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:)

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    $\begingroup$ Can y ou map a half-plane conformally to the unit disk with appropriate normalisation? Can you map a quadrant conformally to a half-plane? $\endgroup$ – Daniel Fischer Oct 27 '13 at 13:30
  • $\begingroup$ No I cannot. I have already asked how? Dear @DanielFischer $\endgroup$ – Nrsnr Oct 27 '13 at 13:40
  • $\begingroup$ Please can you help me? In fact I have its answer. But I dont understand anything. Can I post it? @DanielFischer $\endgroup$ – Nrsnr Oct 27 '13 at 14:44
  • $\begingroup$ Sure. If you want the mapping explained, I'll see what I can do. $\endgroup$ – Daniel Fischer Oct 27 '13 at 14:48
  • $\begingroup$ Okay I posted it. Please can you explain this more clear. I cannot understand such things :( @DanielFischer $\endgroup$ – Nrsnr Oct 27 '13 at 14:59

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.)

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    $\begingroup$ Thank you so so much:))) I am starting to study this step by step right now:) $\endgroup$ – Nrsnr Oct 27 '13 at 16:18
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    $\begingroup$ If you want further elaboration on some points, don't hesitate to ask. $\endgroup$ – Daniel Fischer Oct 27 '13 at 16:25
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    $\begingroup$ Dear Fischer, I dont understand a thing. How do you decide the components of the composition of conformal maps? For example, how to decide the first quadrant $\to$ half plane $\to$ unit disc. And I found a similar question to this. I solved it. Please can you check this? I did that accourding to the way you teached.math.stackexchange.com/questions/542713/… $\endgroup$ – Nrsnr Oct 28 '13 at 12:09
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    $\begingroup$ There is no clear rule how to choose the steps. Anything that works is correct, and anything that you believe will take you closer to the target is something you try. As always, with experience, you can see the way often immediately, because you've already seen so many similar situations. $\endgroup$ – Daniel Fischer Oct 28 '13 at 12:41
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    $\begingroup$ Okay I understand :) $\endgroup$ – Nrsnr Oct 28 '13 at 13:33

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