I was given that range of $f$ lies on a cirlce, and $f$ is analytic on $D$. I want to show that $f$ is constant.
This is my approach: I suppose that $f$ lies on a circle $|w-P|=R$, where $P,R$ are constants. Then consider $|f(z)-P|=R \longrightarrow |f(z)-P|^2=f(z)^2+P^2=R^2$, which means $f(z)=\sqrt{R^2-P^2}$, so $f(z)$ is a constant.
In the above approach, I did not use the fact that $f$ is analytic on $D$. Is it okay to use the argument that I used.