What are the possible expansions of f(z) = $\frac{e^{1/z}}{z}$ about z = i I am trying to find all the possible expansions of f(z) = $\frac{e^{1/z}}{z}$ about  = i, and I got something but I am not sure if my reasoning is right.
We know we have a singularity at z = 0, so we are going to have two domains.
The first one is |z| < 1, while the second one is |z| > 1.
If  understand it correctly, the first domain (inside the disk of radius 1), will be expanded with a Taylor series, due to the absence of singularities, while the second one (|z| > 1) will need a Laurent series to be computed, as we have a singularity at z = 0.
This is what I have done so far:
$$\frac{e^{1/z}}{z} = \frac{e^{-i}}{z} + \frac{e^{-i}(z-i)}{z} + \frac{(e^{-i}+2)(z-i)^2}{2z} + . . .$$
But I am not sure if that is correct, and I am not sure how to account for the different domains.
 A: Unfortunately, what you have written so far involves both factors of $z$ and $z-i$ and is therefore neither a Taylor series nor a Laurent series. It is a convergent sum of analytic functions, but not a specialized series. In general, a function can have many many representations as integrals, various types of series, expressions of other functions, etc. Taylor/Laurent series are special involving specifically terms of $a_k(z-c)^k$ for some chosen $c$; they cannot involve other factors.
Lets correct your domains. Your center is $i$ and your essential singularity is at $0$. Hence, your two domains are $|z-i|>1$ and $|z-i|<1$. Now that we know the Laurent domains, let us write out the Laurent series; in general, given ANY Jordan curve within an annulus we have,
$$f(z) = \sum_{k=-\infty}^\infty a_k\cdot(z-i)^k = \sum_{k=-\infty}^\infty\left(\frac{1}{2\pi i} \int_\gamma \frac{e^{1/w}/w}{(w-i)^{k+1}} dw\right)\cdot (z-i)^k$$
The simplification of the coefficients appears to be non-trivial involving series with incomplete gamma function calculations. We can certainly calculate them though; define a Jordan curve for some $r$ centered at $i$,
$$\gamma(t) = i+re^{i t},\ \ \ t\in[0,2\pi)$$
we then get with u-substitution,
$$a_k = \frac{1}{2\pi i} \int_\gamma \frac{e^{1/w}/w}{(w-i)^{k+1}} dw = 
\frac{1}{2 \pi} \int_0^{2\pi} \frac{e^{1/(re^{t i}+i)}}{\left(re^{t i}+i\right)\left(re^{i t}\right)^k} dt $$
going to wolframalpha, we can calculate these, if you let $0<r<1$ then these coefficients will exhibit a Taylor series, with values,
$$a_k=0\ (k<0),\ \ \ \ \ a_k=-ie^{-i},\ (1-i)e^{-i},\ \left(2+\frac{i}{2}\right)e^{-i},\ \ldots\ (k\ge 0)$$
if you let $r>1$ then these coefficients will exhibit a Laurent series, with values,
$$a_k=1,\ 1-i,\ -\frac{1}{2}-2i,\ \ldots\ (k<0),\ \ \ \ \ a_k=0\ (k\ge 0)$$
If we want better insight to why this happens, we can use the Laurent series of $e^{1/z}/z$ centered at $0$,
$$a_k = \frac{1}{2\pi i} \int_\gamma \frac{e^{1/w}/w}{(w-i)^{k+1}} dw = \frac{1}{2\pi i} \int_\gamma \frac{\sum_{n=0}^\infty \frac{1}{n!w^{n+1}}}{(w-i)^{k+1}} dw = \sum_{n=0}^\infty \frac{1}{n!} \left( \frac{1}{2\pi i} \int_\gamma \frac{1}{w^{n+1}(w-i)^{k+1}} dw \right) $$
This allows us to calculate each $a_k$ using a series of rational integrals. Moreover, we can see now that when $0<r<1$ the rational functions have one residue inside the Jordan curve, but when $r>1$ the rational functions have two residues inside the Jordan curve.
