Suppose $f$ is an entire function satisfying any one of the following conditions for all $z\in \mathbb C$

(1) im$f(z)$ has no zeros

(2)$|f(z)|\geq 1$.

Then f is constant.

My thought: For(2) since $|f(z)|\geq 1$ ,then $f$ has no zero in $\mathbb C$.

Define $g=1/f$, then $g$ is bounded entire function implies $g$ is constant implies $f$ is constant, am I right ? I have no idea about (1), please give some hints. Thanks.

  • 1
    $\begingroup$ use picard's theorem $\endgroup$
    – neelkanth
    Nov 18 '15 at 16:11
  • $\begingroup$ If an entire function omits two values ,then f is constant.Is it? $\endgroup$
    – user291341
    Nov 18 '15 at 16:14
  • $\begingroup$ yes its picard theorem... $\endgroup$
    – neelkanth
    Nov 19 '15 at 3:47

For $1$ you can try this :

Let $f(z)=u+iv$ then since $Im f\neq 0$ either $v>0$ or $v<0$.

Case $1$: $v>0$ .Consider $|e^{if}|=|e^{iu-v}|=\dfrac{1}{e^v}<1$

Case $2$: $v<0 $ Consider $|e^{-if}|=e^v<1$

In both cases we have $f$ constant by Liouville's theorem


Your proof for case $(2)$ is correct. Concerning case $(1)$:

Hint #1:

$f(\mathbb C)$ is connected, therefore $(1)$ implies that $f(\mathbb C)$ is contained in either the upper or the lower halfplane.

Hint #2:

$T(z) = (z-i)/(z+i)$ maps the upper halfplane onto the unit disk.


Hints: 1) Show that ${\rm Im}(f(z))>0$ for all $z$, or ${\rm Im}(f(z))<0$ for all $z$; (if ${\rm Im}(f(z_1))>0$ and ${\rm Im}(f(z_2))<0$, put $g(t)={\rm Im}(f((1-t)z_1+tz_2))$ for $t\in [0,1]$; $g$ is clearly continuous.)

2) If you are in the first case, and if $h(z)=f(z)+i$, what can you say of $|h(z)|$ ?


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