Example 9. Find the conditions under which the fractional-linear function (6)

$$
w=\frac{a z+b}{c z+d}
$$

maps the upper half-plane $\operatorname{Im} z>0$ onto the upper half-plane $\operatorname{Im} w>0$.
Solution. Under this mapping, it is required that the boundary of the region $\operatorname{Im} z>0$ - the $0 x$ axis, traversed from left to right, is mapped to the boundary of the region $\operatorname{Im} w>0$, i.e., to the $O u$ axis, also traversed from left to right. Thus, for any real values of $z$, the values of $w$ must also be real. This, obviously, is possible only for real values of the numbers $a, b, c, d$. Furthermore, each $z=x+i y$, where $y>0$, must correspond to a $w=u+i v$ with $v>0$. Substituting $z=x+i y$ into formula (6), we get

$$
w=u+i v=\frac{(a x+b)(c x+d)+a c y^{2}}{(c x+d)^{2}+y^{2}}+i \frac{(a d-b c) y}{(c x+d)^{2}+y^{2}}
$$

from which

$$
v=\frac{(a d-b c) y}{(c x+d)^{2}+y^{2}}
$$

Since here $y>0$ and the denominator is positive, for $v$ to be positive, it is necessary and sufficient that the condition $a d-b c>0$ be satisfied. This is the required condition.

## Properties of Fractional Linear Transformations

1. Circular Property. A fractional linear transformation maps a circle to a circle. (A straight line is considered a circle of infinite radius.)
2. Symmetry Property. Two points $z_{1}$ and $z_{2}$, symmetric with respect to a circle $C$, are mapped to points $w_{1}$ and $w_{2}$, symmetric with respect to a circle $\Gamma$, to which the circle $C$ is mapped.

Corollary. If under a fractional linear mapping $w=f(z)$ a straight line or circle $\gamma$ is mapped to a circle $\Gamma$ and one of two points symmetric with respect to $\gamma$ is mapped to the center of the circle $\Gamma$, then the other point necessarily maps to the infinitely distant point.

3. There exists a unique fractional linear function that maps three given points $z_{1}, z_{2}, z_{3}$ in the $z$-plane to three given points $w_{1}, w_{2}, w_{3}$ in the $w$-plane. It has the form

$$
\frac{w-w_{1}}{w-w_{2}} \cdot \frac{w_{3}-w_{2}}{w_{3}-w_{1}}=\frac{z-z_{1}}{z-z_{2}} \cdot \frac{z_{3}-z_{2}}{z_{3}-z_{1}}
$$