# Constructing a new conformal map, given two conformal maps.

Let's say I have two conformal maps $f_1, f_2$ such that $f_j:\Omega\to D_j$, where $\Omega, D_j$ are open subsets of $\Bbb{C}$. Then my question is whether there is a common technique to obtain a conformal map that maps $\Omega$ to $D_1\cap D_2$ ? My question is inspired from the following attempt:

Let $g(z) = \dfrac{1}{2}(z+\dfrac{1}{z})$ be the Joukowski map that conformally maps the complement of the open unit disk $\Bbb{D}$, $\Bbb{C}\setminus\Bbb{D}$, to $\Bbb{C}\setminus_{[-1,1]}$. Then, one can set $h(z) = ig(z)$ so that $h:\Bbb{C}\setminus\Bbb{D}\to\Bbb{C}\setminus_{[-i, i]}$. My goal is to conformally map $\Bbb{C}\setminus{\Bbb{D}}$ to $\Bbb{C}\setminus_{[-1,1]\cup[-i,i]}$.

So I am wondering if there is way to get such a map using above $h$ and $g$ ?

• No, conformal maps don't work like that. – user147263 May 8 '16 at 1:25
• If there were such a map, what would it do in the case where the intersection is empty? – John Hughes May 8 '16 at 1:54
• @JohnHughes I mean that case is clearly not of interest, but I now see that the answer to my question is negative, in general. However, I am still in need and vain attempt to find a conformal map between two domains in the 2nd paragraph. – dezdichado May 8 '16 at 2:00
• And what I meant was that if there was a general solution to the problem you asked, it would have to provide a map from a nonempty domain to an empty one, so there cannot be such a solution. Best of luck with finding that map, however. I suspect you'll need a rather different approach. – John Hughes May 8 '16 at 3:02