0
$\begingroup$

I'm confused about how we are supposed to use Laurent expansion of a function to tell types of its singularity.

For example, the book says if $f(z)$'s Laurent expansion about $z_0$ has only finite terms in the principal part, then $z_0$ is a pole.

However, if we expand $\frac{1}{1-z}$ around $z=0$ on the annulus $|z|>1$, we will have infinite terms: $\sum_{n=0}^{\infty}-\frac{1}{z^n}$. The book then moves on to say $z=1$ is an essential singularity which is apparently wrong!

So does the typical "Laurent expansion and singularity relation" only holds when we expand function around an isolated singularity?

$\endgroup$
0
$\begingroup$

Your Laurent series expansion of $\displaystyle \frac{1}{1-z}$ around $z=0$ is correct which you have written.

$\bullet$ But , at $z=0$ , $\displaystyle \frac{1}{1-z}$ has NO singularity, it is analytic at $z=0$ ; so the question of any type singularity does not arise.

$\bullet$ $z=1$ is a pole if $\displaystyle \frac{1}{1-z}$ , NOT an essential singularity ; because Laurent series expansion of $\displaystyle \frac{1}{1-z}$ around $z=1$ is itself $\displaystyle \frac{1}{1-z}=-\frac{1}{z-1}$. Clearly it has only one term in principal part , around z=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.