easier way to find an integral?

Question

Is there an easy way for : $$\int \frac{e^{\frac{1}{z}-2}}{z+2}dz$$ on$|z|=3$ ? thanks for any hint.

(I try it by series expansion , but someone told me there is an easy way to find it )

Answer 1

The integral is $i 2 \pi$ times the sum of the residues at the poles inside $|z|=3$, i.e., at $z=-2$ and $z=0$.

At $z=-2$, we have a simple pole and the residue is $e^{-1/2-2}$.

At $z=0$, we have an essential singularity. Here, we take the Laurent expansion of the integrand at $z=0$ and look for the coefficient of $z^{-1}$. Note that

$$\frac1{2+z} = \frac12 \frac1{1+\frac{z}{2}} = \frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2^k} z^k $$

Further,

$$e^{1/z} = \sum_{m=0}^{\infty} \frac1{m! z^m} $$

Thus the residue is

$$\frac1{e^2}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{k+1} (k+1)!} = -\frac1{e^2}\sum_{k=1}^{\infty} \frac{(-1)^k}{2^k k!} = \frac1{e^2} \left (1-\frac{1}{\sqrt{e}}\right )$$

Thus, the integral is

$$\oint_{|z|=3} dz \frac{e^{1/z-2}}{z+2} = i 2 \pi \left (\frac1{e^2}-\frac{1}{e^2 \sqrt{e}} + \frac1{e^2 \sqrt{e}} \right ) = \frac{i 2 \pi}{e^2}$$