Find a linear transformation such that it maps the disk onto

[Shackleford]

Homework Statement

Find a linear transformation w = f(z) such that it maps the disk Δ(2) onto the right half-plane {w | Re(w) > 0} satisfying f(0) = 1 and arg f'(0) = π/2

Homework Equations

$w = f(z) = \frac{az+b}{cz+d}$

$z = f^{-1}(w) = \frac{dw-b}{-cw+a}$

The Attempt at a Solution

[/B]
$z = f^{-1}(1) = \frac{d(1)-b}{-c(1)+a} = \frac{d-b}{-c+a} = 0$ ⇒ d=b

I don't think I'm quite yet finished. I've seen another method online that would start with the observation that the boundary of the open disk would necessarily map to the boundary of the right half-plane which is the imaginary axis. What's the best way to approach this?

 

[RUber]

You are saying that if the point on the edge of the disk, say point $(r, \theta)$ will necessarily have real part = 0, and as you close in on zero, you will be moving at toward real part = 1, and in some way you should push out to infinity.
Start by thinking about which parts will map to which locations, then find the function that meets those needs.

 

[Ray Vickson]

Homework Statement

Find a linear transformation w = f(z) such that it maps the disk Δ(2) onto the right half-plane {w | Re(w) > 0} satisfying f(0) = 1 and arg f'(0) = π/2

Homework Equations

$w = f(z) = \frac{az+b}{cz+d}$

$z = f^{-1}(w) = \frac{dw-b}{-cw+a}$

The Attempt at a Solution

[/B]
$z = f^{-1}(1) = \frac{d(1)-b}{-c(1)+a} = \frac{d-b}{-c+a} = 0$ ⇒ d=b

I don't think I'm quite yet finished. I've seen another method online that would start with the observation that the boundary of the open disk would necessarily map to the boundary of the right half-plane which is the imaginary axis. What's the best way to approach this?

Please do not call these transformations "linear"---they are not. They are "fractional linear" or maybe "linear fractional", but that is usually much different from straight "linear" (the exception being when the denominator is constant).

 

[Shackleford]

Please do not call these transformations "linear"---they are not. They are "fractional linear" or maybe "linear fractional", but that is usually much different from straight "linear" (the exception being when the denominator is constant).

I've stated the problem verbatim, but I do understand the distinction.

 

[Shackleford]

Okay. I used another method.

[z, 0, 2, -2] = [w, 1, i, 0]

$\frac{(w-q_1)(r_1-s_1)}{(w-s_1)(r_1-q_1)} =\frac{(z-q)(r-s)}{(z-s)(r-q)}$

Plugging in the respective coordinates, I get w = $\frac{2i(2+z)}{(-2zi + 4i -4z)}$