How to prove that composition of two conformal functions is conformal 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 $U \subset \mathbb{C}$?
 A: 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. 
A: 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 $U$.
