Problem on entire function which is reduced to constant 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.

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 $|h(z)|$ ? 
