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I have the following problem:

Problem. Find a conformal mapping from $\{|z-8|<16\}\setminus\{|z-3|<9\}$ onto $\{t<|z|<1\}$. Find $t$.

I know a conformal mapping is one whose derivative doesn't vanish, but nothing more. How should I be thinking about this problem? Is there a standard way of solving it? And what does "find $t$" mean? Is it only possible to find the conformal mapping with one particular value of $t$ and not with other values?

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1 Answer 1

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A conformal mapping of a domain $\Omega$ in the $z$-plane onto a domain $\Omega'$ in the $w$-plane is provided by an analytic function $z\mapsto w:=f(z)$ defined in $\Omega$ and mapping $\Omega$ bijectively onto $\Omega'$. From general properties of such functions it follows that $f'(z)\ne0$ in $\Omega$.

In the example at hand there are several circles involved, in the $z$-plane as well as in the $w$-plane. Therefore we are lead to the conjecture that the required mapping can be realized as a Moebius transformation. So let's use that as a working hypothesis. Such transformations have the form $$f(z)={az+b\over cz+d}$$ and map circles (:= circles or lines) onto circles to start with. The real axis in the $z$-plane is a circle that intersects both circles $|z-8|=16$ and $|z-3|=9$ orthogononally; therefore its image is a circle that intersects both circles $|w|=1$ and $|w|=t>0$ orthogonally, i.e., is a line through $0$ in the $w$-plane. We may as well assume that this image is the real axis; so $a$, $\ldots$, $d$ can be assumed to be real. One more point: Since the given circles are not concentric, but the image circles are, the map $f$ cannot be a similarity. Therefore $c\ne0$, and we may as well assume $c=1$.

So we are now looking for a function $$f(z)={az+b\over z+d}\ , \qquad a,b,d\in{\mathbb R}$$ satisfying $$f(-8)=-1,\quad f(24)=1,\quad f(-6)=- f(12)\ .$$ Solving for $a$, $b$, $d$ we find two such functions, only one of which could possibly solve our problem, namely $$f(z)={2z\over 24+z}\ .$$ It follows from general properties of Moebius transformations that this is indeed the solution of the given problem and that the inner radius $t$ of the image annulus is given by $t=f(12)={2\over3}$.

Is this map essentially the only solution of the given problem? If there were another function $g$ mapping the given domain $\Omega$ onto an annulus $t'<|w|<1$ with $t' > t$ then $h:=g\circ f^{-1}$ would be a conformal map of an annulus onto an inherently thinner annulus. It is a deep theorem of complex analysis that this is impossible. Similarly, if $t'=t$, one can prove that $h$ has to be a rotation $w\mapsto e^{i\alpha} w$. This means that $f$ and $g$ are essentially the same.

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Thank you! I started reading about Mobius transormations and I think I understand the idea. I'm not sure about one thing, still. The choice of Mobius transformations as the class of functions you're condidering seems arbitrary. And the $t$ you obtained was due to this choice. Can we say it's a unique $t$ for all possible conformal maps or just for Mobius transformations? –  Bartek Feb 20 '12 at 0:29
@Bartek: See my edit. –  Christian Blatter Feb 20 '12 at 9:30
Thanks! What is the name of the theorem you are referring to? –  Bartek Feb 20 '12 at 10:29
@ Bartek: The theorem has no name. It is a consequence of the Schwarz lemma (see Ahlfors' "Complex analysis") applied to the universal covering $\tilde\Omega$ of the annulus in question. –  Christian Blatter Feb 20 '12 at 19:27

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