Finding conformal map verification I would like to know if this reasoning is correct:
I want to find a conformal map from $\mathbb D$ to $\mathbb C$.
My reasoning:
First we seek a conformal map from $\mathbb H$ to $\mathbb D$ which is easy to find: $$A=\begin{pmatrix}1&-i\\1&i\end{pmatrix}\leftrightarrow \phi_A(z)=\frac{z-i}{z+i}\text{ and invert it to obtain :}\,\phi_A^{-1}(z)=-i\frac{z+1}{z-1}$$
such that $\phi_A^{-1}:\mathbb D\rightarrow\mathbb H$ is conformal.
Now we seek to extend $\mathbb H$ to $\mathbb C$. FOr this, we consider doubling the argument thus considering the map:
$$\phi_{\mathbb D\rightarrow\mathbb C}(z)=(\phi_A^{-1})(z)^2=\left(-i\frac{z+1}{z-1}\right)^2=-\frac{(z+1)^2}{(z-1)^2}=-\frac{z+1}{z-1}$$
which is a conformal map from $\mathbb D$ to $\mathbb C$.
Thanks in advance.
NB: $\mathbb D=D(0,1)$, $\mathbb H$ is the upper open half plane.

While you used some very good ideas in this attempt, it turns out to be incorrect (and ultimately doomed):

*

*In the last equality, you stripped the squares away incorrectly (nothing cancels there). Indeed, your proposed map is a fractional linear transformation which will take the unit disk to a half-plane, not the entire plane.

*Assuming we're talking about open sets here (the open unit disk, the open half plane), then the squaring map on the half plane is not surjective—its image does not contain the positive real axis. So the second-to-last expression is also incorrect.

*Finally, there is no conformal map from the unit disk (or any bounded set) onto the entire complex plane! If there were, its inverse would be a bounded entire function, hence constant.

