# if f is analytic and it's mapping a region onto a circle

If $f$ is analytic on a domain $D$ and it's mapping the region onto a circle, then what can be said about $f$?

I tried to write $f(z)$ as $(r\cos\theta+a)+(r\sin\theta+b)i$, and by using the Cauchy Riemann equations I get that it must be a unit circle if the function is analytic since cosθ=rcosθ and sinθ=-(-rsinθ)

I think by using the open mapping theorem, one can easily state that $f$ is constant because a circle is not an open set.

It that right? But what can I do if I don't the theorem? Need some hint.

• How do you feel about the maximum modulus principle? Commented Feb 13, 2014 at 15:43

Since $$f$$ maps $$D$$ onto a circle, $$|f (z)|=k ,\forall z\in D$$, where $$k$$ is a constant.
Let $$f=u+iv$$. Then $$|f (z)|=k\implies {[u(x,y)]}^2+{[v (x,y)]}^2=k^2\tag1.$$
If $$k=0$$, $$f$$ is identically $$0$$ on $$D$$. So assume $$k\neq0.$$ Then taking partial derivative on both sides of $$(1)$$ w.r.t. both $$x$$ and $$y$$ will give us $$\begin{cases}uu_x+vv_x=0\\uu_y+vv_y=0\end{cases}.$$ Using C-R equations, we can rewrite it as $$\begin{cases}uu_x-vu_y=0\\uu_y+vu_x=0\end{cases}.$$ Solving for $$u_x$$ and $$u_y$$ gives $$u_x=\frac0 {u^2+v^2}=0$$ and $$u_y =\frac0 {u^2+v^2}=0.$$ Since both partial derivatives are zero and $$D$$ is connected, $$f$$ is constant on D.