# polynomials on the complex numbers

Let p be a complex polynomial $p\left( {a + bi} \right) = p\left( z \right):{\Bbb C} \to {\Bbb C}$ How can I prove the following? $$\lim_{\lVert z\rVert\to\infty} \Vert p(z)\rVert = \infty.$$

I can't use any important result about complex numbers, only the definition, and properties of $\mathbb{R}$ , how can I prove it?

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Well, it is possible to show it for $p(z)=z^n$, and $$a_0+a_1 z + a_2 z^2 + \dots + a_n z^n = z^n \left( {a_0 \over z^n}+\dots+{a_{n-1}\over z} + a_n\right)$$
Of course you have to assume that $p$ is not constant. If $p(z)=\sum_{k=0}^na_kz^k$ where $a_n\neq 0$ and $n\geq 1$, then using the inequality $|a-b|\geq |a|-|b|$ for complex numbers we have $$|p(z)|\geq |a_n||z|^n-\sum_{k=0}^n|a_k||z|^k=|z|^n\left(|a_n|-\sum_{k=0}^{n-1}|a_k||z|^{k-n}\right),$$ and for $|z|\geq R$ where $R>0$ is such that $\sum_{k=0}^{n-1}|a_k||z|^{k-n}\leq \frac{|a_n|}2$ get
$$|p(z)|\geq |z|^n\frac{|a_n|}2,$$ so the result follows.