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chiph588@ · #2 January 12th, 2009, 09:55 AM

Consider the Mobius transformation $T(z) = \frac{az+b}{cz+d}$ that maps $Im(z) \geq 0$ onto the circle $|w| \leq 1$ . Note that $T(z)$ maps $Im(z) = 0$ onto the circle $|w| = 1$ .

Now choose an $\alpha$ , where $Im(\alpha) > 0$ such that $T(\alpha) = 0$ or equivalently $\frac{a\alpha+b}{c\alpha+d} = 0 \Rightarrow \alpha = -\frac{b}{a}$ .

Note that $\overline{\alpha}$ is the inverse of $\alpha$ with respect to the real axis. So $0$ and $T(\overline{\alpha})$ must be inverses of each other with respect to $|w| = 1$ . Hence $T(\overline{\alpha}) = \infty$ or equivalently $\frac{a\overline{\alpha}+b}{c\overline{\alpha}+d} = \infty \Rightarrow \overline{\alpha} = -\frac{d}{c}$ .

$T(z) = \frac{az+b}{cz+d} = \frac{a}{c} \; \frac{z+\frac{b}{a}}{z+\frac{d}{c}} = \frac{a}{c} \; \frac{z-\alpha}{z-\overline{\alpha}}$

When $x \in \mathbb{R} \; \; |T(x)| = 1$ . Also notice that $|x-\alpha| = |\overline{x-\alpha}| = |x-\overline{\alpha}| \Rightarrow \frac{|x-\alpha|}{|x-\overline{\alpha}|} =1$ . Therefore $\left|\frac{a}{c}\right| = 1$ .

By converting $\frac{a}{c}$ into polar coordinates, the above equality yields $\frac{a}{c} = e^{i\beta}$ .

Therefore $T(z) = e^{i\beta} \frac{z-\alpha}{z-\overline{\alpha}}$ .

So notice that only two points were chosen to map the upper-half plane onto the unit disk and certain constraints were given to a third point but no third point was chosen explicitly, therefore it makes sense that there is not a unique solution to do this.

I hope this answers your question.

-Chip