Example 12. Map the upper half-plane $\operatorname{Im} z>0$ onto the unit disk $|w|<1$ so that the point $z=i$ (where $\operatorname{Im} z>0$) is mapped to the center $\boldsymbol{w}=0$ of the disk.
Solution. Since the point $z_{0}$ is mapped by the sought fractional-linear function $w=w(z)$ to the center of the circle, i.e., $w\left(z_{0}\right)=0$, then the conjugate point $\bar{z}_{0}$ must be mapped to the point $w:=\infty$ (by the property of symmetry). Next, we use formula (8) and obtain

$$
w=K \frac{z-z_{0}}{z-\bar{z}_{0}}
$$

where $K$ is a constant multiplier. For any $K$, this function maps the upper half-plane to some circle centered at the point $w=0$. We choose $K$ so that the circle is the unit circle. For this, it is sufficient to require that the point $z=0$ (a boundary point of the region $\operatorname{Im} z>0$) is mapped to a point on the unit circle $|w|=1$. Then

$$
1=|w|=|K| \cdot\left|\frac{z_{0}}{\bar{z}_{0}}\right|, \quad \text { from which } \quad|K|=1, \quad \text { so that } \quad K=e^{i \alpha},
$$

where $\alpha$ is any real number. Thus,

$$
w=e^{i \alpha} \frac{z-z_{0}}{z-\bar{z}_{0}}
$$

Remark. Let's find the derivative $\boldsymbol{w}^{\prime}$ at the point $z_{0}=a+i b(b>0)$.

$$
w^{\prime}\left(z_{0}\right)=-\frac{i e^{i \alpha}}{2 b}, \quad \text { or } \quad w^{\prime}\left(z_{0}\right)=\frac{1}{2 b} e^{i \alpha} \cdot e^{-i \pi / 2}=\frac{1}{=} \frac{e^{i(\alpha-\pi / 2)}}{2 b}
$$

Thus, arg $\boldsymbol{w}_{?}^{\prime}\left(z_{0}\right)=\alpha-\frac{\pi}{2}$, so by the geometric meaning of the derivative, the angle of rotation of curves at the point $z_{0}$ is $\alpha-\frac{\pi}{2}$.

By the Riemann mapping theorem, there exists a unique mapping $w=w(z)$ of the half-plane $\operatorname{Im} z>0$ onto the circle $|w|<1$ which has the form (9).