Some examples of Möbius transformation

[skrat]

Homework Statement

Find Möbius transformation that maps:
a) circle $|z+i|=1$ into line $Im(z)=2$
b) circle $|z-i|=1$ into line $Im(z)=Re(z)$
c) line $Re(z)=1$ into circle $|z|=2$

Homework Equations

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

The Attempt at a Solution

a) Firstly to move the circle into the origin $f_1=z+i$ than to map it into a line $f_2=\frac{1-z}{z+1}$ than rotate it for $pi/2$ with $f_3=iz$ and lastly move it upwards for $2i$ with $f_4=z+2i$

So $f=f_4\circ f_3\circ f_2\circ f_1=f_4(f_3(\frac{1-z-i}{z+1+i}))=\frac{1-z-i}{z+1+i}i+2i=\frac{3i+iz-1}{1+z+i}$

Is that ok?

b) To find a,b,c and d I determine that $f(0)=0$ and $f(2i)=\infty$ and $f(-1+i)=i$ which gives me $f_2=\frac{-z}{z-2i}$

Finally I have to rotate the line for $pi/4$ therefore the answer should be

$f(z)=\frac{-z}{z-2i}e^{-i\pi /4}$

c) Well, I know that $\frac{1-z/2}{1+z/2}$ maps circle (with radius 2) into right half-plane.

So I guess $f(z)=\frac{2-z}{2+z}+1=\frac{4}{z+2}$

Now the inverse transformation is also the answer to part c): $f(z)=\frac{4-2z}{z}$

What do you think?

 

[haruspex]

You can fairly easily check your answers. Just plug in a couple of different values for z. For (a), start with z = 0.
The way I find much easier is more geometric. If a circle passes through the origin then a simple inversion z→1/z will give you a straight line (and v.v.). The line will be orthogonal to the line joining the origin to the centre of the circle. In (a), this immediately gives you a line parallel to the desired one. Just need to multiply by a suitable real factor.