Example 1. Find the function $w=f(z)$ that conformally maps the upper half-plane $\operatorname{Im} z>0$ onto the region

$$
0<\arg w<\alpha \pi, \quad \text { where } \quad 0<\alpha<2
$$

of the $w$-plane.
Solution. Since the given region is a polygon with vertices $\overline{A_{1}(w=0)}$ and $A_{2}(w=\infty)$, the solution can use the Christoffel-Schwarz integral, which defines the desired function as

$$
w=f(z)=C_{1} \int_{0}^{z}\left(\tau-a_{1}\right)^{\alpha-1} d \tau+C_{2}
$$

Assume that on the $O x$ axis of the $z$-plane, the vertices $A_{1}$ and $A_{2}$ correspond to points $a_{1}$ and $a_{2}$ such that

$$
a_{1}(z=0) \rightarrow A_{1}(w=0), \quad a_{2}(z=\infty) \rightarrow A_{2}(w=\infty)
$$

Then

$$
w=f(z)=C_{1} \int_{0}^{z} \tau^{\alpha-1} d \tau+C_{2}=\frac{C_{1}}{\alpha} z^{\alpha}+C_{2}
$$

From the condition $w=0$ when $z=0$, we get $C_{2}=0$. The function $w=\frac{C_{1}}{\alpha} z^{\alpha}$ is defined up to a constant factor, which determines a similarity transformation. This arbitrariness is associated with the fact that condition (1) specifies the correspondence of only two boundary points, while the function performing the conformal mapping is uniquely determined by the correspondence of three pairs of boundary points. If we require the correspondence of one more pair of points, for example, $w=1$ and $z=1$, then from this condition we find $1=\frac{C_{1}}{\alpha} 1^{\alpha}$, from which $C_{1}=\alpha$.

Thus, the function $w=z^{\alpha}$ conformally maps the upper half-plane $\operatorname{Im} z>0$ onto the given region of the $w$-plane.