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...
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...
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 $f$.
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.
There are probably much better ways of doing this.