I'm not sure how to approach this question. Can someone please give me a hint?
If $h(z)$ is analytic on $\Omega:=\{z:\Re(z) < 1\}$ and $\exp(h(z))=1-z$, $\forall z \in \Omega$ and $h(0) = 0$. Prove that $|\Im(h(z))| < \pi/2$, $\forall z \in \Omega$.
Thank you.
Clearly the function $h$ defined in my comment is the inverse of $f \circ \exp$ where $f(z)=1-z$. It is easy to see that it is continuous as well. Moreover, $(f \circ \exp)'(z)=-\exp(z)$, and this is nonzero for all $z$ whose imaginary part is in between $-\pi/2$ and $\pi/2$. Thus by the inverse function theorem $h$ is analytic.
You cannot choose other ranges of $\theta$ as you asked, because that does not define analytic $h$. (Be careful when choosing the range.) For instance, if you choose $0\leq \theta<\pi/2$, $2\pi/2 <\theta <2\pi$, then that function is not continuous at $1-z=1$ (that is, $\theta$ jumps from $0$ to $2\pi$).