7
$\begingroup$

Let $f$ be a non-constant and analytic on a neighborhood of closure of the unit disk such that $|f(z)|=\text{constant}$ for $|z|=1$. Prove $f$ has at least one zero inside unit disk.

I thought of using Rouche's somehow. Using $f(z)-z$, and taking constant is less than $1$, I can actually conclude from Rouche's theorem that the equation $f(z)-z=0$ have a fixed point inside the unit disk. I am stuck for other constants greater than equals to one and exactly zero. I hope there should be a little trick I am missing here. It will be awesome to see if maximum principle can be applied to conclude the result.

$\endgroup$
1
7
$\begingroup$

Hint: if $f$ has no zero inside the disk, consider $1/f$ and use the maximum modulus principle.

$\endgroup$
9
  • 1
    $\begingroup$ I see, that is sweet. This is going to give you a contradiction that f is constant inside the disk. humm. Thanks @Robert Israel. $\endgroup$ – Deepak Dec 23 '12 at 6:31
  • $\begingroup$ I don't if this is the way you ask about your confusion in other solution given. I think it is pretty similar problem and I am having a hard time to understand the last line of the solution give n by @Montez on this problem math.stackexchange.com/questions/147202/an-entire-function. I would appreciate if you could explain that to me. at Robert Israel $\endgroup$ – Deepak Dec 23 '12 at 6:36
  • $\begingroup$ Oops I can not tag two persons at a time. @Robert Israel, would you please check my previous comment. $\endgroup$ – Deepak Dec 23 '12 at 6:41
  • $\begingroup$ I don't know what you mean by "I don't if this is the way you ask about your confusion in other solution given." That is the only "confusion" I have at the moment. $\endgroup$ – Robert Israel Dec 23 '12 at 7:08
  • $\begingroup$ Opps, I am sorry, my English sucks sometime. I mean to say "I don't know if this is the way someone ask about their confusion in other solution given" I was trying to understand the solution given in above link, and I tried to talk with the original author, he/she did not responded. That is why I feel like I should ask with you because the problem we are doing is kind of similar to the problem I have a confusion. I don't know if this English make sense now :). @Robert Israel $\endgroup$ – Deepak Dec 23 '12 at 7:10
1
$\begingroup$

If you want to use Rouche's Theorem, you can consider $g(z)=f(z)-f(w)$ for some $w\in D$ where $D$ is unit disc. Then on $\partial D$ we have $|g(z)-f(z)|=|f(w)|<|f(z)|$, since by maximum modulus principle the maximum of $|f(z)|$ can only happens on $\partial D$ and also we know that $|f(z)|$ is constant on $\partial D$. Then by Rouche's Theorem we know that number of zeros of $f$ equals to it of $g$ on $D$. As $g$ at least has one zero $z=w$, $f$ at least has one zero.

$\endgroup$

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.