From Remmert's Theory of complex functions chapter 9, page 269

Let p(z)$\in\mathbb{C}$[z] be a non constant polynomial, using the growth lemma and open set mapping theorem (but not the fundamental theorem of algebra) show that p($\mathbb{C}$)=$\mathbb{C}$

I have no clue how to approach it, I thought that the growth lemma gives me that p($\mathbb{C}$\R)=$\mathbb{C}$\R but im not sure of it either. R is as given in growth lemma below:

growth lemma

  • $\begingroup$ What is $R$ here? $\endgroup$ – Lada Dudnikova Apr 22 at 9:09
  • $\begingroup$ @LadaDudnikova I added in the question $\endgroup$ – Roni Ben Dom Apr 22 at 9:16

Growth lemma: If $P\colon\mathbb C\longrightarrow\mathbb C$ is a non-constant polynomial and if $a$ is the coefficient of its leading term, then, if $M$ is large enough,$$\lvert z\rvert>M\implies\frac12\lvert az\rvert^n<\bigl\lvert P(z)\bigr\rvert<\frac32\lvert az\rvert^n.$$

By the open mapping theorem, $P(\mathbb C)$ is an open subset of $\mathbb C$. It is clearly non-empty. Now, I shall prove that $P(\mathbb C)$ is also a closed subset of $\mathbb C$. Let $(z_n)_{n\in\mathbb N}$ a sequence of elements of $P(\mathbb C)$ which converges to some $z\in\mathbb C$; I will prove that $z\in P(\mathbb C)$. For each $n\in\mathbb N$, there is some $w_n\in\mathbb C$ such that $P(w_n)=z_n$. The sequence $(w_n)_{n\in\mathbb N}$ must be bounded; otherwise, by the growth lemma, there would be a subsequence $(w_{n_k})_{k\in\mathbb N}$ such that $\lim_{k\to\infty}\lvert z_{n_k}\rvert=\infty$. But then $(w_n)_{n\in\mathbb N}$ has a convergent subsequence $(w_{n_k})_{k\in\mathbb N}$. If $w=\lim_{k\to\infty}w_{n_k}$, then$$z=\lim_{k\to\infty}P(w_{n_k})=P(w)$$and therefore $z\in P(\mathbb C)$.

So, I proved that $P(\mathbb C)$ is open, closed, and non-empty. Since $\mathbb C$ is connected, this proves that $P(\mathbb C)=\mathbb C$.


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