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Problem: Suppose that $f:D(z_o,R)\to C$ is analytic and has a zero of order $m\ge1$ and that $g:D(z_o,R)\to C$ is analytic and has a pole of order $l\le m$ at $z_o$. Prove that $fg$ has a removable singularity at $z_o$.

My approach: In order to show that $fg$ has a removable singularity at $z_o$, we can show that $fg$ is bounded as $z \rightarrow z_o$, or $|f(z_o)g(z_o)|\lt n \in N$. But because $f(z_o)=0$ with order $\ge 1$, we only have to show that $g(z)$ does not approach infinity as $z \rightarrow z_o$.

My question: How do I show that $g(z)$ is bounded as $z \rightarrow z_o$? I know that $g$ has a pole of order $l\le m$ at $z_o$, so $ \frac{1}{g} $ has a zero of order $l\le m$ at $z_o$. It'd be easy to show that $ \frac{1}{g} $, and therefore $ \frac{f}{g} $, has a removable singularity, but I'm stuck on proving it for $fg$.

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  • $\begingroup$ Can you not just use a Laurent series for each of the functions at $x_0$? $\endgroup$ – Zach L. Oct 22 '12 at 2:07
  • $\begingroup$ You mean g is analytic on the punctured disc with $z_o$ removed, right? $\endgroup$ – BobaFret Oct 22 '12 at 2:22
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I'd try to use the following result:

$f$ has a removable singularity at $z=a$ if and only if $\lim_{z \to a} (z-a)f(z)=0$.

Since $f$ has a zero of order $m$ at $z=z_o$ it can be expressed as $f(z)=(z-z_o)^m \alpha (z)$; $g$ has a pole order $l$ at $z=z_o$ so it can be expressed as $g(z)=\frac{ \beta (z)}{(z-z_o)^l}$.

Then $f(z)g(z)=(z-z_o)^{m-l} \alpha(z) \beta (z)$, multiply by $z-z_o$, and then take a limit.

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  • $\begingroup$ So if i have $\frac{f(z)}{(z-a)^n}$ with f having zero @ a of order n then $z=a$ is a removable singularity because of the above?? $\endgroup$ – d13 May 6 '13 at 18:43

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