# Evaluate the integral $\int\exp(\frac{z^2-3}{z^3})$

evaluate the integral $\int\exp(\frac{z^2-3}{z^3})$ the unit circle oriented counterclockwise.

I think that we use the following theorem:

Suppose that $f$ is holomorphic in an open set containing a circle $C$ and its interiror, except for poles at the points $z_1,...,z_N$ inside $C$. Then $$\int_{C}f(z)dz=2\pi i\sum_{k=1}^{N}Res(f,z_k)$$

we have only pole $z_0=0$ for $f(z)=\exp(\frac{z^2-3}{z^3})$ and

$$Res(f,0)=\frac{1}{2}\frac{d^2}{dz^2}\lim_{z\to0}z^3f(z)$$

Is there another way to evaluate this integral? Thanks.

Hint: Several ways to go, but you may e.g. write: $$\exp (\frac{z^2-3}{z^3}) = \exp(\frac{1}{z}) \exp(\frac{-3}{z^3})$$ Do a Laurent series expansion (if you know about those?) and pick up the coefficient to $1/z$. [it is in fact quite simple]