# For an analytic function $f(z)$, $|f(z)^2-1|<1$ implies $\Re f(z)>0$ or $\Re f(z)<0$?

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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Is $\Omega$ connected? – anon Feb 3 '12 at 7:22
@anon Yes, $\Omega$ is assumed to be connected. – Dedede Feb 3 '12 at 7:24
Dear Dedede: +1 for your perfect solution and above all for your sense of self-criticism.You are right that your proof only uses that $Re(f)=u$ is continuous and that $\Omega$ and thus its image inder $u$ are connected, so that both interpretations of the question are equivalent. Since you don't use analyticity, I conjecture that your proof is cleverer than the author's! Who was that, by the way? – Georges Elencwajg Feb 3 '12 at 7:29
Dear @Georges, thanks for the upvote. The author is Lars Ahlfors, this is just problem 3 on page 72 of his Complex Analysis. Do you mind saying in a bit more detail how the two interpretations are equivalent? – Dedede Feb 3 '12 at 7:35
Ahlfors, eh? Now you can boast that your solution is better than that of the first ever Fields medalist ! – Georges Elencwajg Feb 3 '12 at 8:46

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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Thanks anon. I think I see what you're getting at, but I don't know how to express it formally. I visualize this in $\mathbb{R}^3$, identifying $\Omega$ as some subset of the plane, and the image of $\Re f$ on $\Omega$ as the image $(x,y,0)\mapsto (x,y,\Re f(x,y))$ when $z=x+iy$. Since $\Re f(z)<0$ the path starts out at $z$ underneath $\Omega$, and since $\Re f(w)>0$, the path ends above $\Omega$ so it must go through $\Omega$ at some point, in which case $\Re f=0$ at some point, contradiction. How can I say this better? – Dedede Feb 3 '12 at 7:43
Actually, Georges points out $\operatorname{Im}f$ need not be continuous at all and this still works. – anon Feb 3 '12 at 7:44
Dedede: Identify the path with an interval on the real line and invoke the intermediate value theorem from real calculus. – anon Feb 3 '12 at 7:46
If I identify the path as some function $\gamma$ on an integer $[a,b]$, with $\gamma(a)=\Re f(z)$ and $\gamma(b)=\Re f(w)$, then there is some $c$, $a,c<b$ such that $\gamma(c)=0$ by IVT. By how do I know $\gamma(c)=\Re f(t)$ for some $t\in\mathbb{C}$ to get the desired contradiction? – Dedede Feb 3 '12 at 7:54
@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 $$\operatorname{Re}f\circ\gamma:[a,b]\to\mathbb{R}.$$ – anon Feb 3 '12 at 8:02

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 $D'$. – Georges Elencwajg Feb 3 '12 at 11:14