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Doing a bit of self study, and I'm unsure about a problem. It says,

Suppose $f(z)$ (a complex valued function) is analytic and satisfies the condition $|f(z)^2-1|<1$ in a region $\Omega$. Show that either $\Re f(z)>0$ or $\Re f(z)<0$ throughout $\Omega$.

I write $f=u+iv$ and suppose to the contrary that $\Re f(z)=0$ at some point $z_0$. Then $f(z_0)^2=-v(z_0)^2$. But $v$ is real valued, and so $ |f(z_0)^2-1|=|-v(z_0)^2-1|\geq 1 $ a contradiction.

What makes me uneasy is I don't see if I used that fact that $f$ is analytic. Did I interpret the question correctly, or did it mean that $\Re f(z)>0$ on all of $\Omega$ or $\Re f(z)<0$ on all of $\Omega$, but doesn't take both positive and negative values? Thanks.

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    It might be a more interesting problem if you changed the assumption to $|f(z)^2-1| \le 1$, because $\{w: |w^2-1|\le 1\}$ is connected. The conclusion in this case is that one of the three possibilities \Re f(z)<0 or \Re f(z) > 0 or $f(z) = 0$ is true throughout $\Omega$. This one needs a bit more than just continuity: e.g. the Open Mapping Theorem will help. But maybe that comes after page 72.2012-02-03

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I don't think $f$ even needs to be analytic - only continuous on $\Omega$. In any case, I think the latter interpretation you pose is the correct one: the problem wants an either-or on all of $\Omega$, i.e.

$\left(\;\forall z\in\Omega:\operatorname{Re} f>0 \;\right)\text{ or }\left(\;\forall z\in\Omega:\operatorname{Re} f<0 \;\right).$

This isn't too much more work than what you've already done. You've shown the real part can't be zero; now assume there are two arguments $z$ and $w$ in $\Omega$ with $\operatorname{Re} f(z)<0<\operatorname{Re}f(w)$. Since $\Omega$ is connected, there is a path going from $z$ to $w$ contained in $\Omega$. Consider how $\operatorname{Re}f$ looks on this path...

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    @Dedede: The path will have a *parametrization* $\gamma:[a,b]\to\text{path}$ with $\gamma(a)=z,\gamma(b)=w$. Then use IVT on the real function $\operatorname{Re}f\circ\gamma:[a,b]\to\mathbb{R}.$2012-02-03
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Let $D$ be the open disk centered at $1$ and with radius $1$, and let D'=\{w:w^2\in D\}. Since $f(z)^2\in D$ for all $z\in\Omega$, f(z)\in D'. What do we know about D'? The following two facts are easy to prove:

  1. D' is symmetric with respect to the imaginary axis;
  2. no point in the imaginary axis is in D'.

Thus, \{w\in D':\Re z>0\}\ and \{w\in D':\Re z<0\}\ are disconnected; since $f(\Omega)$ is connected, it must be contained in one of them.

In fact, D' is the interior of a lemniscate.

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    +1: the description of $D'$ as the interior of a lemniscate is nice and interesting. You are describing how the covering map $\mathbb C^*\to \mathbb C^*: z\mapsto z^2$ is trivial over the disc $D$ by exhibiting the two disjoint pieces of its inverse image $D'$.2012-02-03