Example 16. Find the function that maps the upper half of the circle $|z|<1$, $\text{Im } z>0$, onto the upper half-plane $\text{Im } w>0$.
Solution. The given region represents a biangle with vertices at points $z_{1}=-1$ and $z_{2}=1$ and an angle at the vertex $\alpha=\frac{\pi}{2}$ (Fig. 20,a).

The auxiliary function $t=\frac{1+z}{1-z}$ performs a conformal mapping of this biangle onto the first quadrant of the $t$-plane (Fig. 20,b). The function $w=t^{2}$ or $w=\left(\frac{1+z}{1-z}\right)^{2}$ provides the desired mapping (Fig. 20,c).

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-134.jpg?height=270&width=398&top_left_y=1239&top_left_x=138)
a)

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-134.jpg?height=266&width=287&top_left_y=1239&top_left_x=579)
b)

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-134.jpg?height=254&width=242&top_left_y=1243&top_left_x=879)
c)

Fig. 20

## 2. Radical

The function $w=\sqrt[n]{z}$, which is the inverse of the power function $z=w^{n}$, is $n$-valued, i.e., each $z=\rho e^{i \varphi}(z \neq 0$ and $z \neq \infty)$ corresponds to $n$ values of $w$ given by the formula

$$
w_{k}=\sqrt[n]{\rho}\left(\cos \frac{\varphi+2 k \pi}{n}+i \sin \frac{\varphi+2 k \pi}{n}\right), \quad k=0,1, \ldots, n-1
$$

Each of the functions $w_{k}$ is a branch of the multivalued function $w=\sqrt[n]{z}$. The point $z=0$ is a branch point of this function:

On the extended $z$-plane with any cut from $z=0$ to $z=\infty$, in particular, with a cut along the positive real axis, one can distinguish $n$ single-valued branches $w_{k}$. These branches map the extended plane with a cut along the positive real axis one-to-one onto sectors

$$
\frac{2(k-1) \pi}{n}<\arg w<\frac{2 k \pi}{n} \quad(k=1,2, \ldots, n)
$$