[Complex Analysis] Singularity in product of analytic functions

[ludwig1]

Suppose f,g:ℂ→ℂ are analytic with singularities at z=0. I was wondering whether f(z)^2 or f(z)g(z) will have a singularity at z=0? For each, can you give me a proof or a counterexample?

 

[S.G. Janssens]

Suppose f,g:ℂ→ℂ are analytic with singularities at z=0. I was wondering whether f(z)^2 or f(z)g(z) will have a singularity at z=0? For each, can you give me a proof or a counterexample?

I think the most staightforward (but perhaps not the most handy?) approach would be to just consider Laurent series for $f$ and $g$ on the same annulus centered at the origin. What do you know about the coefficients in the product of two such series?

Assuming that "singularity" means either "pole" or "essential singularity", you could start by considering the case that $f$ and $g$ have poles of (finite) orders $n$ and $m$ in $\mathbb{N}$, respectively. Next, you could look at the case that one or both of $f$ and $g$ have essential singularities (i.e. infinitely many non-zero terms occur in the principal parts of their Laurent series).

 

[math771]

Yes, both f(z)^2 and f(z)g(z) will have singularities at z=0. This can be seen by considering the definition of a singularity - a point at which the function is not defined or becomes infinite.

For f(z)^2, we can see that at z=0, f(z)^2 will become undefined if f(0)=0, as any real number squared is still defined, but any complex number squared will have an imaginary component. Therefore, f(z)^2 will have a singularity at z=0.

For f(z)g(z), we can consider the case where f(z)=1/z and g(z)=z. Both of these functions have singularities at z=0, and when multiplied together, f(z)g(z)=1, which is defined at z=0. However, if we take the limit as z approaches 0, we can see that the function will become infinite, indicating a singularity at z=0.

In general, if both f(z) and g(z) have singularities at z=0, then any function formed by combining them (such as f(z)^2 or f(z)g(z)) will also have a singularity at z=0. This can be proven using the definition of a singularity and the properties of analytic functions.

Therefore, both f(z)^2 and f(z)g(z) will have a singularity at z=0.