On Cauchy-Riemann equations

Question

Given $f:\mathbb C\to \mathbb C$ is a non-constant entire function, which of the following is possible?

1. Re $f(z)=$ Im $ f(z)$,

2. Im$\,f(z)<0$,

3. Re$\,f(z)$ is bounded,

4. $f(z)\neq 0,$ for all $z\in \mathbb C$.

Definitely, 4. is correct as $f(z)=e^z$ is an example. Also by using Cauchy-Riemann equations it can be shown that 1. and 3. are wrong as those options lead to constant $f$. But why is 2. wrong? Please help!

Answer 1

Only 4. is possible. For example $\mathrm{e}^z\ne 0$, for all $z\in\mathbb C$.

1. If $f=u+iv$, and $u=v$, then $f=(1+i)u$. But $g=(1+i)f$ is also entire and
   $g(z)=(1-i)f(z)=2u$, i.e., it is real-valued, and hence constant, which in turn implies that $f$ is constant.

2. If Im$\,f<0$, then $g(z)=\exp\big(-i f(z)\big)$ is bounded, since $$ \lvert g(z)\rvert =\exp\big(\mathrm{Re}\,(-if(z))\big)=\exp(\mathrm{Im}\,f(z))<\exp(0)=1. $$ Thus $g$ is bounded, and hence constant, and so is $f$.

3. If $\mathrm{Re}\, f(z)\le M$, then as in 2. $g=\exp(f)$ is bounded and thus constant, and so it $f$.

Answer 2

Hint: For $2$, consider

$$ \exp (-if)$$