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If $f:\mathbb{C}\rightarrow\mathbb{C}$ is analytic and $Im(f(z))\neq 0$ whenever $|z|\neq 1$, show that $f$ is a constant.

It sounds familiar but not so trivial at all...

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up vote 7 down vote accepted

Pick a disk $D=D_r(0)$ with $r>1$, since $Im(f(z))\neq 0$ on $\partial D$, we can assume that it's positive there, but since it's compact we have $Im(f(z))\geq c_r>0$ on $\partial D$, now the maximum principle applied to the harmonic function $-Im(f(z))$ gives that $Im(f(z))\geq c_r>0$ in $D$. Since $r$ was arbitrary, we get that $Im(f)>0$ in $\mathbb{C}$ and so $f$ maps $\mathbb{C}$ into the upper half plane $\mathbb{H}$. Now, composing with the Möbius tranformation $g=\frac{z-i}{z+i}$, we get that $g\circ f$ is a bounded entire function since $g$ maps $\mathbb{H}$ onto the unit disk, so $g\circ f$ is constant, and then so is $f$.

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If $f$ is nonconstant, its image is either $\mathbb{C}$ or $\mathbb{C} - \{pt\}$ by Picard's little theorem. That means that all but at most one point in the real axis is in the image of $f$. But your hypothesis implies that only the circle $|z| = 1$ can map into $\mathbb{R}$. The circle is compact, and hence its image under $f$ is compact. In particular, its image can't be $\mathbb{R}$ or $\mathbb{R} - \{pt\}$. It follows that $f$ must be constant.

Edit: To do this without using Little Picard. We will prove the statement with a few observations. Assume $f$ is nonconstant.

  1. First note that the assumption about the map $f$ implies that $f$ maps the unit disk $\mathbb{D}$ into either the upper half plane or the lower half plane. Similarly, $f$ must map the exterior of $\overline{\mathbb{D}}$ into either the upper or lower half plane.
  2. $f$ cannot map both $\mathbb{D}$ and $\mathbb{C} - \overline{\mathbb{D}}$ to the same half plane. Indeed, if they both mapped to the same half plane $H$, then the image of $f$ would be contained in $\overline{H}$, contradicting Liouville's theorem.
  3. From this it follows that $f$ maps the circle $\partial\mathbb{D}$ to the real axis $\mathbb{R}$, since the image of $f$ crosses from one half plane to the other at $\partial\mathbb{D}$.
  4. The function $u(z):=Im\,f(z)$ is therefore a harmonic function which takes the value $0$ on $\partial\mathbb{D}$. It follows that $u\equiv 0$ on $\mathbb{D}$, i.e., $f(\mathbb{D})\subseteq\mathbb{R}$. This contradicts your assumption on $f$.

There are probably much better ways of doing this.

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Can you avoid using strong results like Picard's Theorems? – iloveinna May 9 '12 at 1:35
@Warwicker: Added a solution without Picard. Hopefully low-tech enough :) – froggie May 9 '12 at 1:51
Not too sure why it contradicts with Liouville. But I derive the same conclusion by considering the connectedness of $D$ and $\mathbb{C}-\overline{D}$. – iloveinna May 9 '12 at 2:11

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