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

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  • $\begingroup$ grazie per la correzione $\endgroup$ – Federica Maggioni Nov 22 '12 at 8:36
  • $\begingroup$ 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) $\endgroup$ – Georges Elencwajg Nov 22 '12 at 8:54
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Use Casorati-Weierstrass Theorem

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  • $\begingroup$ $z_0$ 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? $\endgroup$ – Federica Maggioni Nov 21 '12 at 16:17
  • $\begingroup$ 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. $\endgroup$ – user50228 Nov 21 '12 at 16:31
  • $\begingroup$ 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? $\endgroup$ – Federica Maggioni Nov 21 '12 at 18:54
  • $\begingroup$ Use that $P$ is a continous function. $\endgroup$ – user50228 Nov 21 '12 at 19:25
  • $\begingroup$ ok thank you for help $\endgroup$ – Federica Maggioni Nov 21 '12 at 20:11
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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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  • $\begingroup$ 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. $\endgroup$ – Federica Maggioni Nov 22 '12 at 8:56
  • $\begingroup$ +1 for giving a complete proof, including the fact that under a surjective continuous map dense sets are sent onto dense sets. $\endgroup$ – Georges Elencwajg Nov 22 '12 at 9:01
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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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