Introductory Material

Let the polynomial

be schlicht (univalent) in the unit circle $\left|z\right| < 1$. Then $n \left|a_{n}\right| \leq 1$.

Let the function $w = f(z)$ effect a schlicht mapping of the unit circle $\left|z\right| < 1$ onto a region (non-empty connected open set) $\mathfrak {R}$ and let $\varphi (w)$ be schlicht in $\mathfrak {R}$. Then $\varphi [f(z)]$ is schlicht in $\left|z\right| < 1$.

Let the function $f(z)$ be schlicht in the unit circle $\left|z\right| < 1$ and let $f(0) = 0$. Then the function

with a fixed branch of the square root, is also schlicht in $\left|z\right| < 1$. A similar result holds for $\sqrt[n]{f(z^{n})}$, where $n$ is a positive integer.

The function $\varphi (z)$ in problem 79 is the most general odd function that induces a schlicht mapping of the unit circle $\left|z\right| < 1$. More precisely: If $\varphi (z)$ is an odd function schlicht for $\left|z\right| < 1$, then there exists a function $f(z)$ which is schlicht for $\left|z\right| < 1$ from which $\varphi (z)$ is derived as in problem 79.

Let the open circular disc $\mathfrak {C}$ be mapped one-to-one and conformally onto the region $\mathfrak {R}$. In particular, let the concentric circular disc $\mathfrak {c}$ contained in $\mathfrak {C}$ be mapped onto the sub-region $\mathfrak {r}$ of $\mathfrak {R}$. Denoting the respective areas by $\left|\mathfrak {R}\right|, \left|\mathfrak {C}\right|, \left|\mathfrak {c}\right|, \left|\mathfrak {r}\right|$, we have

Let the open circular disc $\mathfrak {C}$ be mapped one-to-one and conformally onto the region $\mathfrak {R}$. In particular let the center $\mathfrak {c}$ of $\mathfrak {C}$ be mapped into the point $\mathfrak {r}$ of $\mathfrak {R}$. Denote by $a^{2}$ the area enlargement ratio at $\mathfrak {c}$ (see Volume I, p. 117). Then we have (using the notation of problem 81)

Let the open annular region $\mathfrak {A}$ between two concentric circles be mapped one-to-one and conformally onto the doubly connected region $\mathfrak {S}$. Let the smallest open circular disc which contains $\mathfrak {A}$ be $\mathfrak {C}$. Those points of $\mathfrak {C}$ not belonging to $\mathfrak {A}$ form a closed circular disc $\mathfrak {c}$. Similarly let $\mathfrak {R}$ be the smallest simply connected region containing $\mathfrak {S}$, and let $\mathfrak {r}$ be the set of all points of $\mathfrak {R}$ which do not belong to $\mathfrak {S}$ ($\mathfrak {r}$ is closed). Finally let a circle concentric with $\mathfrak {c}$ and $\mathfrak {C}$ contained in $\mathfrak {A}$ and its image curve in $\mathfrak {S}$ be traversed in the same sense. Then we have the relation between the areas