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It's an exercise in the book complex analysis of Ahlfors,but I can't work it out.Can somebody give me some hints please. Suppose that $f(z)$ is analytic and satisfies the condition$\left | f(z)^{2}-1 \right | <1$ in a region $\Omega$. Show that either $Ref(z)>0$ or $Ref(z)<0$ through-out $\Omega$. Thank you.

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It's a simple continuity argument. Can you see that $f$ cannot vanish on the domain? –  Zhen Lin Aug 3 '11 at 3:02
    
It seems like it is Rouche's theorem . –  gary Aug 3 '11 at 3:09
    
Open mapping theorem. In other words $z$ can't be on the $y$-axis or we could find an open ball about it of radius $\epsilon$ and that gives you points both in the left side and right side of the $y$-axis. –  user786 Aug 3 '11 at 3:28
    
@user786: I don't understand. The reason $f(z)$ can't be be on the $y$-axis has nothing to do with the open mapping theorem, but follows directly from the inequality. –  Jonas Meyer Aug 3 '11 at 3:32
    
@Jonas, you're most likely right. I answered extremely fast since I did a similar exercise in Conway, I didn't take the time to read all of molan's question. :) –  user786 Aug 3 '11 at 3:36
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up vote 5 down vote accepted

Here is an outline of a method similar to what Zhen Lin suggests.

  • Show that if $\mathrm{Re}(w)=0$, then $|w^2-1|\geq 1$.
  • Note that $\mathrm{Re}f$ is a continuous function with connected domain, and with range contained in $\mathbb{R}\setminus\{0\}$.
  • Use what you know about continuous images of connected sets.
  • Use the fact that connected subsets of $\mathbb R$ are intervals.

(Note that all that is needed in addition to the inequality is that $f$ is continuous and its domain is connected.)

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Thank you.I have worked it out. In fact,I don't konw how to use the condition that $f(z)$ is analytic first. But I found that all I need is that $f(z)$ is continuous later.Thank you all the same. –  molan Aug 3 '11 at 10:33
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Suppose $\mathrm{Re}f$ is $0$ for some $z_0\in\Omega$. Then let $\mathrm{Im}f(z_0)=y$, and you get $|(iy)^2-1|=1+y^2<1$, which is impossible and hence a contradiction. By analyticity, $f$ is continuous, and I assume the region $\Omega$ is at least connected, so that if $\mathrm{Re}f$ takes on two different signs at (say) $z=a$ and $z=b$, there must be a path between them and by an adapted intermediate value argument $\mathrm{Re}f$ is $0$ somewhere along the way, another contradiction.

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