Residue of $e^{1/z}$?

Question

I have a function that diverges like $e^{1/z}$ in the origin, and I'm integrating the function a closed contour in the complex plane. The point $z=0$ lies inside the contour, so if I integrate by residues the singularity at $z=0$ (a pole?) must be taken into account. How would you do that?

Answer 1

Hint. It's neither a pole nor a removable singularity. Write the Laurent series expansion of $e^{1/z}.$ This is a classic example where a function has an essential singularity (at $z=0.$)

Answer 2

$$e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdot\cdot\cdot, -\infty<z<\infty$$

This may also be written in the following manner.

$$e^z=\sum_{j=0}^\infty \frac{{z}^j}{j!}$$

Now, for $f(z)=e^{\frac{1}{z}}$ (which happens to have a singularity at $z=0$) we can write:

$$e^{\frac{1}{z}}=\sum_{j=0}^\infty \frac{(\frac{1}{z})^j}{j!}=\sum_{j=0}^\infty \frac{1^j}{j!\cdot z^j}$$

We know the residue would be the coefficient of the term containing the $-1^{th}$ power of $(z-z_o)$.

Coefficent of $j^{th}$ term would be:

$$\frac{1^j}{j!}$$

Clearly, the term we're looking for can be obtained at $j=1$.

So,

$$Res(e^{\frac{1}{z}})=1$$