1. ## Complex Mappings

Can anyone help me with these problems I am stuck on?

1. Consider the complex mapping
z -> f(z) = (1+z)/(1-z)
Find the images of i and 1 − i. What are the images of the real and the
imaginary axes?

2. Find Mobius transformations f such that
(a) f(0) = i, f(1) = 1, f(−1) = −1;
(b) f(0) = infinity, f(1) = −1, f(infinity) = 1.

3. Find a Mobius transformation mapping {z : Imz > 2} onto the disc
{w : |w − 2| < 3}.

2. Originally Posted by alawrie
Can anyone help me with these problems I am stuck on?

1. Consider the complex mapping
z -> f(z) = (1+z)/(1-z)

$\frac{z+1}{-z+1}=\frac{-|z|^2+1+2Im(z)\,i}{|-z+1|^2}$ , and now just substitute

Find the images of i and 1 − i. What are the images of the real and the
imaginary axes?

2. Find Mobius transformations f such that
(a) f(0) = i, f(1) = 1, f(−1) = −1;
(b) f(0) = infinity, f(1) = −1, f(infinity) = 1.

A Moebius transformtion is of the form $z\mapsto \frac{az+b}{cz+d}\,,\,a,b,c,d\in\mathbb{C}\,,\,\,a d-bc\neq 0$ , so choose the coefficients accordingly. For example , for (a):

$\frac{b}{d}=i\,,\,\,\frac{a+b}{c+d}=1\,,\,\,\frac{-a+b}{-c+d}=-1$ $\Longrightarrow b=di\,,\,a+di=c+d\Longrightarrow a-c=d(1-i)\,,\,\,-a+di=c-d\Longrightarrow$ $a+c=d(1+i)\longrightarrow \frac{a-c}{1-i}=\frac{a+c}{1+i}\Longrightarrow c=ai$,

and together with $a+di=c+d$ we get $a+di=ai+d\Longrightarrow (d-a)i=d-a\Longrightarrow a=d=-bi=-ci$ , and the transf. is $z\mapsto\frac{z+i}{iz+1}$

Tonio

3. Find a Mobius transformation mapping {z : Imz > 2} onto the disc
{w : |w − 2| < 3}.