Existence of the Mapping Function

In the mapping induced by the formula

to a point in the $z$-plane there correspond in general two points in the $w$-plane. Which $z$-points form an exception to this?

Let $\mathfrak {R}$ be a simply-connected region in the $z$-plane, which contains the point $z = 0$ and is contained in the circle $\left|z\right| < 1$. Let $a$, $\left|a\right| < 1$, be the boundary point of $\mathfrak {R}$ closest to the point $z = 0$. By means of the mapping of problem 88 there correspond to the region $\mathfrak {R}$ two regions of the $w$-plane, of which one $\mathfrak {R}^{*}$ contains the point $w=0$ and the other $\mathfrak {R}^{**}$ does not. $\mathfrak {R}^{*}$ and $\mathfrak {R}^{**}$ have no common points; however they have at least one common boundary point and both lie in the interior of the unit circle.

Investigate the Koebe image region of the slit region which arises from the unit circle $\left|z\right| < 1$ by deleting (``cutting along'') the line segment $a \leq z \leq 1$, $0 < a < 1$. In particular determine the boundary point which is nearest to the origin.

Investigate the Koebe image region of the circle $\left|z\right| < a$, $0 < a < 1$ and determine the boundary points which are nearest to and furthest from the origin.

Every point of the Koebe image region of $\mathfrak {R}$ is at a greater distance from the origin than the corresponding point of $\mathfrak {R}$.

Let $\mathfrak {R}_{1}$ be the Koebe image region of $\mathfrak {R}$, $\mathfrak {R}_{2}$ that of $\mathfrak {R}_{1}$, $\mathfrak {R}_{3}$ that of $\mathfrak {R}_{2}, \ldots$. Denote by $a, a_{1}, a_{2}, a_{3}, \ldots$ the nearest point from the origin of $\mathfrak {R}, \mathfrak {R}_{1}, \mathfrak {R}_{2}, \mathfrak {R}_{3}, \ldots$ respectively. Then we have

(Continued from problem 93.) By the Koebe mapping $\mathfrak {R}$ is mapped one-to-one and conformally onto $\mathfrak {R}_{1}$, $\mathfrak {R}_{1}$ onto $\mathfrak {R}_{2}, \ldots , \mathfrak {R}_{n-1}$ onto $\mathfrak {R}_{n}$. Let the resulting mapping of $\mathfrak {R}$ onto $\mathfrak {R}_{n}$ be expressed by the analytic function $f_{n}(z)$ ($z$ runs through the points of $\mathfrak {R}$). Then $f_{n}(0) = 0$, $0 < f_{n}'(0) < 1/\left|a\right|$. Express $f_{n}'(0)$ in terms of $a, a_{1}, a_{2}, \ldots , a_{n-1}$.

Denoting by $a, a_{1}, a_{2}, \ldots , a_{n}, \ldots$ the points introduced in problem 93, prove by means of problem 94 that

(Continuation of problems 93, 94.) The limit $\lim \limits_{n \to \infty } f_{n}(z) = f(z)$ exists at every point $z$ of $\mathfrak {R}$ and the convergence is uniform in every interior domain of $\mathfrak {R}$. The limit function $f(z)$ maps $\mathfrak {R}$ schlicht onto the interior of the unit circle.