Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

2 Answers 2

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 $f$.

share|improve this answer

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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.