1
$\begingroup$

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".

$\endgroup$
7
  • 1
    $\begingroup$ Didn't you mean in (i) that $\;|z|\to \infty\;$ ? $\endgroup$
    – DonAntonio
    Jan 12 '19 at 14:06
  • $\begingroup$ Do you know the extreme value theorem? $\endgroup$ Jan 12 '19 at 14:08
  • 1
    $\begingroup$ @DonAntonio the exercise doesn't use absolute value $\endgroup$
    – user619755
    Jan 12 '19 at 14:09
  • $\begingroup$ 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? $\endgroup$
    – DonAntonio
    Jan 12 '19 at 14:14
  • 1
    $\begingroup$ @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 ...! $\endgroup$
    – DonAntonio
    Jan 12 '19 at 14:26
1
$\begingroup$

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 $\min(M,N)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.