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$-1 \mapsto i,$

$\infty \mapsto 1,$

$i \mapsto 1+i$

This is driving me mad! Should be easy but I keep getting tied up in knots. Any help much appreciated!

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up vote 2 down vote accepted

Suppose the Möbius transformation is that $M(z)=\frac{z+a}{z+b}$, and then by $M(-1)=i,M(i)=1+i$, we can get that $a=2+i,b=2-i$, so the desired Möbius transformation is that $$M(z)=\frac{z+2+i}{z+2-i}.$$

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It might be worth saying that the form of $M(z)$ you assume comes from the fact that $M(\infty) = 1$. – t.b. Apr 9 '12 at 15:50
Thanks! I hadn't thought of simplifying the original form by dividing the top and bottom by the coefficient of $z$. Lesson learned. – ABB Apr 9 '12 at 16:04
yes, $M(\infity)=1$ is very important! – Riemann Apr 9 '12 at 16:07

$$f(z)=\frac{az+b}{cz+d}$$ $$f(\infty)=\frac{a}{c}=1$$ $$f(-1)=\frac{-a+b}{-c+d}=i$$ $$f(i)=\frac{ai+b}{ci+d}=1+i$$ We have got four variables, three equations, all you have to do is to solve them, you will can choose one variable freely

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We can find $\omega(z)=\dfrac{az+b}{cz+d}$ using formula $$\frac{z-z_1}{z-z_2}:\frac{z_3-z_1}{z_3-z_2}=\frac{\omega-\omega_1}{\omega-\omega_2}:\frac{\omega_3-\omega_1}{\omega_3-\omega_2}$$ where $(z_1,z_2,z_3)=(-1,\infty,i)$ and $(\omega_1,\omega_2,\omega_3)=(i,1,1+i).$

In our case we have $$\frac{z+1}{1}:\frac{i+1}{1}=\frac{\omega-i}{\omega-1}:\frac{1}{i},$$ where we change $z-z_2$ with $1$ because $z_2=\infty,$ etc.

And now we find that $\omega=\omega(z)=\dfrac{z+2+i}{z+2-i},$ like previous solutions.

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