Chapter 2 UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

2.1 Univalent conformal mapping of a doubly connected domain onto an annulus

2.1.1 page 206

Let us now begin with the simplest case of the problem posed, namely, the case of doubly connected domains. Let us show that every doubly connected domain can be mapped univalently onto some circular annulus, whose boundary circles may degenerate to points.

Let \(B\) denote a doubly connected domain in the \(z\)-plane.

If it has an isolated boundary point \(z_0\), then, by adjoining it to the domain \(B\), we obtain a simply connected domain that we can then map univalently either onto the disk \(I (I < I)\) or onto the \((-\)plane with the point \((= \infty)\) excluded. We can do this in such a way that the point \(z = z_0\) is mapped into \((= 0)\). The domain \(B\) is thus univalently mapped either onto the annulus \(0 < |z| < I\) or onto the annulus \(0 < |z| < \infty\).

Suppose now that the boundary of the domain \(B\) consists of two continua \(K_1\) and \(K_2\). One of these, let us say \(K_1\) is necessarily bounded. The complement of \(K_1\) in the \(z\)-plane is an open set consisting of two disjoint domains. One of these domains, let us say \(B_1\) contains the domain \(B\). The domain \(B_1\) is simply connected. Therefore, it can be mapped conformally onto the disk \(|z'| < I\). Under this mapping, the continuum \(K_2\) is mapped into a continuum \(K\) contained in \(|z' | < I\), and the domain \(B\) is mapped into a domain \(B'\).

Let us now map whichever of the simply connected domains complementary to \(K\) contains \(B'\) onto the domain \(|z''| > I\) in such a way that \(z' = \infty\) is mapped into \(z'' = \infty\). Under this mapping, the circle \(|z'| = I\) is mapped into an analytic Jordan curve contained in the domain \(|z''| > I\) and the domain \(B'\) is mapped into a doubly connected domain \(B''\) that does not include \(\infty\) and that is bounded by this curve and by the circle \(|z''| = I\).

The composite of these two mappings constitutes a univalent mapping of the domain \(B\) onto the domain \(B''\). Furthermore, on the basis of \(\S 3\) of Chapter II, we conclude that this mapping sets up a one-to-one correspondence between the prime ends of the domain \(B\) and the boundary points of the domain \(B''\). Here, just as in \(\S 3\) of Chapter II, by the prime end of the domain \(B\) that corresponds to the point \(z\) on the boundary of the domain \(B''\), we mean the set of all cluster points of all sequences of points of the domain \(B\) that approach the boundary point \(z\).

2.2 Univalent mapping of a multiply connected domain onto a plane with parallel rectilinear cuts.

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Let us now investigate univalent conformal mapping of arbitrary multiply connected domains onto canonical domains of various kinds. The simplest such domain is a plane with parallel rectilinear cuts. In this case, our investigation will be based on the solution of certain extremal problems.

Lemma 2.1 Among all functions \[F(z) = z + \frac{a_1}{z}+\frac{a_2}{z^2} \cdots\] that are univalent in \(|z| > R\), the quantity \(\Re(e^{-2i\theta}a_1)\), where \(\theta\in \mathbb{R}\) is maximized by a function that maps the domain \(|z| > R\) onto the plane with a rectilinear cut making an angle \(\theta\) with the real axis. Furthermore, this maximum is achieved only by that specific function. For this function, we have \(\Re(e^{-2i\theta}) = R^2\).

Recall the Area theorem,

Theorem 2.1 (The Area Theorem) Suppose that the function \[F(\xi) = \xi+\frac{b_1}{\xi}+\cdots=\xi + \sum_{k=1}^{\infty} \frac{b_k}{ \xi^k}\] is regular in the finite plane (Complex Plane \(\mathbb{C}\)), has a pole at \(\xi = \infty\), and is univalent in the domain \(|\xi| > 1\). Then, \[ \sum_{n=1}^{\infty}n|b_n|^2\leq 1\]

Proof. We use area therom for the following functions, \[G(\xi)=\frac{1}{R}F(R\xi)=\xi+\frac{a_1}{R^2\xi}+\cdots\] that for all \(|a_1|\leq R^2\),
equality holds only when \(F(z) = z + \frac{R^2 e^{i\theta}}{z}\).

Note that \(G(\xi)\) is univalent in \(|\xi|>1\) beacuse \(F(z)\) is univalent in \(|z|>R\). \[\frac{\Re(a_1)}{R^2}=\frac{\Re(a_1)}{\Re(R^2)}=\Re\left(\frac{a_1}{R^2}\right)=\Re(b_1)\leq \cdot |b_1|\leq |b_1|^2 \leq \sum_{n=1}^\infty n|b_n|\leq 1\] Thus, \[{\Re(a_1)}\leq {R^2}\] Note that, \[\Re(e^{-2i\theta})=\cos(2\theta)\leq 1,\quad \forall \theta \in \mathbb{R}\] \[\begin{eqnarray*} \Re(e^{-2i\theta}a_1)=\Re(e^{-2i\theta})\Re(a_1)\leq \Re(a_1) \end{eqnarray*}\] By inequity in (*), \[ \Re(e^{-2i\theta} a_1) \leq \Re(a_1)\leq R^2 \] Therfore, \[\Re(e^{-2i\theta} a_1) \leq R^2\] with equality holding only for the function \(F(z) = z + \frac{R^2 e^{2i\theta}}{ z}\), which maps the domain \(|z| > R\) onto the plane with a cut at an angle \(\theta\) to the real axis.

Lemma 2.2 If the function \(\xi=F(z) := z + a_0 + \sum_{n=1}^{\infty} a_n z^n\) is univalent in \(|z| > R\), then \(|F(z) - a_0| < 2|z|\) in the domain \(|z| > R\), and the entire boundary of the image of the domain \(|z| > R\) under the mapping \(F(z)\) is contained in the disk \(|\xi-a_0| \leq 2R\).

Proof. If \(|z_0|>R\) then the function, \[\xi=F_1(z):=\frac{1}{z_0}F(z_0z)-\frac{a_0}{z_0}=\frac{1}{z_0}\left( z_0z+ a_0+ \frac{a_1}{z_0z}+\cdots\right)-\frac{a_0}{z_0}=z+\frac{a_1}{z_0^2z}+\cdot\] is univalent in \(|z|>1\). (One to one propetry is obvious,)

Recall the follwing therom. (This appears on the Goluzin page 50)
Theorem: If the function \[w=F(\xi)=\xi+a_o+ \frac{a_1}{\xi} +\cdot\] maps the domain \(\xi > 1\) univalently, then the entire boundary of its image is contained in the disk \(|w - a_o |\leq 2\).

By this therom,the entire boundary of the image of the domain \(|z| > 1\) under this function is contained in the disk \(\{|F_1(z)-0|\leq 2\}=\{|F_1(z)|\leq 2\}\). In particular,\(|F_1(1)|\leq 2\). \[\begin{eqnarray*} |F_1(1) |&\leq & 2\\ |\frac{F(z_0(1))-a_0}{z_0}| &\leq & 2\\ |F(z_0(1))-a_0| &\leq & 2|z_0| \end{eqnarray*}\] ,where \(|z_0|>R\).

That is,\(|F(z)-a_0|\leq 2|z|\) for \(|z|>R\)

Theorem 2.2 Every domain \(B\subseteq \mathbb{C}\) can be mapped univalently onto a domain \(B'\subseteq \hat{\mathbb{C}}\) such that an arbitrary continuum in the complement of the domain \(B'\) with respect to the plane is a straight line segment of given inclination \(\theta\) to the real axis. Furthermore, this mapping is such that a given point \(a\) of the domain \(B\) is mapped into \(\infty\), and the expansion of the mapping function about \(z = a\) is of the form:

\[ \frac{1}{z - a} + a_1(z- a) +\ldots ~\text{ or }~ z+\frac{a_1}{z}+\ldots\]

where \(z\) is finite or infinite.