How to correctly determine the mobius transformation that maps three distinct points to 0,1,$\infty$

Question

I was trying to find the mobius transformation that maps the point $i,-i,\infty$ to $0,1,\infty$. following the formula.

$M(z) = K \frac{z-i}{z+i}$. $M(i) = 1 \implies 1 = K\frac{i-i}{i+i} \implies \frac{K(-2i)}{0} = 1$. How should I proceed to find the mobius transformation? (I remember that we adopt certain convention when dealing with $\infty$ and $0$, but I couldn't find the related information).

Also, when I was observing the example from the book where the book was trying to find the mobius transformation that maps $\infty, i , 2$ to $0,1,\infty$ . it gave the first step as :

$M(z) = \frac{K}{z-2}$. shouldn't it be $K\frac{z-\infty}{z-2} $? (I know I am probably wrong but I am just not clear with the concept)

Thank you all for your response and help.

Answer 1

The point that maps to $\infty$ through a Mobius transformation is the root of the denominator (and if the denominator is constant, then it is $\infty$ that maps to $\infty$). So, if we want $\infty$ to map to $\infty$, the Mobius transformation is going to be a linear function $M(z)=az+b$. Then we have the equations $0=ai+b$ and $1=-ai+b$, so $b=1/2$ and $a=-1/(2i)=i/2$, giving $M(z)=(iz+1)/2$.

We can check that $M(i)=0$, $M(-i)=1$, and $M(\infty)=\lim_{z\to\infty}\frac{1}{2}(iz+1)=\infty$.

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With that example, the $z-2$ in the denominator is from wanting $2$ to map to $\infty$. Sending $\infty$ to $0$ takes a bit more finesse than putting $z-\infty$ in the numerator, since Mobius transformations need actual elements of $\mathbb{C}$ in the expression. To do this, you just need the formula $\lim_{z\to\infty}\frac{az+b}{cz+d}=\frac{a}{c}$. Thus, if $\infty$ is mapped to zero, the numerator must be a constant.

Answer 2

See the first lines, and especially closed form given by formula (1) (that I hadn't seen elsewhere), in my recent answer math.stackexchange.com/q/3231106.

Check that the different conditions are fulfilled...

The rest of the upsaid answer can interest you as well : it takes into account the isomorphism between Möbius functions and $2 \times 2$ matrices "up to a multiplication" ; see https://en.wikipedia.org/wiki/Modular_group

About your "Also" paragraph, @Kyle Miller has well answered it ; I will add that you cannot have an algebraic expression such as $z-\infty$ because $\infty$ is not a number.