# Thread: isolated singularity

1. ## isolated singularity

Locate each of the isolated singularities of the function:

(e^z - 1) / z

How do I do this? Thanks in advance...

2. The answer is very easy: the function...

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

... is analytic on the whole complex plane...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by jzellt
Locate each of the isolated singularities of the function:

(e^z - 1) / z

How do I do this? Thanks in advance...
z = 0 is a removable singularity: Removable singularity - Wikipedia, the free encyclopedia

4. Originally Posted by chisigma
The answer is very easy: the function...

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

... is analytic on the whole complex plane...

Kind regards

$\chi$ $\sigma$

I'm new to this stuff, but I wanted to check myself. I think it goes something like this:

$f(z)$ clearly has a singularity at $z=0$. I mean, it's undefined there so $f(z)$ itself is not what is holomorphic, right? But because 0 is a removable singularity it is possible to define a function $\hat{f}(z)$ where the new function is everywhere equal to the original except at the singularity, where we can define $\hat{f}(0)$ to be the limit of $f(z)$ as $z\to 0$, which is legitimate because this limit exists and is finite. Have I understood this correctly?

Also, what is the justification that 0 is a removable singularity for this function? Is it, in fact, because $\lim_{z\to 0} f(z)$ exists and is finite? If so, would this need to be shown for completeness, or could it be taken as obvious enough?