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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.

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Deepak
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  • See also http://math.stackexchange.com/questions/1565376/if-f-is-a-non-constant-analytic-function-on-b-such-that-f-is-a-constant – Martin Sleziak Dec 11 '15 at 10:00

2 Answers2

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

Robert Israel
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    I see, that is sweet. This is going to give you a contradiction that f is constant inside the disk. humm. Thanks @Robert Israel. – Deepak Dec 23 '12 at 06:31
  • 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 http://math.stackexchange.com/questions/147202/an-entire-function. I would appreciate if you could explain that to me. at Robert Israel – Deepak Dec 23 '12 at 06:36
  • Oops I can not tag two persons at a time. @Robert Israel, would you please check my previous comment. – Deepak Dec 23 '12 at 06:41
  • 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. – Robert Israel Dec 23 '12 at 07:08
  • 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 – Deepak Dec 23 '12 at 07:10
  • Is there an example of such a function $f$ that there is more than one zero inside the disk? – Syoung Jul 28 '16 at 22:01
  • $f(z) = \dfrac{(z-a)(z-b)}{(1-\overline{a} z)(1-\overline{b}z)}$ for $a, b \in D$. – Robert Israel Jul 28 '16 at 22:09
  • Thanks, it is a good example. – Syoung Jul 28 '16 at 22:42
  • @Syoung In general, any product of finitely many functions of the form $\dfrac{z-a}{1-\bar{a}z}$ for $a\in\mathbb{D}$ will have this property. These are called finite Blaschke products. – Trevor J Richards Apr 25 '17 at 03:19
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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.

Syoung
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