Multiplication of a complex function with essential singularity with another complex function with a pole at the same point

Question

Im trying to proof or disprove the following claim:

If $f(z)$ and $g(z)$ are holomorphic in an annulus $0 < |z − z(\beta)| < R$ and $f$ has an essential singularity at $z(\beta)$ and $g$ has a pole of order $n$ at $z(\beta)$, then $f(z) · g(z)$ has an essential singularity at $z(\beta)$.

The only thing I can think of is that if I multiply the power serieses (Laurent serieses) around $z(\beta)$ of both $f$ and $g$ then I would still have an infinite number of cooeficients. Is is sums the whole claim? because it seems to simple, am I missing something?

Thanks

Answer 1

If $g(z)$ has a pole of order $n$ at $z(\beta)$, then $(z-z(\beta))^n g(z)$ is holomorphic and nonzero near $z(\beta)$. Now suppose the claim is false. Then $f(z) g(z)$ has at worst a pole of order (say) $m$ at $z(\beta)$, which means that $(z-z(\beta))^m f(z) g(z)$ extends smoothly across $z(\beta)$. Dividing by the nonzero smooth function $(z-z(\beta))^n g(z)$ tells you that $(z-z(\beta))^{m-n} f(z)$ extends smoothly across $z(\beta)$; i.e. $f(z)$ must have only a pole, contradiction.

Answer 2

The difficulty I can see is in proving that there are infinitely many coefficients (with negative exponent) in the product $fg$ that do not cancel out.

I would prove the statement it by contradiction. Suppose that the product $h=fg$ does not have an essential singularity at $\beta$, so it is holomorhpic or has a pole of order $k$ at $\beta$. We also know that $1/g$ has a root of degree $n$ at $\beta$. Then the product of the two Laurent series, $f=h\cdot \frac1g$ does not contain any term with degree below $n-k$, so $f$ has at most a pole, contradicting that $f$ has an essential singularity.

Answer 3

WLOG $z(\beta)= 0.$ Fact: If $h$ is analytic in some $\{0<|z|<r\},$ then $h$ has an essential singularity at $0$ iff

$$\limsup_{z\to 0} |z|^p|h(z)| = \infty$$

for every $p>0.$ So now take the given $f,g$ and let $p> 0.$ Because $g$ has a pole at $0,$ $|g(z)|\ge 1$ near $0.$ So for any $p>0,$

$$\limsup _{z\to 0}|z|^p|f(z)g(z)| \ge \limsup _{z\to 0}|z|^p|f(z)| = \infty.$$

Hence $fg$ has an essential singularity at $0.$