Compute the following integral, where $C$ is the circle $|z|=3$

Question

Evaluate:$$\int_{C} (1 + z + z^2)(e^\frac{1}{z}+e^\frac{1}{z-1}+e^\frac{1}{z-2}) dz $$ where $ C$ is a circle $|z|=3$ and $z \ \epsilon \ \mathbb{C}$

The function that is being integrated has singularities at $0, \ 1, and \ 2$

Therefore I know we must consider the residue.

Answer 1

Only contributions from the $z^{-1}$, $(z-1)^{-1}$, $(z-2)^{-1}$ terms contribute.

$$\oint_C dz \, (1+z+z^2) e^{1/z} = \sum_{k=0}^{\infty} \frac1{k!} \oint_C dz \, (1+z+z^2) \frac1{z^k} = i 2 \pi \left (1+\frac1{2!}+\frac1{3!} \right ) = i \frac{10 \pi}{3}$$

Rewrite the quadratic as

$$z^2+z+1 = (z-1)^2 + 3(z-1) + 3 \implies \\\oint_C dz \, (1+z+z^2) e^{1/(z-1)} = i 2 \pi \left (3+\frac{3}{2!}+\frac1{3!} \right )=i \frac{28 \pi}{3}$$

$$z^2+z+1 = (z-2)^2 + 5(z-1) + 2 \implies \\\oint_C dz \, (1+z+z^2) e^{1/(z-2)} = i 2 \pi \left (2+\frac{5}{2!}+\frac1{3!} \right )=i \frac{28 \pi}{3}$$

The integral is then the sum of these terms, or $i 22 \pi$.

Answer 2

More short way:

Let , $f(z)=(1 + z + z^2)(e^\frac{1}{z}+e^\frac{1}{z-1}+e^\frac{1}{z-2})$.

Find $$Res(f,\infty)=-\lim_{z\to 0}\frac{1}{z^2}f(\frac{1}{z})$$

Then $$\int_{|z|=3}f(z)\,dz=2\pi i[Res(f,0)+Res(f,1)+Res(f,2)]$$

$$=-2\pi iRes(f,\infty)$$