If f has an essential singularity, does $P\circ f$ have the same type of singulaity?

Question

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 $0<|z-z_0|<r$, 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.

Answer 1

Use Casorati-Weierstrass Theorem

Answer 2

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.

Answer 3

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 $h\circ f$.