Schlicht Mappings

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

A one-to-one conformal mapping which maps an open circular disc onto itself, with the center a fixed point, is a rotation.

Let two open annular regions each bounded by two concentric circles have the same outer boundary and be related by a one-to-one and conformal mapping. Also let the mapping be such that a circle of one region concentric with the boundary corresponds to a curve of the other region traversed in the same sense. Then the two annular regions are identical and the conformal mapping is a rotation.

There do not exist two distinct functions $w = f(z)$ which map the unit circle $\left|z\right| < 1$ in a schlicht manner onto a given region $\mathfrak {R}$, where the origin $z = 0$ is mapped onto a given point $w = w_{0}$ of $\mathfrak {R}$ and $f'(0) > 0$.

The only functions which map the unit circle $\left|z\right| < 1$ in a schlicht manner onto itself are the bilinear functions $e^{i\alpha } \dfrac {z-z_{0}}{1-\overline{z_{0}}z}$, $\alpha$ real, $\left|z_{0}\right| < 1$.

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.

Calculate the inner radius $r_{a}$ and the outer radius $\bar{r}$ of the circle $\left|z\right| = \rho$, $\rho > 0$, $\left|a\right| < \rho$.

Calculate the inner radius $r_{a}$ of the exterior of the circle $\left|z\right| > \rho$, $a$ finite, $\left|a\right| > \rho$ and the limit $\lim_{a \to \infty } r_{a}$.

Calculate $r_{a}$ for the sector $0 < \arg z < \vartheta_{0}$, $0 < \vartheta_{0} \leq 2\pi$.

Under a similarity transformation $z' = hz+k$ with an enlargement ratio $\left|h\right|$ different from zero the inner and the outer radius are magnified in the ratio $1:\left|h\right|$. I.e. if the region $\mathfrak {R}$ goes over into $\mathfrak {R}'$ and the finite image point $a$ into $a' = ha+k$ then between the inner radius $r_{a'}$ of $\mathfrak {R}'$ with respect to $a'$ and the inner radius $r_{a}$ of $\mathfrak {R}$ with respect to $a$ there exists the following relation:

There is a similar relation for the outer radius.

The outer radius of a line segment of length $l$ is

The outer radius of an ellipse whose axes are of length $2a$ and $2b$, respectively, is

Calculate the inner radius of an open half-plane with respect to a point lying within it whose distance from the boundary is $d$.

Let $r_{a}$ be the inner radius of a region $\mathfrak {R}$ of the $z$-plane, and let the schlicht mapping

map $\mathfrak {R}$ onto a region $\mathfrak {R}'$ of the $z'$-plane. Denoting by $r_{b}'$ the inner radius of the region $\mathfrak {R}'$ with respect to the image point $z'=b$, we have

A similar result holds for the outer radius in the case of schlicht mappings of the form

Calculate the outer radius $\bar{r}$ of the curve which is formed by adding to the circle $\left|z\right|=1$ the two real line segments $1 < z \leq a_{1}$, $-a_{2} \leq z < -1$, $a_{1} > 1$, $a_{2} > 1$.

Calculate the inner radius of an infinite parallel strip of width $D$ with respect to a point lying within it whose least distance from the boundary is $d$, $2d \leq D$.

Let $a$ be an arbitrary finite point of the region $\mathfrak {R}$ in the $z$-plane. The rational linear transformation $z' = \dfrac {1}{z-a}$ maps $\mathfrak {R}$ into a region $\mathfrak {R}'$ containing the point at infinity. The inner radius of $\mathfrak {R}$ with respect to $a$ is equal to the reciprocal of the outer radius of the complement of $\mathfrak {R}'$.

Consider the region $\mathfrak {R}$ which is obtained from the unit circle $\left|z\right| < 1$ by removing the two real line segments $b_{1} \leq z < 1$, $-1 < z \leq -b_{2}$, $0 < b_{1} < 1$, $0 < b_{2} < 1$. Calculate the inner radius of $\mathfrak {R}$ with respect to the origin.

The inequalities

determine a domain $\mathfrak {C}$ in the $z$-plane bounded by two circular arcs and with two corners at $z_{1}$ and $z_{2}$ at which the sides meet at an angle of $\vartheta_{2}-\vartheta_{1}$. What is the outer radius of $\mathfrak {C}$? Note the special cases $\vartheta_{1}+\vartheta_{2} = 2\pi$; $\vartheta_{1} = \vartheta_{2}$; $\vartheta_{1} = \pi /2$, $\vartheta_{2} = \pi$.

Let $a$ and $b$ be points of the region $\mathfrak {R}$ and let $f(z)$ denote the normed mapping function associated with the central point $b$. Then

Derive the normed mapping function of the whole plane slit along the real segment $\tfrac {1}{4} \leq z < +\infty$ associated with the origin and investigate the variation of the inner radius $r_{a}$ as $a$ runs through the real values from $-\infty$ to $\tfrac {1}{4}$.

Let $L$ be an analytic curve (i.e. let the normed mapping function associated with an arbitrary interior point of $L$ admit continuation across $L$ and assume different values for distinct points of $L$). If the central point $a$ in the interior of $L$ converges to a point of $L$, then the inner radius converges to $0$.

Let $a$ be a point of the region $\mathfrak {R}$ such that at $a$ we obtain a relative maximum of the inner radius $r_{a}$. Then the normed mapping function associated with $a$

is such that $c_{2} = 0$.

What is the locus of those points $a$ of the upper half-plane $\Im z > 0$ with respect to which the inner radius $r_{a}$ of $\Im z > 0$ is constant?

Let $\mathfrak {R}$ be a finite region in the $z$-plane, $a$ an arbitrary point of $\mathfrak {R}$, and $n$ a positive integer. By means of the transformation $z' = \sqrt[n]{z-a}$ there corresponds to $\mathfrak {R}$ a region $\mathfrak {R}'$ ``$n$-fold symmetric'' with respect to the origin. A point $z'$ belongs to $\mathfrak {R}'$ if and only if $z'^{n}+a$ is a point of $\mathfrak {R}$. Between the inner radii $r_{a}$ of $\mathfrak {R}$ with respect to $a$ and $r_{0}'$ of $\mathfrak {R}'$ with respect to $z'=0$ we have the relation

A similar result holds for the outer radii in the case of the transformation $z' = \sqrt[n]{z}$ if the origin belongs to the complement of $\mathfrak {R}$.

Let $n$ be a positive integer. Cut the plane along the $n$ rays

and calculate $r_{a}$ for the resulting slit region. To what limiting value does $r_{a}$ tend when $a$ approaches along the line segment $0 \leq \left|z\right| < \sqrt[n]{\tfrac {1}{4}}$, $\arg z = \dfrac {2\pi \nu }{n}$ the end point of the corresponding cut?

Calculate the outer radius of the slit region whose boundary consists of the two straight lines $-\alpha \leq z \leq \alpha$, $-\beta \leq iz \leq \beta$; $\alpha > 0$, $\beta > 0$.

Let $\mathfrak {R}$ be an arbitrary simply-connected region of the $z$-plane, and $a$ a finite point of $\mathfrak {R}$. The normed mapping function $f(z)$ associated with $a$ and the inner radius $r_{a}$ have the following minimum property: Of all the functions of the form

which are regular in $\mathfrak {R}$, $\left|F(z)\right|$ has the least maximum value in $\mathfrak {R}$. This ``minimum maximorum'' is equal to $r_{a}$. More precisely: If $M$ is the supremum of $\left|F(z)\right|$ in $\mathfrak {R}$ then

The equality sign occurs if and only if $F(z) = f(z)$.

Let $\mathfrak {R}$ be a region symmetric with respect to the real axis, $a$ real. Then the power series expansion of the normed mapping function associated with $a$ in powers of $z-a$ has only real coefficients. A similar result holds for the normed mapping function associated with the point at infinity as central point, if $\mathfrak {R}$ contains the point at infinity.

Let $\mathfrak {R}$ be a region which is symmetric with respect to the origin. Then the power series expansion of the normed mapping function associated with the origin contains only odd powers of $z$. A similar result holds for the normed mapping function associated with the point at infinity, if $\mathfrak {R}$ contains the point at infinity.

Let $\mathfrak {R}^{*}$ be a proper sub-region of $\mathfrak {R}$ and $r_{a}$ and $r_{a}^{*}$ the inner radii of $\mathfrak {R}$ and $\mathfrak {R}^{*}$ respectively, associated with the point $a$ of $\mathfrak {R}^{*}$. Then

A similar result holds for the outer radius.

Let $\mathfrak {R}$ be an arbitrary region, $a$ a finite point of $\mathfrak {R}$, $r$ the radius of the largest open circular region with center $a$ contained in $\mathfrak {R}$ and $R$ the radius of the smallest open circular region with center $a$ containing $\mathfrak {R}$. Then $r > 0$, $R \geq r$, $R$ finite or infinite, and

The equality sign holds only if $\mathfrak {R}$ is an open circular region with center $a$.

Let the region $\mathfrak {R}$ contain the point at infinity, and let $\mathfrak {D}$ be the complement of $\mathfrak {R}$. We denote by $r$ the radius of the largest closed circular region contained in $\mathfrak {D}$ and by $R$ the radius of the smallest closed circular region containing $\mathfrak {D}$. Then $r \geq 0$, $R \geq r$, $R$ finite. Further

The equality sign holds only if $\mathfrak {D}$ is a closed circular domain (closure of a non-empty connected open set).

Let $l$ denote the length of a closed continuous curve $L$ without double points. Then

where $a$ denotes an arbitrary point in the interior of $L$, $r_{a}$ the corresponding inner radius of $L$ and $\bar{r}$ the outer radius of $L$. In both cases equality is attained only for a circle, and in the first case only for a circle with center $a$.

Let $l$ denote the length of the perimeter of an ellipse with unequal semi-axes $a$ and $b$. Give two different proofs (with and without the use of conformal mapping) for the inequality

Let $l$ denote the length of the perimeter of an ellipse with unequal semi-axes $a$ and $b$. Give two different proofs (with and without the use of conformal mapping) for the inequality

Let the inner measure of the region $\mathfrak {R}$ be $\left|\mathfrak {R}\right|_{i}$, and let $a$ be an arbitrary finite point of $\mathfrak {R}$. Then

The equality sign holds only for the open disc $\left|z-a\right| < r_{a}$.

Let the complement $\mathfrak {D}$ of the region $\mathfrak {R}$ containing the point at infinity be of outer measure $\left|\mathfrak {D}\right|_{e}$. Then

The equality sign holds only if $\mathfrak {D}$ is a circular domain.

Let the outer radius and the inner radius (with respect to an arbitrary interior point $a$) of a simple closed continuous curve $L$ be $\bar{r}$ and $r_{a}$ respectively. Then

The equality sign holds if and only if $L$ is a circle and $a$ lies at the center of $L$.

Let $L$ be a simple closed continuous curve containing the origin, $\bar{r}$ the outer radius of $L$ and $r_{0}$ the inner radius of $L$ associated with the origin. Let $P(z)$ be a polynomial whose lowest order term is $a_{k}z^{k}$ and whose highest order term is $a_{n}z^{n}$, $P(z) = a_{k}z^{k}+a_{k+1}z^{k+1}+\cdots +a_{n}z^{n}$, and let $M$ denote the maximum of $\left|P(z)\right|$, as $z$ describes the curve $L$. Then

The equality sign holds only if $L$ is a circle with center at the origin and the polynomial in question is a power of $z$ multiplied by a constant.

Let the function $f(z)$ be regular and have positive real part in the interior of the simple closed continuous curve $L$, and be continuous in the closed interior of $L$. If the real part of $f(z)$ vanishes on an arc $L'$ of $L$ then the imaginary part of $f(z)$ varies always in the same sense, namely monotonically decreasing as $z$ describes the arc $L'$ in the positive sense.

Map the strip $0 < \Im z < D$ onto the circle $\left|w\right| < 1$ in such a way that the point $z=i$ corresponds to the center of the circle $w=0$ ($D > 1$). How large is the arc of the boundary of the circle $\left|w\right|=1$ which in this mapping corresponds to the real axis $\Im z = 0$? (Obvious for $D=2$ and $D=\infty$.) How does the arc in question vary as $D$ increases?

Let two simple closed continuous curves $L_{1}$ and $L_{2}$ have a finite number of common arcs and let the interior of $L_{1}$ be contained in the interior of $L_{2}$. ($L_{1}$ consists of an even number of arcs which run alternately in the interior and on the boundary of the region enclosed by $L_{2}$.)

Map first the interior of $L_{1}$ and then the interior of $L_{2}$ onto the interior of one and the same circle, so that in both cases the same point $O$ in the interior of $L_{1}$ goes over into the center of the circle. The two mappings associate with the distinct segments of $L_{1}$ and $L_{2}$ well-defined circular arcs as their images (see the last remark preceding problem 97).

Prove that the length of the image of any common arc of $L_{1}$ and $L_{2}$ will be smaller in the mapping of the smaller region (bounded by $L_{1}$) than in the mapping of the larger region (bounded by $L_{2}$). Example: problem 130.

Find an interpretation in electrostatics of problem 131.

Let two simple closed, continuous curves $L_{1}$ and $L_{2}$ have only finitely many common points, and let the two regions enclosed by them have a common region $\mathfrak {T}$. Denote by $O$ a point of $\mathfrak {T}$ and by $\mathfrak {T}^{*}$ that connected part of $\mathfrak {T}$ which contains $O$. Such arcs of $L_{1}$ and $L_{2}$ as belong to the boundary of $\mathfrak {T}^{*}$ are termed visible'' from $O$, such arcs as do not belong to the boundary of $\mathfrak {T}^{*}$ are termed hidden'' from $O$. (The terms visible'' and hidden'' have their usual meaning if $L_{1}$ and $L_{2}$ are star-shaped with respect to $O$; see Problem 109 of Part III.)

Map first the interior of $L_{1}$ and then the interior of $L_{2}$ onto the interior of the unit circle, in both cases in such a manner that the center of the circle is the image of $O$. The images of the visible'' parts of $L_{1}$ take up a larger percentage of the boundary of the circle than the images of the hidden'' parts of $L_{2}$. (To visualize this imagine $L_{2}$ as a circle with centre $O$.)

Let $\mathfrak {D}$ be a domain, $\zeta$ an interior point of $\mathfrak {D}$, and $\mathfrak {B}$ the set of the boundary points of $\mathfrak {D}$ whose distance from $\zeta$ is not greater than $\rho$. Let those arcs of the circle of radius $\rho$ and with centre $\zeta$ that do not belong to $\mathfrak {D}$ have total length $\rho \Omega$.

Let the function $f(z)$ be regular and single-valued in the interior of $\mathfrak {D}$ and continuous on the boundary of $\mathfrak {D}$. Further, let $\left|f(z)\right| \leq a$ at points of $\mathfrak {B}$, $\left|f(z)\right| < A$ at the remaining boundary points of $\mathfrak {D}$, $a < A$. Then

The simply connected region $\mathfrak {R}_{n}$ lying in the $z$-plane is subject to the following conditions:

1. $\mathfrak {R}_{n}$ is contained in the circle $\left|z\right| < a$, $a > 1$.

2. $\mathfrak {R}_{n}$ contains the circle $\left|z\right| < 1$.

3. $\mathfrak {R}_{n}$ contains the arc determined by the relations

   $$\left|z\right| = 1, \qquad -\alpha _{n} < \arg z < \alpha _{n}, \qquad 0 < \alpha _{n} < \pi .$$

4. The arc on the periphery of the unit circle complementary to the arc in (3), for which

   $$\left|z\right|=1, \qquad \alpha _{n} \leq \arg z \leq 2\pi -\alpha _{n},$$

   belongs to the boundary of $\mathfrak {R}_{n}$. Let $\mathfrak {R}_{n}$ be mapped one-to-one and conformally onto the unit circle by means of $w=f_{n}(z)$, and in such a manner that $f_{n}(0)=0$, $f_{n}'(0) > 0$.

If the unit circle is gradually cut off from the remaining part of $\mathfrak {R}_{n}$ as $n$ increases, i.e. if $\lim_{n\to \infty } \alpha_{n} = 0$, then for $\left|z\right| < 1$

independently of the behaviour for large $n$ of that remaining part of $\mathfrak {R}_{n}$.

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.)