Evaluate the line integral $\oint_C \frac{e^z}{(z+1)^2}\, dz$

Question

$$\oint_C \frac{e^z}{(z+1)^2}\, dz,\quad C: |z-1|=3$$

My attempt: I think that this integral $\neq 0$ because the curve is a disc. If it were a circle, then by Cauchy's Integral Theorem this integral would be $=0$. I think the next step would be to find a way to describe this disc in term of a parameter $t$. But I'm not sure.

Answer 1

The symbol $\oint_C$ is not referring to a surface integral. We're not integrating over a disc. In the context of your question, $\oint_C$ is referring to a closed path integral along the path named $C$.

Your integrand has a singularity only at $z=-1$, which is inside the region bounded by $C$. If you have yet to learn the Residue Theorem, you may apply Cauchy's Integral formula instead:

$$\oint_C \frac{e^z}{(z+1)^2}\, dz = \frac{2\pi i}{1!}\frac{d}{dz}e^z\bigg|_{z=-1}= 2\pi ie^{-1}$$

Answer 2

Observe that the only singularity of $\;f(z)=\frac{e^z}{(z+1)^2}\;$ within the domain enclosed by the given contour is a double pole at $\;z=-1\;$ , so

$$\oint_{|z-1|=3}f(z)\,dz=2\pi i\cdot\text{Res}(f)|_{z=-1}\;,\;\;\;\text{and}$$

$$\text{Res}(f)|_{z=-1}=\frac d{dz}\left((z+1)^2f(z)\right)'=\left.e^z\right|_{z=-1}=\frac1e$$

so

$$\oint_{|z-1|=3}f(z)\,dz=\frac{2\pi i}e$$