Example 13. Map the unit circle $|z|<1$ onto the unit circle $|\boldsymbol{w}|<1$.
Solution. Let the desired linear fractional transformation $w=w(z)$ map the point $z_{0}$, located inside the circle $|z|<1$, to the center of the circle $|w|<1$, so that $w\left(z_{0}\right)=0$. Then the point $z_{0}^{*}=\frac{1}{\bar{z}_{0}}$, symmetric to the unit circle $|z|=1$, will map to the point $\infty$, i.e., $w\left(z_{0}^{*}\right)=\infty$. Then, by formula (8), we get

$$
w=K \frac{z-z_{0}}{z-\frac{1}{\bar{z}_{0}}}, \quad \text { or } \quad w=K_{1} \frac{z-z_{0}}{1-z \bar{z}_{0}}
$$

where $K_{1}=-K \bar{z}_{0}$ is some constant complex number.

We will choose the constant $K_{1}$ so that the circle in the $w$-plane is the unit circle. For this, it is sufficient to require that the point $z=1$ maps to a point on the unit circle $|w|=1$. Then we get

$$
1=|\boldsymbol{w}|=\left|K_{1}\right| \cdot\left|\frac{1-z_{0}}{1-\bar{z}_{0}}\right|
$$

from which $\left|K_{1}\right|=1$, because $\left|1-z_{0}\right|=\left|1-\bar{z}_{0}\right|$. Therefore, $K_{1}=e^{i \alpha}$, where $\alpha$ is any real number.

Thus,

$$
w=e^{i \alpha} \frac{z-z_{0}}{1-z \bar{z}_{0}}
$$

where $\left|z_{0}\right|<1, \alpha$ is any real number.

Remark. Since

$$
w^{\prime}(z)=\frac{e^{i \alpha}}{1-z \bar{z}_{0}}, \text { then } \quad w^{\prime}\left(z_{0}\right)=\frac{e^{i \alpha}}{1-\left|z_{0}\right|^{2}},
$$

i.e., arg $w^{\prime}\left(z_{0}\right)=\alpha$. This means that the angle of rotation of curves at the point $z_{0}$ under the mapping (10) is $\alpha$.

According to the Riemann mapping theorem, there exists a unique mapping $w=w(z)$ of the unit circle $|z|<1$ onto the unit circle $|w|<1$ that satisfies the conditions $w\left(z_{0}\right)=0$, arg $w^{\prime}\left(z_{0}\right)=\alpha$.

Therefore, any linear fractional mapping of the unit circle $|z|<1$ onto the unit circle $\boldsymbol{w}<1$ has the form (10).