Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $ B(0,1) = \{ z\in \mathbb{C} | |z|<1\} $ and $ f $ be an holomorphic function on $ B(0,1) $ such that $ f(z)\in\mathbb{R} \iff z\in\mathbb{R} $

Prove: $ f $ has at most 1 root in $ B(0,1) $

i think this exercise requires rouche theorem or the argument principle theorem but i cant see how to use it

share|cite|improve this question
up vote 3 down vote accepted

Let $g:(-1,1)\to\mathbb R$ be the restriction of $f$. All zeros of $f$ are zeros of $g$. If $g$ has two distinct zeros, then $g'(c)=0$ for some $c\in(-1,1)$ by Rolle's theorem. This implies that $f'(c)=0$, and that $h(z)=f(z)-f(c)$ has a zero of order at least $2$ at $c$. The argument principle implies that $h(z)$ takes on real values at least $4$ times as you go around a small enough circle centered at $c$, and only $2$ of these have real inputs.

share|cite|improve this answer
Would anyone care to elaborate on the last sentence ? How does the argument principle imply that $h(z)$ assumes real values at least 4 times ? – Teddy Aug 1 '12 at 7:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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