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Suppose $f(z)$ and $g(z)$ are entire functions such that $(f\circ g)(z)$ is a non constant polynomial. Show that both $f(z)$ and $g(z)$ are polynomials.

I tried by contradiction: Ruling out $f(z)$ and $g(z)$ one by one by assuming not a polynomial. And the hint given in the book is to use the Casorati-Weierstrass Theorem. "The analytic function with isolated singularity at $a$ maps a punctured neighbourhood of $a$ to a dense set in $\mathbb{C}$

I am not sure I am on the right way or not?

I need help. Please some one suggest me.

Hagen von Eitzen
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    You are on the right way. If $f\circ g$ is a non-constant polynomial, then $(f\circ g)(z_n) \to \infty$ for all sequences $z_n\to \infty$. And if $h$ has an essential singularity at $\infty$, then there is a sequence $w_n\to \infty$ with $h(w_n) \to 0$ by Casorati-Weierstraß. Set $h = g$ to show $g$ must be a polynomial, then $h = f$ to show that $f$ also must be one. – Daniel Fischer Oct 26 '14 at 15:58
  • Thank you , I got this idea. – 101672964 Oct 26 '14 at 16:09
  • entire = holomorphic ? – idm Oct 26 '14 at 16:14
  • @idm "entire = holomorphic on all of $\mathbb{C}^n$ (here, $n = 1$)" (Note that one also speaks of entire meromorphic functions or entire harmonic functions, but without further qualification, it means "entire holomorphic" in every instance I have come across so far.) – Daniel Fischer Oct 26 '14 at 16:23

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We can use the following fact:

An entire function is polynomial if and only if $\infty$ is not an essential singularity.

By assumption, $\infty$ is not an essential singularity of $f\circ g$. Hence by Casaroti-Weierstrass, for a suitable neighbourhood $U=\{\,z\in \mathbb C:|z|>r\,\}$ of $\infty$, the image $f(g(U))$ is not dense in $\mathbb C$, say it avoids a disk $B(z_0,\rho)$. If $g(U)$ is dense, this implies that $f(\mathbb C)$ avoids $B(z_0,\rho)$ as well, $\frac1{f(z)-z_0}$ is bounded, hence constant, hence $f$ finally and $f\circ g$ are constant, contradiction. We conclude that $g(U)$ is not dense, i.e. $\infty$ is not an essential singularity of $g$, so $g$ is a nonconstant polynomial. Consequently, $g(U)$ is a neighbourhood of $\infty$ and its image under $f$ is not dense, i.e. $f$ has no essential singularity at $\infty$ so that $f$ is polynomial.

Hagen von Eitzen
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