How to find a conformal mapping of the first quadrant. Find a conformal mapping of the first quadrant onto the unit disc mapping the points $1+i$ and $0$ onto the points $0$ and $i$ respectively. 

I think that i need to use "the change of variables $w=z^k$" but how? And why do we apply this?  
Please can someone explain thisstep by step? 
Thanks alot:) 

 A: The composition of conformal maps is conformal, so to obtain a conformal map between two domains, we can - if it seems more simple - map the first domain conformally to a simpler intermediate domain, and then map that intermediate domain conformally to the target (perhaps with more intermediate steps).
One needs to know some standard conformal maps of course. A well-known family of conformal maps are the Möbius transformations. These allow us to map any (open) haf-plane conformally to any (open) disk, mapping any prescribed point in the interior of the half-plane to the centre of the disk.
Knowing that, we need a conformal map from the quadrant to a half-plane. The boundary of the quadrant has a vertex where the two straight half-lines making up the boundary meet at a right angle. The boundary of a half-plane has no vertex, in the plane, it is a straight line, in the sphere, a circle. So we need something that straightens the right angle of the boundary of the quadrant.
Now one should remember that the power maps $z \mapsto z^k$ multiply angles at $0$ by $k$ - writing $z = \rho e^{i\varphi}$, we have $z^k = \rho^k e^{ik\varphi}$ - so to straighten the right angle at $0$ of the boundary of the quadrant, we need $k = 2$ (generally, to straighten an angle $\alpha$, we need the power $\pi/\alpha$ [which need not be an integer]). So the first part of our map is
$$s \colon Q \to \mathbb{H};\quad z \mapsto z^2$$
that maps the first quadrant $Q = \{ x+iy \in \mathbb{C} : x > 0, y > 0\}$ conformally to the upper half-plane $\mathbb{H} = \{ z \in \mathbb{C} : \operatorname{Im} z > 0\}$.
Then we need a conformal map $T \colon \mathbb{H} \to \mathbb{D}$ from the upper half-plane to the unit disk, that maps $s(1+i) = (1+i)^2 = 2i$ to $0$ and $s(0) = 0$ to $i$.
A Möbius transformation mapping $2i$ to $0$ and the real line (the boundary of the upper half-plane) to the unit circle is
$$T_0 \colon z \mapsto \frac{z-2i}{z+2i}.$$
That does not yet quite do what we want, since $T_0(0) = \frac{-2i}{2i} = -1$, so we compose it with a rotation that takes $-1$ to $i$, and that is multiplication by $-i$, so we get
$$T\colon z \mapsto -i\frac{z-2i}{z+2i}$$
for our conformal map from the upper half-plane to the unit disk. Composing the two conformal maps, we get $f = T \circ s \colon Q \to \mathbb{D}$, given by
$$f(z) = -i\frac{z^2-2i}{z^2+2i}.$$
(Note: That is the only map with the required properties; if $g \colon Q \to \mathbb{D}$ is conformal with $g(1+i) = 0$ and $g(0) = i$, then $g\circ f^{-1}$ is an automorphism of $\mathbb{D}$ that fixes $0$, hence a rotation, and also fixes $i$, hence the rotation must be the identity.)
