# Show that the function |P|: $\mathbb{C}\to\mathbb{R^+_0}$ has a minimum

I can't figure out the following question and I was hoping that somebody could help me. Thank you in Advance.

$$P(z)$$ is a non-constant polynomial with complex coefficients and is defined by $$P: \mathbb{C}\to\mathbb{C}$$

The exercise requires me to prove the following:

(i) $$\lim \limits_{z \to \infty} P(z)=\infty$$

(ii) The function $$|P|: \mathbb{C}\to\mathbb{R^+_0}$$ has a minimum. (Hint:Show that $$|P|$$ has a minimum on every closed disk and use (i) )

I have proven (i) but I don't know how to go on from there. I know that the function in (ii), as it is taking the absolute value and the range being all positive real numbers with 0, must have a minimum, however I am confused how to use the "hint".

• Didn't you mean in (i) that $\;|z|\to \infty\;$ ? Jan 12 '19 at 14:06
• Do you know the extreme value theorem? Jan 12 '19 at 14:08
• @DonAntonio the exercise doesn't use absolute value Jan 12 '19 at 14:09
• Fine, @SVL . Then how did you manage to prove (i)? What does $\;z\to\infty\;$ means at all? That the real part tends to infinity, the imaginary part...both? Jan 12 '19 at 14:14
• @Yanko Well, that assumption is now, after the OP changed the question, pretty trivial. First, he wrote that $\;P\;$ is "a function". Then, he changed that to $\;P\;$ is a polynomial ...! Jan 12 '19 at 14:26

If $$P$$ is a polynomial it is also continuous and so it has a minimum on every closed disk. Let $$M$$ be the minimum on the disk of radius $$1$$.
On the other hand you know that $$\lim_{z\rightarrow\infty} P(z) = \infty$$ (note that as DonAntonio comment you may want to take $$|z|\rightarrow\infty$$ but I consider these two notions equivalent).
This means that if $$z$$ lies outside of a sufficiently large disk (say of radius $$r$$) then $$|P(z)|>M$$. Let $$N$$ be the minimum over the disk of radius $$r$$.
Then $$\min |P(z)|$$ is the minimum between three terms (The disk of radius $$1$$, the disk of radius $$r$$, and outside of the disk of radius $$r$$). But by the choice of $$r$$ the minimum can't be outside of the disk of radius $$r$$ and so the minimum is $$\min(M,N)$$.