# Complex analysis: show for non constant polynomial that p(C)=C

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

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

## 1 Answer

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 $$P(\mathbb C)=\mathbb C$$.