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!
Best Answer
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 $$T(z) = \frac{\frac{1+2i}{5}z + \frac{6-3i}{5}}{z-i} = \frac{(1+2i)z + 6-3i}{5z-5i}$$