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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.

Dedede
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  • Is $\Omega$ connected? – anon Feb 03 '12 at 07:22
  • @anon Yes, $\Omega$ is assumed to be connected. – Dedede Feb 03 '12 at 07:24
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    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 03 '12 at 07:29
  • 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 03 '12 at 07:35
  • Dedede: I believe Georges simply means they are logically equivalent (one implies the other in light of the background facts in play here). Clearly the latter implies the former, and how the former implies the latter is what I touch on in my answer. – anon Feb 03 '12 at 07:41
  • @anon Oh ok, I see now. Thanks both. – Dedede Feb 03 '12 at 07:44
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    Ahlfors, eh? Now you can boast that your solution is better than that of the first ever Fields medalist ! – Georges Elencwajg Feb 03 '12 at 08:46
  • 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. – Robert Israel Feb 03 '12 at 09:18

2 Answers2

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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...

anon
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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 03 '12 at 07:43
  • Actually, Georges points out $\operatorname{Im}f$ need not be continuous at all and this still works. – anon Feb 03 '12 at 07:44
  • Dedede: Identify the path with an interval on the real line and invoke the intermediate value theorem from real calculus. – anon Feb 03 '12 at 07:46
  • 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 03 '12 at 07:54
  • @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 03 '12 at 08:02
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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.

Julián Aguirre
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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 03 '12 at 11:14