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Let $z_0\in\mathbb C$, $f$ a function having an essential singularity at $z_0$ and $P$ a non-constant polynomial. Show that the composite $P\circ f$ has an essential singularity at $z_0$.

I tried to solve it looking at Laurent series expansion. Let $f(z)=\sum_{-\infty}^{\infty}a_n (z-z_0)^n$ the Laurent expansion of $f$ for $0<|z-z_0|, for some $r>0$. Let $P(z)=\sum_{n=0}^{n=M}b_nz^n$ the polynomial. So we get $(P\circ f)(z)=b_0+b_1 f(z)+\ldots +b_M(f(z))^M$ I think the RHS is well defined, since sums, products and powers of power series are well defined. Now I would like to show that RHS contains an infinite number of negative powers of $(z-z_0)$, but i don't know the way.

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    Per dire la verità, non ho saputo resistere alla tentazione di scrivere $q$ualche parola nella Sua bellissima lingua (e ho cancellato il commento precedente che non serve più a niente)2012-11-22

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We can extend your question to any non-constant entire function $h(z)$ (instead of a polynomial $P$). From the Casorati-Weierstrass Theorem, we know that for every $w\in\mathbb{C}$ there is a sequence of numbers $z_n$ such that $\lim_{n\rightarrow\infty}z_n=a$ and in addition: $\lim_{n\rightarrow\infty}f(z_n)=w$ Because $h$ is non-constant, we can pick $w_1,w_2$ such that $h(w_1)\neq h(w_2)$, and because of the previous statement we can pick sequences $z_n\rightarrow a,w_n\rightarrow a$ such that: $\lim_{n\rightarrow\infty}f(z_n)=w_1,\lim_{n\rightarrow\infty}f(w_n)=w_2$ And therefore, we get: $\lim_{n\rightarrow\infty}h\circ f(z_n)\neq \lim_{n\rightarrow\infty}h\circ f(w_n)$ However, if $a$ was a removable singularity of $h\circ f$ then these two limits would have to be equal, and if $a$ was a pole then these two limits would have to be infinity. Therefore, by the classification of isolated singularities, $a$ must be an essential singularity of $h\circ f$.

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Use Casorati-Weierstrass Theorem

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    You're welcome. If you need some more help with that, just ask here again.2012-11-21
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i argued as follows: by fundamental theorem of algebra i get that every non constant complex polynomial is surjective. Moreover, every polynomial is obviously continous. By general topology, we can characterize dense subsets of a topological space as those subsets which intersect non trivially any non empty open subset. Now, if $f:X\rightarrow Y$ is a continous surjective map of topological space, then it maps dense subsets of $X$ to dense subsets of $Y$, since: take $E$ dense in $X$,take $B$ open non-empty in $Y$, then $f^{-1}(B)$ is open (by continuity)and non-empty (by surjectivity) in $X$, hence $f^{-1}(B)\cap E$ is non-empty by the density of $E$ in $X$, hence $B\cap f(E)$ is non-empty. Applying this to $f=P$, the polynomyal, i get that $(P\circ f)(D(z_0,\epsilon)-\{z_0\})$ is dense in $\mathbb C$. Now, i know that if $z_0$ is essential, then $(P\circ f)(D(z_0,\epsilon)-\{z_0\})$ is dense in $\mathbb C$, but now i should prove the reverse. I have that $(P\circ f)(D(z_0,\epsilon)-\{z_0\})$ is dense in $\mathbb C$ and i must prove that $z_0$ is essential. It's definitely clear that $z_0$ cannot be removable, and intuitively i can understand it is not a pole, but i can't exclude this second case with a formal proof.

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    +1 for giving a complete proof, including the fact that under a *surjective* continuous map dense sets are sent onto dense sets.2012-11-22