Solution. Sector $00$ (Fig. 19, b). The point $z_{1}=e^{i \pi / 8}$ will map to the point $t_{1}=z_{1}^{4}=i$, and $z_{2}=0$ will map to the point $t_{2}=0$.

Then we map the upper half-plane $\operatorname{Im} t>0$ to the unit disk $|w|<1$ such that the point $t_{1}=i$ maps to the center of the disk (Fig. 19, c). Using formula (9), we get

$$
w=e^{i \varphi} \frac{t-i}{t+i}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-134.jpg?height=380&width=1076&top_left_y=145&top_left_x=92)

Fig. 19

The requirement that the point $t_{2}=0$ maps to the point $w_{2}=1$ gives $e^{i \varphi}=-1$. Substituting $e^{i \varphi}=-1$ and $t=z^{4}$ into the expression for $w$, we finally get

$$
w=-\frac{z^{4}-i}{z^{4}+i}
$$
Example 15. Map the sector $0<\arg z<\frac{\pi}{4}$ onto the unit disk $|w|<1$ such that the point $z_{1}=e^{i \pi / 8}$ is mapped to the center $w_{1}=0$, and the point $z_{2}=0$ is mapped to the point $w_{2}=1$.