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.

  • 3
    $\begingroup$ How do you feel about the maximum modulus principle? $\endgroup$
    – Erick Wong
    Commented Feb 13, 2014 at 15:43

1 Answer 1


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.

PS: This is not the easiest method, but the above proof only uses the C-R equations and a bit of real analysis.


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