Taylor Series of a Complex Function with a Singular Point

Question

I need to find the Taylor series of the function $e^{\frac{1}{1-z}}$ at $0$. Clearly, this function has a singular point at $z = 1$. I also know that the Taylor series representation for the function $e^z$ at $0$ is $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$.

I was studying this example here:

$e^{1/z}$ and Laurent expansion

and from this example I see that we are looking at a deleted neighborhood around the singular point. So we consider the series to be analytic outside the deleted neighborhood and hence is defined outside the deleted neighborhood.

I am finding this topic to be rather confusing, so I want to make sure that I am understanding this concept correctly.

You say you're looking for the Taylor series at $0$ but what you've written down is a Laurent series at $1$. Which one are you looking for? — Qiaochu Yuan, Oct 10 '20 at 03:28
@QiaochuYuan I obtained that series by substituting $1/(1-z)$ into the Taylor series of $e^z$ at $0$. — , Oct 10 '20 at 12:42

score 0 · Accepted Answer · answered Oct 10 '20 at 02:57

We say a function $f:U \to \Bbb C$ is holomorphic(analytic) at a point $p \in U$ if the there exists a power series $\sum_{k=0}^{\infty} a_k(x-p)^{k}$, that is convergent in a neighborhood of radius $r$ of $p$ in side $U$.

In your situation, to be more precise, you can write down the Taylor series of $e^{z}$ in a neighbor of $1$ and then substitute it with $\frac{1}{1-z}$. Outside the punctured $\Bbb C$, the series is convergent. So This series is convergent in an annulus of outer radius $\infty$ and inner radius $0$. And by the uniqueness of the Laurent series, it is the desired series we were looking for.

Taylor Series of a Complex Function with a Singular Point

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