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Two domains \(G\) and \(G^{*}\) are said to be _conformally equivalent_ if there is a one-to-one and onto (i.e., bijective) conformal map
\[f:G\to G^{*}.\]
If such function \(f\) exists, then its inverse function will be holomorphic as well (Figure 3.24).
Figure 3.23: Non-conformality of \(z\to z^{2}\) at 0In some cases when a bijection exists it can be given by explicit formulas. For example, in the following we will see that the upper half-plane can be mapped by a holomorphic bijection to the disc, and this is given by a linear fractional transformation. If two domains are conformally equivalent, then it is not possible to distinguish between them. The main theorem regarding conformability is the following Riemann Mapping Theorem.
**Definition 106**.: A domain is called _simply connected_ if every closed path \(\gamma\) in \(G\) is homotopic to a null path in \(G\). Intuitively, this means if every closed path \(\gamma\) in the domain can be continuously shrunk to a point.
**Theorem 127**.: _(The Riemann Mapping Theorem) Two simply connected domains, neither of which is equivalent to \(\mathbb{C}\), are conformally equivalent._
This theorem, even though it is a corner stone result, is an existence result; it does not produce for us the desired function \(f\). In practice, to find a conformal mapping from a domain \(G\) onto a simpler domain \(G^{*}\), we first find the conformal map from \(G\) onto the open unit disc \(|w|<1\). We then apply the theorem once again to find a mapping \(g\) from \(G^{*}\) onto the open disc \(|w|<1\); since the theorem guarantees that \(g\) is one-to-one, it has a inverse \(g^{-1}\) that maps the open disc onto \(G^{*}\). The desired mapping from \(G\) onto \(G^{*}\) is then given by \(w=(g^{-1}\circ f)(z)\), i.e., to conformally relate the two domains, we are composing various maps to the disc and inverse maps to the disc. Note that it is not obvious that it is possible to construct a conformal mapping from a region with complicated boundary onto a nice region like the disc or vice versa. The Riemann Mapping Theorem is very striking in this context; it simply states that if \(G\) is a simply connected region with \(G\neq\mathbb{C}\), then there exists a one-to-one conformal mapping \(f\) from \(G\) onto the unit disc \(\mathbb{D}\) with \(f^{-1}:\mathbb{D}\to G\) also conformal. For the proof of the Riemann Mapping Theorem we refer the reader to [5], pp. 221-227.
Figure 3.24: Conformally equivalent domains
### The Disc and Upper Half-Plane
The unit disc denoted by \(\mathbb{D}\) is defined as \(\mathbb{D}=\{z\in\mathbb{C}:\ |z|<1\}\). The upper half-plane, which we denote by \(\mathbb{H}\), consists of those complex numbers with positive imaginary parts; that is,
\[\mathbb{H}=\{z\in\mathbb{C}:\ \mathrm{Im}(z)>0\}.\]
**Theorem 128**.: _The map \(F:\mathbb{H}\to\mathbb{D}\) defined by \(F(z)=\dfrac{i-z}{i+z}\) is a conformal map with inverse \(G:\mathbb{D}\to\mathbb{H}\) defined as \(G(z)=i\dfrac{1-w}{1+w}\)._
Proof
: Observe that \(F\) maps \(\mathbb{H}\) to \(\mathbb{D}\) because \(|i-z|<|i+z|\) and thus \(|F(z)|<1\). It is also easy to see that \(G\) maps into the upper half-plane. Take \(w=u+iv\) in \(\mathbb{D}\) and observe that we have
\[\mathrm{Im}(G(w))=\mathrm{Re}\left(\dfrac{1-w}{1+w}\right)=\dfrac{1-u^{2}-v^{ 2}}{(1+u)^{2}+v^{2}}>0.\]
Lastly,
\[G(F(z))=i\dfrac{1-\dfrac{i-z}{i+z}}{1+\dfrac{i-z}{i+z}}=z,\]
and similarly \(F(G(w))=w\).
**Remark 72**.: The above theorem at first seems surprising in the sense that \(\mathbb{H}\) is an unbounded set which is conformally equivalent to the unit disc. This is one reason why complex analysts work on studying function theory on the disc; as seen above, knowledge about the disc can be carried out to knowledge about any simply connected region. However in the theory of several complex variables, the situation is different. There are many simply connected domains in \(\mathbb{C}^{n}\) that are not conformally equivalent.
### Harmonic Functions and Conformal Maps
Laplace’s equation
\[u_{xx}+u_{yy}=0\]
is used often in electrostatics, heat conduction, and fluid flow. Harmonic functions which we introduce briefly below are used in the mathematical modeling of such phenomena. A classical problem involving Laplace’s equation is called _the Dirichlet Problem_. Suppose \(G\) is a domain in \(\mathbb{R}^{2}\) and that \(h\) is a function defined on the boundary of \(G\). The problem of finding a function \(u(x,y)\) which satisfies the Laplace’s equation in \(G\) and which equals \(h\) on \(\partial G\) (the boundary of \(G\)) is called a Dirichlet problem. Such problems arise often in the two-dimensionalmodeling of heat flow, fluid flow, and electrostatics. Solving a Dirichlet problem by conformal mapping is often used. Roughly speaking, suppose that we have a region \(G\) (a nonempty open connected set in \(\mathbb{C}\)), a one-to-one conformal mapping \(g\) from \(G\) onto the much simpler region \(D(0,1)\)
\[g:\overline{G}\to\overline{D(0;1)},\]
and \(g\) maps the boundary \(\partial G\) onto the unit circle. Since we have both \(g\) and \(g^{-1}\) are conformal (conformality of \(g^{-1}\) is a consequence of the inverse function theorem), it turns out that they transfer harmonicity backward and forward. Hence the Dirichlet problem for \(G\) is converted to a Dirichlet problem for \(D(0,1)\). After solving the Dirichlet problem in \(D(0,1)\), one transfers the solution to \(G\).
**Definition 107**.: A real-valued function \(u(x,y)\) of two variables \(x\) and \(y\) that has continuous first and second order partial derivatives in a domain \(D\) and satisfies Laplace’s equation is said to be _harmonic_ in \(G\).
The following theorem shows that real and imaginary parts of a holomorphic functions are necessarily harmonic.
**Theorem 129**.: _Let \(f(z)=u(x,y)+iv(x,y)\) be holomorphic in an open set \(G\). Then the functions \(u(x,y)\) and \(v(x,y)\) are harmonic in \(G\)._
Proof
: Since \(f\) is holomorphic it is infinitely differentiable. Thus \(u(x,y)\) and \(v(x,y)\) have partial derivatives of all orders and so the partials are continuous. In particular \(u\) and \(v\) have continuous second order partials. Since \(f\) is holomorphic, the Cauchy-Riemann equations are satisfied at every point \(z\). By differentiating the both sides of \(u_{x}=v_{y}\) and \(u_{y}=-v_{x}\) with respect to \(x\) and \(y\) we find
\[u_{xx}=v_{yx}\quad\text{and}\quad u_{yy}=-v_{xy}.\]
Since we assumed continuity, the mixed partials \(v_{xy}\) and \(v_{yx}\) are equal. Hence adding the two equations above we obtain
\[u_{xx}+u_{yy}=0.\]
This shows \(u(x,y)\) is harmonic. Similarly by taking \(u_{x}=v_{y}\), differentiating both sides with respect to \(y\) and differentiating both sides of \(u_{y}=-v_{x}\) with respect to \(x\) gives
\[u_{xy}=v_{yy}\quad\text{and}\quad u_{yx}=-v_{xx}.\]
Subtracting the last two equations will yield \(v_{xx}+v_{yy}=0\).
Now the question is if you are given a harmonic function \(u(x,y)\) in \(G\), is it possible to find another harmonic function \(v(x,y)\) so that \(u\) and \(v\) satisfy the Cauchy-Riemann equations throughout the domain \(G\)? In this case we call \(v(x,y)\) a _harmonic conjugate_ of \(u(x,y)\). By combining the functions \(u(x,y)\) and \(v(x,y)\) as \(u(x,y)+iv(x,y)\) we obtain a holomorphic function in \(G\).
**Example 102**.: The functions \(u(x,y)=x^{2}-y^{2}\) and \(v(x,y)=2xy\) are harmonic conjugates of each other, since they are the real and imaginary parts of \(f(z)=z^{2}\).
**Example 103**.: Given one of the two conjugate harmonic functions, the Cauchy-Riemann equations can be used to find the other. Consider \(u(x,y)=y^{3}-3x^{2}y\). We observe that \(u(x,y)\) is harmonic because \(u_{x}=-6xy\), \(\ u_{xx}=-6y\), \(u_{y}=3y^{2}-3x^{2}\,,\ \ u_{yy}=6y\) and hence \(u_{xx}+u_{yy}=0\). In order to find the harmonic conjugate \(v(x,y)\), note that
\[u_{x}=-6xy=v_{y}.\]
Partial integration of \(-6xy=v_{y}\) with respect to \(y\) will give \(v(x,y)=-3xy^{2}+\phi(x)\) and since \(v_{x}=-u_{y}\), we also have
\[-3y^{2}+\phi^{\prime}(x)=-(3y^{2}-3x^{2}).\]
So we obtain \(\phi^{\prime}(x)=3x^{2}\) and thus \(\phi(x)=x^{3}+C\), where \(C\) is a real constant. Therefore, the harmonic conjugate of \(u(x,y)\) is
\[v(x,y)=-3xy^{2}+x^{3}+C.\]
The corresponding function \(f=u+iv\) is \(f(z)=(y^{3}-3x^{2}y)+i(x^{3}-3xy^{2})+iC\), and one can verify that \(f(z)=i(z^{3}+C)\).
### Linear Fractional Transformations
A map of the form \(f(z)=az+b\) where \(a\) and \(b\) are complex numbers with \(a\neq 0\) is called a _linear transformation_. It is easy to see that every non-constant complex linear transformation can be described as a composition of three basic types of motion. Namely, a translation, a rotation and a magnification. To see this all we need to do is to write \(a=\rho\,e^{i\phi}\). Then
\[w=f(z)=\rho\,e^{i\phi}z+b\]
and \(e^{i\phi}\,z\) is a rotation, \(\rho\,e^{i\phi}z\) is a magnification, and \(\rho e^{i\phi}z+b\) is a translation. For example the linear transformation
\[w=f(z)=2iz+1+i=2e^{i\frac{\pi}{2}}\,z+(1+i)\]
is a rotation by \(\frac{\pi}{2}\), magnification by \(2\) and a translation by \(1+i\).
An _inversion_ is a transformation defined by
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