Find a Mobius Transformation that carries the points $ -1, i, 1+i$ to the following:

Question

My goal is to find a Mobius transformation that transforms $-1, i, 1+i$ onto the points

a) $0, 2i, 1-i$

b) $i, \infty, 1$

For part a, I know that the Mobius transformation $M$ will be such that $M(-1) = 0$. So, this means that it should have the form

$$M(z) = \frac {z+1} {az+b}$$

Therefore, what should happen is that $M(i) = 2i$. That is,

$$M(i) = \frac {i+1} {ai+b}$$

Multiplying by the conjugate of the denominator, we obtain $$M(i) = \frac {-(a+b)+(a-b)i} {-(a^2+b^2)} = 2i $$

Now, this means that $a = -b$, because the real part of this has to be $0$. Therefore, with this substitution, we can say that

$$M(i) = \frac {2ai} {-2a^2}= \frac {-i} {a} = 2i$$

Therefore, $ a = -\frac {1}{2}, b = \frac {1}{2}$.

However, when I use these values for the third point I do not find that $M(1+i) = 1-i$ is true. Is there something I'm missing here?

Also, for part b, I know the Mobius transformation has to be of the form

$$M(z) = \frac {az+b}{z-i} $$

because $0$ is the denominator whenever $z=i$, yielding a value of $\infty$ for $M$. However, I eventually run into similar problems like I'm having for part a). Can somebody tell why this isn't working, whether it's an algebraic mistake, or if I'm missing something more important? Thank you!

Answer 1

A Mobius transformation is an automorphism of the Riemann sphere $S = \Bbb C \cup \{ \infty \}$. It comes from a rational fraction of degree $1$ on $\Bbb C$, which is a map of the form $z \mapsto \frac {az+b}{cz+d}$ where $a,b,c,d$ are complex numbers.

In order for the map to be a bijection you need a condition on $a,b,c,d$, namely that $ad-bc \neq 0$ (if it is zero then $f$ is a constant map instead).
Even though you need $4$ numbers, multiplying all of them by a nonzero complex number doesn't change the map itself, so really the "space" of Mobius transformations has dimension $3$. This means that usually, you need $3$ data points to determine $f$. In our case, we have the image of $3$ distinct points, so this should be enough to determine $a,b,c,d$ up to a multiplicative constant.

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Given a Mobius transformation $f$, the graph of $f$ is $\{(z,w) \mid w = f(z)\} = \{(z,w) \mid czw+dw-az-b = 0\}$, that is, the graph of $f$ in $\Bbb C^2$ is given by a linear relationship between $1,z,w,$ and $zw$. If you think about this long enough, this shows that if you have three points $(z_i,w_i)$ on the graph of $f$, then the relationship between $z$ and $w$ is described by the equation :

$$ 0 = \begin{vmatrix} z_1w_1 & z_1 & w_1 & 1 \\ z_2w_2 & z_2 & w_2 & 1 \\ z_3w_3 & z_3 & w_3 & 1 \\ zw & z & w & 1 \end{vmatrix}$$

For case a), plugging in the values of $(z_i,w_i)$ you get

$$ 0 = \begin{vmatrix} 0 & -1 & 0 & 1 \\ -2 & i & 2i & 1 \\ 2 & 1+i & 1-i & 1 \\ zw & z & w & 1 \end{vmatrix}$$

Next we do some row manipulations on the top $3$ rows to simplify it

$$\begin{vmatrix} 0 & -1 & 0 & 1 \\ -2 & i & 2i & 1 \\ 2 & 1+i & 1-i & 1 \\ zw & z & w & 1 \end{vmatrix} = \begin{vmatrix} 0 & -1 & 0 & 1 \\ 0 & 1+2i & 1+i & 2 \\ 2 & 1+i & 1-i & 1 \\ zw & z & w & 1 \end{vmatrix} = \begin{vmatrix} 0 & -1 & 0 & 1 \\ 0 & 0 & 1+i & 3+2i \\ 2 & 0 & 1-i & 2+i \\ zw & z & w & 1 \end{vmatrix} = \begin{vmatrix} 0 & -1 & 0 & 1 \\ 0 & 0 & 1+i & 3+2i \\ 2 & 0 & 0 & 4i \\ zw & z & w & 1 \end{vmatrix} $$

Now we divide row 3 by $2$, then we can develop this on the bottom row, to get the equation (getting the signs right is the hardest part)

$((-1)(1+i)(2i))zw + (1.(1+i).1)z + (1.(-1)(3+2i))w - (1.(-1).(1+i)) = 0$
This is equivalent to $2(1-i)zw + (1+i)z + (-3-2i)w + (1+i) = 0$
and to $M(z) = w = \frac {(1+i)(1+z)} {(-2+2i)z+(3+2i)}$

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As for case b), since we got an $\infty$ in there, I can't exactly plug $w_2 = \infty$.
However, this is still equivlent to a linear relationship between $a,b,c,d$ : $f(z_2)=\infty \iff cz_2+d = 0$.
Or we can replace $z$ and $w$ with projective coordinates. Replace $z$ with $z/t$ and $w$ with $w/v$. While before we had a row $(zw : z : w : 1)$, now doing the replacement and multiplying it by $vt$ we get a row $(zw : zv : tw : tv)$, and now we can plug $\infty$ be putting $w/v = 1/0$

Anyway, you get :

$$ 0 = \begin{vmatrix} -i & -1 & i & 1 \\ i & 0 & 1 & 0 \\ 1+i & 1+i & 1 & 1 \\ zw & z & w & 1 \end{vmatrix} = \begin{vmatrix} 1-i & -1 & 0 & 1 \\ i & 0 & 1 & 0 \\ 1 & 1+i & 0 & 1 \\ zw & z & w & 1 \end{vmatrix} = \begin{vmatrix} -i & -2-i & 0 & 0 \\ i & 0 & 1 & 0 \\ 1 & 1+i & 0 & 1 \\ zw & z & w & 1 \end{vmatrix} = \begin{vmatrix} -i & -2-i & 0 & 0 \\ i & 0 & 1 & 0 \\ 0 & 3i & 0 & 1 \\ zw & z & w & 1 \end{vmatrix} $$

We develop the bottom row and we get

$-(-2-i)zw + (-i)z + i(-2-i)w - (-i)(3i) = 0$
then $(2+i)zw - iz + (1-2i)w -3 = 0$
Which means $M(z) = w = \frac {iz+3}{(2+i)(z-i)}$

Answer 2

For part a, one way to solve it is using the formula:

$$ \frac{w-w_1}{w-w_2} : \frac{w_3-w_1}{w_3-w_2} = \frac{z-z_1}{z-z_2} : \frac{z_3-z_1}{z_3-z_2} $$ This defines a unique Mobius transformation $T$ such that $T(z_k)=w_k$ for $k=1,2,3$

Here we have, $z_1=-1, \space z_2=i, \space z_3=1+i, \space w_1=0, \space w_2=2i, \space w_3=1-i$

Plugging in we get: $$\frac{w}{w-2i} : \frac{1-i}{1-3i} = \frac{z+1}{z-i}:\frac{2+i}{1}$$ There's a bit of simplification but eventually you should get $$w = \frac{2z+2}{4iz+5-i}$$

For part b, I think you were on the right track.
Plugging in $i \to \infty$ you get $$T(z) = \frac{az+b}{z-i}$$ Then from $-1 \to i$, we get $$\frac{-a+b}{-1-i} = i \implies b = 1-i+a$$ Plugging this back in $T$, we get $$T(z) = \frac{az+1-i+a}{z-i}$$ Now $1+i \to 1$ gives $$\frac{a(1+i)+1-i+a}{1}=1 \implies a=\frac{1+2i}{5}$$ which gives $b=1-i+a=\frac{6-3i}{5}$ And so $$T(z) = \frac{\frac{1+2i}{5}z + \frac{6-3i}{5}}{z-i} = \frac{(1+2i)z + 6-3i}{5z-5i}$$