Evaluate $\int_{\mathcal{C}}\frac{e^z}{z^2}\,\mathrm{d}z$

Question

How can I solve this problem using Cauchys integral formula $$\int_{\mathcal{C}}\frac{e^z}{z^2}\,\mathrm{d}z$$ When $$C:|z| = 1$$ I believe I should get something like this$$\int_{|z|=a} \frac{f(z)}{z-a}dz = 2\pi i f(a) $$ So it would end up looking like this $$2\pi ie^z(1)$$ But I dont know if thats right, please help me, Im trying to learn how to solve Cauchy Integrals.

Answer 1

You need to use the extending version: Suppose the curve $\mathcal C \subset \Omega \subset \mathbb C$ and let $f: \Omega \to \mathbb C$ be analytic. Then $$f^{(n)}(0)= \frac{n!}{2\pi i } \int_{\mathcal C} \frac{f(z)}{z^{n+1}} \, dz $$ for each $n=0,1,\dots$, see wiki. In your case, $f(z) = e^z$ and $n=1$, so $$\int_{\mathcal C} \frac{e^z}{z^2} \, dz=\frac{2\pi i }{1!} f'(0)=2\pi i. $$

Answer 2

If you expand the holomorphic function $z\to e^z$ around the origin $$ e^z=1+z+\frac{z^2}{2!}+... $$ and divide by $z^2$, you get the Laurent series $$ \frac{1}{z^2}+\frac 1{z}+\frac 1{2!}+\dots $$ The only piece of the series that contributes to the integral is $1/z$. The coefficient is $1$ so the answer is $2\pi i$.