrate this?) Or is it because when we cross multiply the fractions, we get a numerator whose derivative is zero at z=0?
2026-03-29 16:02:52.1774800172
What is $\int_c {e^{z^2}}(1/{z^2} - 1/{z^3})$ when c is the unit circle?
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You need to work out the residue at $0$, i.e. the coefficient of $1/z$. Of course, $$\frac{e^{z^2}}{z^2}=\frac1{z^2}+1+\frac{z^2}2+\cdots$$ and $$\frac{e^{z^2}}{z^3}=\frac1{z^3}+\frac{1}z+\frac{z}2+\cdots.$$ Can you see what the residue of $e^{z^2}(1/z^2-1/z^3)$ is?