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Let $\hat{\mathbb{C}}$ =$\mathbb{C}\cup\{\infty\}$. A theorem from my lecture notes says that a function $f: \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$ is conformal iff f is a linear fractional transformation (that is to say $f(z)=\frac{az+b}{cz+d}, ad-bc \neq 0$).

Assuming $f ,g:\mathbb{D} \rightarrow \mathbb{D}$ are conformal, how can we prove that the composite function $f \circ g$ is conformal? Is it enough to state the theorem above?

Finally, is a conformal function the same as an analytic function in some set $U \subset \mathbb{C}$?

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up vote 2 down vote accepted

Well if $$f = \frac{a_1z+a_2}{a_3z+a_4}, g = \frac{a_5z+a_6}{a_7z+a_8}$$ Then their composition is:

$$f \circ g = \frac{\frac{a_1(a_5z+a_6)}{a_7z+a_8}+a_2}{\frac{a_3(a_5z+a_6)}{a_7z+a_8}+a_4}\cdot\frac{a_7z+a_8}{a_7z+a_8}\\ =\frac{(a_1a_5+a_2a_7)z+(a_1a_6+a_2a_8)}{(a_3a_5+a_4a_7)z+(a_3a_6+a_4a_8)}$$ Which is obviously of the correct form, and $$(a_1a_5+a_2a_7)(a_3a_6+a_4a_8)-(a_1a_6+a_2a_8)(a_3a_5+a_4a_7) \neq 0$$ follows directly from both $ad-bc$ from $f,g$ not being 0.

So yes, their composition is conformal. These classes of functions are what we call Möbius transformations, and we just proved that composition of two Möbius transformations is another one.

For $\Bbb D$ the situation is a lot easier. Not all analytic functions are coformal, but $$f \text{ analytic with derivative non-zero everywhere} \Leftrightarrow f \text{ conformal}$$

and you know compositions of analytic functions are analytic, and the derivative of $f \circ g$ will also be non-zero everywhere, so the composition of conformal functions on $\Bbb D$ is in fact another.

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So you don't think that domain($\mathbb{C}$) and codomain(range)($\mathbb{C}$) in the "theorem" are irrelevant in proving the claim? In other words theorem works also in unit disk? –  laovultai Jan 30 '13 at 15:03
    
@alvoutila I apologize, I misread your question. I added the correct answer. –  Sam DeHority Jan 30 '13 at 16:43
    
So it doesn't matter whether you have unitdisk $\mathbb{D}$ or whole complex plane $\mathbb{C}$ as domain and codomain in proving the claim when you use theorem in the proof? –  laovultai Jan 30 '13 at 20:48
    
Yes. Because if it is analytic and it's derivative is non-zero everywhere the Lagrange inversion theorem gives an inverse and it satisfies the same properties of analyticity and non-zero derivative. This applies for any subset of $\Bbb C$ as domain and codomain, but it is notably different for $\hat {\Bbb C}$, which requires a Mobius transformation. –  Sam DeHority Jan 30 '13 at 20:53
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It isn't enough to state the above theorem, without knowing that $\Bbb D=\hat{\Bbb C}$. You should be able to prove it directly by whatever definition you have for conformal.

It turns out that an analytic function on $U\subset\Bbb C$ is conformal iff its derivative is everywhere non-zero in $U$.

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any sources for this latest claim? from wikipedia for example? –  laovultai Jan 30 '13 at 14:53
    
Here you go. –  Cameron Buie Jan 30 '13 at 22:12
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