1. ## complex analysis

We know that there is only one linear fractional transformation that maps three given points $z_1, z_2, z_3$ to three specified points $w_1, w_2, w_3$. So the mapping $\text{Im} \ z = 0$ onto the circle $|w| = 1$ is uniquely determined if we choose the points (for example): $z = 0, \ z = 1, \ z = \infty$.

Why do we write $w = e^{i \alpha} \frac{z-z_{0}}{z-z_{1}}$. Where did the $e^{i \alpha}$ come from?

2. 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.

-Chip