The Koebe Distortion Theorem and Related Topics

Let the function

be regular for $\left|z\right| > 1$, and map this region (the exterior of the unit circle) schlicht onto a region $\mathfrak {R}$ containing the point at infinity. Then

In particular $\left|b_{1}\right| \leq 1$ and in this last inequality the equality sign holds if and only if $\mathfrak {R}$ is the whole plane cut along a line segment of length $4$.

(Continued from problem 136.) For $\left|z\right| > 1$

The equality sign holds at the point $\dfrac {\rho }{\varepsilon }$, $\left|\varepsilon \right|=1$, $\rho >1$, if and only if $g(z)$ has the form

What is the nature of the image region in this case?

Let the simply connected region $\mathfrak {R}$ contain the point at infinity and be symmetric with respect to a point $P$. Then $P$ is the conformal center of gravity of $\mathfrak {R}$.

(Continuation of problem 136.) If the region $\mathfrak {R}$ does not contain the origin, then the conformal centre of gravity lies inside the circle of radius $2$ centered at the origin, that is $\left|b_{0}\right| \leq 2$. The equality sign holds if and only if $\mathfrak {R}$ is the whole plane cut along a straight line segment of length $4$ starting from the origin.

(continued). The distance $d$ of an arbitrary boundary point of $\mathfrak {R}$ from the conformal center of gravity is at most $2$. In fact $d<2$ except in the case of the particular mapping specified in problem 136.

(continued). The maximum distance $D$ between the boundary points of $\mathfrak {R}$ (diameter of the boundary of $\mathfrak {R}$) lies between the limits $2$ and $4$, i.e.

The equality sign holds for the lower estimate only if $\mathfrak {R}$ is the exterior of a circle of radius $1$, and for the upper estimate only if $\mathfrak {R}$ is the slit region of problem 136.

Of all continuous arcs which connect two fixed points the straight line has the smallest outer radius.

Let the region $\mathfrak {R}$ in problem 136 have the origin as its conformal center of gravity. Then

The equality sign holds only for the mapping $w = z + \dfrac {e^{i\alpha }}{z}$, $\alpha$ real.

(continued). By the mapping in question no point $z$ can be displaced by more than $\dfrac {3}{\left|z\right|}$ from its original position. I.e.

Investigate the displacement (see problem 144) for the mapping of the exterior of the unit circle onto a slit region of a particular kind, which is bounded by the horseshoe-shaped curve that consists of three straight line segments and connects the four points

in that order; $a>0$, $\delta >0$. Note especially the case where $a$ tends to $2$, $\delta$ tends to $0$, and show by means of this example that the constant $3$ in problem 144 cannot be replaced by any smaller constant.

Let the function

be regular and schlicht inside the unit circle $\left|z\right|<1$. Then

The equality sign holds only for functions of the form

Let the function

be schlicht in the interior of the circle $\left|z\right|<R$. If the image region $\mathfrak {R}$ in the $w$-plane does not contain the point $w=\infty$ then it completely contains the open circular disc $\left|w\right|<\dfrac {R}{4}$. In other words, if $d$ denotes the least distance of the boundary of $\mathfrak {R}$ from the origin $z=0$, then $d \geq \dfrac {R}{4}$. Moreover $d > \dfrac {R}{4}$ unless $\mathfrak {R}$ is the whole plane slit along the segment $\arg w = \text{const.}$, $\dfrac {R}{4} \leq \left|w\right| < +\infty$.

We have the following refinement of problem 147: The shortest boundary distance $d$ satisfies the inequality

(Continuation of problem 147.) We term the straight line connecting two boundary points of $\mathfrak {R}$ a principal chord of $\mathfrak {R}$ if it passes through the origin. Every principal chord of $\mathfrak {R}$ has a length of at least $R$. The extreme case occurs if and only if $\mathfrak {R}$ is the whole plane slit along the two straight line segments $w = \pm \left|w\right| e^{i\alpha }$, $\alpha$ real, $\dfrac {R}{2} \leq \left|w\right| < +\infty$. (See problem 116.)

(Continuation of problem 146.) In the interior of the unit circle $\left|z\right|<1$ we have the inequality

The equality sign holds only if the image region is the plane slit along a straight line.

(The Koebe distortion theorem.) Let the function

be regular and schlicht inside the unit circle $\left|z\right|<1$. Also let $r$ be a positive number $r<1$. Then in the circular domain $\left|z\right| \leq r$ we have the inequalities

The equality sign can occur only for the mappings

(continued). In the circular disc $\left|z\right| \leq r$ we have

The equality sign as in problem 151.

We have the following refinement of problem 152: For $\left|z\right| \leq r$ we have

The equality sign as in problem 151. (Generalization of problem 148.)

For odd functions $f(z) = -f(-z)$ we can obtain the following refinement of problem 152:

The equality sign holds only for $f(z) = \dfrac {z}{1-z^{2}}$.

(continued). We have

(Continuation of problem 152.) Let $n$ be a positive integer. There exists a function $\omega_{n}(r)$ of $n$ and $r$ only, $0 \leq r < 1$, such that for any function $f(z)$ of the type mentioned in problem 151 we have in the circular region $\left|z\right| \leq r$ the inequality

(continued). Let $n$ be a positive integer, $n \geq 2$. There exists a constant $\omega_{n}$ depending only on $n$ such that for any function $f(z)$ of the type mentioned in problem 151 we have the inequality

Let $\omega_{n}$ be the smallest constant of this kind. Then

(This bound is less than $\dfrac {e^{2}}{4}n^{2}$.)

(continued). We have

(continued). We have (refinement of problem 157)

For functions

which are schlicht and in absolute value less than $M$ inside the unit circle $\left|z\right|<1$ the estimate of problem 146 can be refined as follows: $M \geq 1$ ($M=1$ only for $f(z)=z$) and

When does the equality sign occur?

Let the function

be regular inside the unit circle $\left|z\right|<1$ and map the unit circle onto a region which is star-shaped with respect to the origin (see Problem 109 of Part III). Then

The equality sign holds if and only if

Let the function

be regular inside the unit circle $\left|z\right|<1$ and map the unit circle onto a convex region (see Problem 108 of Part III). Then

The equality sign holds if and only if

the image region in that case is a half-plane containing the origin, whose boundary line is at a distance $\tfrac {1}{2}$ from the origin.

If the function $f(z)$ is regular and schlicht inside the unit circle $\left|z\right|<1$, then the image of the circular region $\left|z\right|<r$ is a convex region provided $r \leq 2-\sqrt{3} = 0.26\ldots$. This number (the curvature bound) cannot be replaced by any smaller number.