Example 3. Conformally map the upper half-plane $\operatorname{Im} z>0$ onto a polygon in the $w$-plane (Fig. 36) such that

$$
w\left(A_{1}=0, A_{2}=1, A_{3}=\infty\right)
$$

corresponds to

$$
z\left(a_{1}=0, a_{2}=1, a_{3}=\infty\right)
$$
Solution. Consider the given region of the plane $\boldsymbol{w}$ as the interior of a "triangle" with vertices $A_{1}=0, A_{2}=1, A_{3}=\infty$ and angles at these vertices

$$
\alpha_{1} \pi=\frac{3}{2} \pi, \quad \alpha_{2} \pi=\frac{\pi}{2}, \quad \alpha_{3} \pi=-\pi
$$

From this, we have

$$
\alpha_{1}=\frac{3}{2}, \quad \alpha_{2}=\frac{1}{2}, \quad \alpha_{3}=-1 \quad\left(\alpha_{1}+\alpha_{2}+\alpha_{3}=1\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-149.jpg?height=632&width=572&top_left_y=185&top_left_x=417)

Fig: 36

Using these data, we construct the following table:

| $n$ | $A_{n}$ | $a_{n}$ | $\alpha_{n}$ |
| :---: | :---: | :---: | ---: |
| 1 | 0 | 0 | $\frac{3}{2}$ |
| 2 | 1 | 1 | $\frac{1}{2}$ |
| 3 | $\infty$ | $\infty$ | -1 |

In this case, the Christoffel-Schwarz integral will take the form

$$
\begin{aligned}
w & =\tilde{C}_{1} \int_{0}^{2}(\tau-0)^{3 / 2-1}(\tau-1)^{1 / 2-1} d \tau+C_{2}=\tilde{C}_{1} \int_{0}^{2} \tau^{1 / 2}(\tau-1)^{-1 / 2} d \tau+C_{2}= \\
& =\tilde{C}_{1} \int_{0}^{2} \frac{\sqrt{\tau} d \tau}{\sqrt{\tau-1}}+C_{2}=C_{1} \int_{0}^{2} \frac{\sqrt{\tau}}{\sqrt{1-\tau}} d \tau+C_{2} \quad\left(C_{1}=\frac{\widetilde{C}_{1}}{i}\right)
\end{aligned}
$$

By the substitution $\sqrt{\tau}=t, \tau=t^{2}, d \tau=2 t d t$, the obtained integral is reduced to the integral

$$
\begin{aligned}
\int_{0}^{z} \frac{\sqrt{\tau}}{\sqrt{1-\tau}} d \tau & =2 \int_{0}^{\sqrt{2}} \frac{t^{2} d t}{\sqrt{1-t^{2}}}=2 \int_{0}^{\sqrt{2}}\left(\frac{1}{\sqrt{1-t^{2}}}-\sqrt{1-t^{2}}\right) d t= \\
& =2 \arcsin \sqrt{z}-2\left(\frac{1}{2} \sqrt{z-z^{2}}+\frac{1}{2} \arcsin \sqrt{z}\right)=\arcsin \sqrt{z}-\sqrt{z-z^{2}}
\end{aligned}
$$

Thus,

$$
w=C_{1}\left(\arcsin \sqrt{z}-\sqrt{z-z^{2}}\right)+C_{2}
$$

Using the specified correspondence of points, we find the constants $C_{1}$ and $C_{2}$. We have

$$
\left\{\begin{array}{l}
0=C_{1}(\arcsin 0-0)+C_{2} \\
1=C_{1}(\arcsin 1-0)+C_{2}
\end{array}\right.
$$

from which $C_{1}=\frac{2}{\pi}, C_{2}=0$.

In the end, we get

$$
w=\frac{2}{\pi}\left(\arcsin \sqrt{z}-\sqrt{z-z^{2}}\right)
$$