Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
grazie per la correzione –  Federica Maggioni Nov 22 '12 at 8:36
Per dire la verità, non ho saputo resistere alla tentazione di scrivere qualche parola nella Sua bellissima lingua (e ho cancellato il commento precedente che non serve più a niente) –  Georges Elencwajg Nov 22 '12 at 8:54
add comment

2 Answers 2

Use Casorati-Weierstrass Theorem

share|improve this answer
$z_0$ is essential singularity for $f$.Hence, by Casorati-Weierstrass $f(D(z_0,\epsilon))$ is dense in $\mathbb C$ for every $\epsilon>0$. Thus $(P\circ f) (D(z_0,\epsilon))$ is also dense in $\mathbb C$. Therrfore we have $lim_{z\rightarrow z_0}|f(z)|=\infty$ and so $z_0$ is essential for $P\circ f$. is it correct? –  Federica Maggioni Nov 21 '12 at 16:17
The last part isn't, it would mean $z_{0}$ is a pole. Also $z_{0}$ is not in the domain of $f$, since it's an isolated singularity, so you have to take the disc without the center. Casorati-Weierstrass Theorem says that essential singularitys are the only ones satisfying that property, so you are done. –  user50228 Nov 21 '12 at 16:31
yes, i was wrong, now i've understood, but, how can i prove that $f(0<|z-z_0|<\epsilon)$ dense in $\mathbb C$ implies $P(f(0<|z-z_0|<\epsilon))$ is also dense? –  Federica Maggioni Nov 21 '12 at 18:54
Use that $P$ is a continous function. –  user50228 Nov 21 '12 at 19:25
ok thank you for help –  Federica Maggioni Nov 21 '12 at 20:11
show 2 more comments
up vote 2 down vote accepted

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.

share|improve this answer
ok, maybe i can solve it: if $z_0$ were removable for $P\circ f$ then $P\circ f$ would be bounded in a neighborhood of $z_0$, so the image couldn't be dense; if $z_0$ a pole, then $lim_{z\rightarrow z_0} |P\circ f(z)|=\infty$ ans so for any $M>0$ there exists $\epsilon>0$ such that $|P\circ f(z)|>M$ in $D(z_0,\epsilon)$ and again this contradicts the density. –  Federica Maggioni Nov 22 '12 at 8:56
+1 for giving a complete proof, including the fact that under a surjective continuous map dense sets are sent onto dense sets. –  Georges Elencwajg Nov 22 '12 at 9:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.