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Question:

Let $f(z)$ be analytic in an open set $G\subset \mathbb{C}$ except for a pole at $z_0\in G$. $g(z)$ be an entire function that is not a polynomial. how to show that $g(f(z))$ has an essential singularity at $z=z_0$?

My work: I tried to prove by contradiction from that if it is a pole, then $\forall \delta>0 \exists R>0$ such that $\{w:|w|>R\}\subset f(\{z\in\mathbb{C}:|z-z_0|<\delta\})$.

If it is a removable singularity, then $\lim_{z\to z_0}g(f(z))=c$ for some $c\in\mathbb{C}$. If it is a pole, then it is a removable singularity of $\frac{1}{g(z)}$, then I cannot find the contradiction come from?

Could anyone kindly help? Thanks!

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2 Answers 2

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You can expand $f(z)$ in a Laurent series around $z = z_0$, this must be of the form \begin{equation} f(z) = \frac{a_{-n}}{(z - z_0)^n} + \frac{a_{-n+1}}{(z - z_0)^{n-1}} + ... \end{equation} Since $g(z)$ is an entire function that is not a polynomial, we can expand it as a Taylor series around any point, let's say $z = 0$, we have \begin{equation} g(z) = \sum_{k = 0}^{\infty}b_k z^k. \end{equation} Then we can see that $g(f(z))$ can be written as a Laurent series around $z = z_0$ with infinite principal part \begin{equation} g(f(z)) = \sum_{k=0}^\infty b_k (\frac{a_{-n}}{(z - z_0)^n} + \frac{a_{-n+1}}{(z - z_0)^{n-1}} + ...)^k = \sum_{k=-\infty}^{-1}c_k(z - z_0)^k + ... \end{equation} This is because we cannot find $N > 0$ such that $c_k = 0, \forall k < -N$. For each $k$ term we will always get a pole of order $kn$ from expanding $(\frac{a_{-n}}{(z - z_0)^n} + \frac{a_{-n+1}}{(z - z_0)^{n-1}} + ...)^k$ and $k$ running from $0$ to $\infty$ in the summation. Hence an essential singularity at $z = z_0$.

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  • $\begingroup$ Thank you for the answer! Could you help explain how to combine the two series and get the Laurent series $g(f(z))=\sum_{k=-\infty}^{-1}c_k(z-z_k)^k+...$? $\endgroup$
    – Sherry
    Aug 12, 2016 at 1:06
  • $\begingroup$ I have added some more explanation. Does it make more sense now? $\endgroup$
    – user113988
    Aug 12, 2016 at 3:10
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Suppose that function $f(z)$ has the only pole in $z_0$ of degree $n$. Then in some local holomorphic coordinates $\{w\}$ we can write our function as $\frac{1}{w^n}$ where $w(z_0)=0$. The composition of $f$ with $g$ in these coordinates looks very simply: $\frac{1}{g^n}$. As the only entire functions which don't have an essential singularity in $\infty$ are rational functions, $g$ - does have an essential singularity in $\infty$ and so is $g(\frac{1}{w^n})$

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