# Determining transformation sending $|z|=2$ to $|z+1|=1$.

I'm trying to teach myself complex analysis, and am reading about linear fractional transformations.

I want to find the transformation carrying the circle $|z|=2$ into $|z+1|=1$, $-2$ into the origin, and the origin into $i$.

My text states that the transformation is determined by prescribing a point $z_1$ on circle $C$ to correspond to a point $w_1$ on $C'$, and a point $z_2$ not on $C$ to correspond to $w_2$ not on $C'$. Then $z^\ast_2$ corresponds to $w^\ast_2$, the symmetric points of $z_2$ and $w_2$ in $C$ and $C'$, respectively. Then the transformation is obtained from the relation $(w,w_1,w_2,w^\ast_2)=(z,z_1,z_2,z^\ast_2)$ (those are cross-ratios).

I let $C$ be the circle $|z|=1$, and $C'$ be $|z+1|=1$. Then by the notions above $z_1=-2$, $w_1=0$, $z_2=0$, $w_2=i$. I calculate $z^\ast_2=\infty$ and $w^\ast_2=\frac{-1+i}{2}$.

So following what I've read, I have $(w,0,i,(-1+i)/2)=(z,-2,0,\infty)$. I don't understand how to gather the explicit form of the desired linear fractional transformation is from this, since $w$ and $z$ seem unknown to me. Could someone please explain what the transformation is now? Thank you.

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Linear fractional transformations preserve cross-ratios, so if $w$ is an LFT we have $$(a,b,c,d)=(w(a),w(b),w(c),w(d)) \text{ for all }a,b,c,d\in\hat{\mathbb{C}}.$$

The utility of this is that if you know $w(b),w(c),w(d)$ for three known arguments $b,c,d$ as you do, then we can let $a$ be the independent variable $z$ and $w$ the dependent variable $w=w(z)$; solving for the variable $w$ in terms of $z$ will give us the desired transformation in the appropriate form,

$$w=\frac{\square\, z+\square}{\square\, z+\square}.$$

In particular, $(z,-2;0,\infty)=(w,0;i,\frac{-1+i}{2})$ says

$$\frac{(z-0)(-2-\infty)}{(-2-0)(z-\infty)} = \frac{(w-i)\left(0-\frac{-1+i}{2}\right)}{(0-i)\left(w-\frac{-1+i}{2}\right)}$$

$$-\frac{z}{2}=\frac{1+i}{2}\frac{w-i}{w-\frac{-1+i}{2}}.$$

Multiply by the righthand denominator, collect terms and solve for $w$ in a linear equation.

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That's the crucial idea I was missing. Thank again anon. I find $$w=\frac{\frac{-1+i}{2}z+i-1}{z+1+i}.$$ – Dedede Feb 10 '12 at 3:59