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Suppose an entire function $f$ maps the real line onto the circle $C=\{ z:|z|=R\}$, $R>0$.

  1. Show that $f\left(z\right)\neq 0$ for all $z$ in $\mathbb{C}$.
  2. Is 1. still valid if the real line is replaced by an arbitrary line?
  3. Is it possible for an entire function to map a circle onto a line?

For the first question, the function which is $y=Re^{iz}$ satisfies the condition without zero. But I don't know if there exists other function satisfying the condition.

If the function appearing in the first question is unique, then the second question is still valid since the rotation and translation will change nothing. For the third,the unit circle is compact, so it's impossible to construct desired function.

Are all my arguments valid or not?

b00n heT
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Jack
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  • I've added some edits. I hope I made no mistakes (especially in $1$, where I changed $C$ with $\mathbb{C}$) – b00n heT Jul 31 '16 at 11:34

2 Answers2

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Hint: For the first question, use the fact that if $f$ is entire, then $\overline{f(\overline{z})}$ is entire too (see If $f$ is analytic, prove that $\overline{f(\overline{z})}$ is also analytic), and put $g(z)=f(z)\overline{f(\overline{z})}$. What is $g(x)$ for $x\in \mathbb{R}$ ?

Kelenner
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It suffices that an entire function $f$ maps any straight segment $I$ into the unit circle. Then $f$ can not have any zeroes. This is clear when $I=]a,b[$ since the analytic function $\bar{f}(z) f(z)= \overline{f(\bar{z})} f(z)$ equals one on $I$ whence one in the complex plane. In the general case, you find a linear map $\phi$ that maps $]0,1[$ to your segment and then consider $f\circ \phi$. If $f$ maps $I$ into a circle centered at $a$ of radius $R>0$ then $f$ does not take the value $a$.

Your map $f$ may be written as $f(z)=\exp(i \phi(z))$ where $\phi$ is any entire function preserving the real line. So e.g. $\exp( i (z^2-z^3))$ has the same property.

An entire function is continuous and maps a compact set to a compact, whence bounded set.

H. H. Rugh
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