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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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    @Zhen. Thank you.First I didn't konw how to use the C-R equations, but I recogenised that I didn't need it after I saw your hints.2011-08-03

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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.2011-08-03
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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.