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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.27 Conformal maps

Definition 1.27.1. Jacobian matrix of differentiable function.

Let \(U\) be an open subset of \(\R^2\text{,}\) let \(f\colon U\rightarrow \R^2\) be a differentiable function. Writing \(f(x,y)=(u(x,y), v(x,y))\text{,}\) for all \(P=(x_0,y_0)\in U\text{,}\) we define the Jacobian matrix \(J_f(P)\) of \(f\) at \(P\) to be the matrix
\begin{equation*} J_f(P)=\begin{bmatrix} u_x(x_0,y_0)\amp u_y(x_0,y_0) \\ v_x(x_0,y_0)\amp v_y(x_0,y_0) \end{bmatrix}\text{.} \end{equation*}

Definition 1.27.2. Conformal map.

A \(2\times 2\) matrix \(A\) with real coefficients is conformal if its associated linear transformation \(T_A\colon \R^2\rightarrow \R^2\) preserves the oriented angle between any two nonzero vectors.
Let \(D\) be an open subset of \(\R^2\text{.}\) A real differentiable function \(f\colon D\rightarrow \R^2\) is conformal at a point \(P=(x_0,y_0)\in D\) if the Jacobian matrix \(J_f(P)\) is conformal. Given an open set \(U\subseteq D\text{,}\) we say \(f\) is conformal on \(U\) if it is conformal at all points in \(U\text{,}\) and we say \(f\) is conformal if it is conformal on its entire domain.

Remark 1.27.3. Oriented angle.

Mark well the oriented angle detail in this definition. The linear transformation \(T\colon \R^2\rightarrow \R^2\) defined as \(T(x,y)=(x,-y)\) preserves angles, but not oriented angles: the oriented angle from \(\boldv=(1,0)\) to \(\boldw=(0,1)\) is \(\pi/2\text{,}\) but the oriented angle from \(T(\boldv)=(1,0)\) and \(T(0,1)=(0,-1)\) is \(-\pi/2\text{.}\)

Remark 1.27.4. Geometric interpretation.

The chain rule allows us to give a more geometric interpretation of conformality. Assume \(f\) is conformal at the point \(P=(x_0,y_0)\text{,}\) and suppose \(\phi\) and \(\psi\) are two smooth paths defined on an interval \([-a,a]\) satisfying
\begin{align*} \phi(0) \amp = \psi(0)=(x_0,y_0)\\ \phi'(0) \amp \ne \boldzero, \psi'(0)\ne \boldzero\text{.} \end{align*}
Composing \(\phi\) and \(\psi\) yields smooth paths \(f\circ \phi\) and \(f\circ \psi\) intersecting at
\begin{equation*} f(P)=f\circ\phi(0)=f\circ\psi(0)\text{.} \end{equation*}
Using the multivariable chain rule, the tangent vectors of the two curves at \(f(P)\) are given as
\begin{align*} (f\circ\phi)'(0) \amp = J_f(P)\phi'(0) \\ (f\circ \psi)'(0) \amp = J_f(P)\psi'(0) \text{.} \end{align*}
Since \(J_f(P)\) is a conformal matrix, we conclude that the oriented angle between the tangent vectors of the paths \(\phi\) and \(\psi\) at \(P\) is equal to the oriented angle between the tangent vectors of the images of these paths under \(f\) at \(f(P)\text{.}\)

Remark 1.27.5. Conformal matrices.

Let \(A\) be a \(2\times 2\) matrix with real coefficients. The following statements can be shown to be equivalent using some elementary linear algebra.
  1. \(A\) is conformal.
  2. \(A=cQ\text{,}\) for some positive \(c\in \R\) and orthogonal matrix \(Q\) of determinant 1. (Recall that \(Q\) is orthogonal if \(Q^TQ=I\text{.}\))
  3. \(A=\begin{bmatrix} a\amp -b \\ b\amp a \end{bmatrix}\) for some \(a,b\in \R\text{,}\) \((a,b)\ne (0,0)\text{.}\)

Proof.

The key to all the proofs is the fact that a real differentiable function \(f\colon U\rightarrow \R^2\text{,}\) \(f=(u,v)\text{,}\) is complex differentiable if and only if \(u\) and \(v\) satisfy the Cauchy-Riemann equations at all \(P=(x,y)\text{.}\)

Example 1.27.7. Conformal map: squaring function.

Let \(f(z)=z^2\text{,}\)
  1. Show that \(f\) is conformal on \(\C-\{0\}\)
  2. Illustrate with an explicit geometric example that \(f\) does not preserve the angle between curves intersecting at \(z_0=0\text{.}\)
Solution.
  1. Clearly \(f\) is analytic everywhere, and \(f'(z)=2z\) is nonzero at all \(z\ne 0\text{.}\) Thus \(f\) is conformal on \(\C-\{0\}\) by Theorem 1.27.6.
  2. Consider the rays \(R_0\) and \(R_{\pi/4}\text{,}\) which intersect at \(z_0=0\) at an angle of \(\pi/4\text{.}\) However, their images under \(f\) are \(f(R_0)=R_0\) and \(f(R_{\pi/4})=R_{\pi/2}\text{,}\) which intersect at an angle of \(\pi/2\text{.}\)

Definition 1.27.8. Möbius transformation.

Let \(a,b,c,d\in \C\) be complex constants for which the matrix \(A=\begin{bmatrix} a\amp b\\ c\amp d \end{bmatrix}\) satisfies \(\det A=ad-bc\ne 0\text{.}\) The Möbius transformation (or fractional linear transformation) associated to \(A\) is the function
\begin{equation*} f_A\colon \C-\{z\colon cz+d=0\}\rightarrow \C \end{equation*}
defined as
\begin{equation} f_A(z)=\frac{az+b}{cz+d}\text{.}\tag{1.99} \end{equation}
It is clear that a Möbius transformation \(f_A\) as in (1.99) has at most one isolated singularity at \(z_0=-d/c\) (assuming \(c\ne 0\)), where \(f_A\) has a simple pole, and in this case \(\lim\limits_{z\to z_0}f_A(z)=\infty\text{.}\) Furthermore, a straightforward computation reveals that
\begin{equation*} \lim\limits_{z\to \infty}f(z)=\begin{cases} \frac{a}{c}\amp \text{if } c\ne 0 \\ \infty \amp \text{if } c=0 \end{cases}\text{.} \end{equation*}
Let \(C^*=\C\cup \{\infty\}\text{,}\) the Riemann sphere. The observations allow us to extend \(f_A\) to a function \(f_A\colon \C^*\rightarrow \C^*\) by defining
\begin{equation*} f_A(z_0)=\lim\limits_{z\to z_0}f_A(z)\text{.} \end{equation*}
Note that for all \(z_0\in \C-\{z\colon cz+d=0\}\) this formula agrees with \(\frac{az_0+b}{cz_0+d}\) since the original function \(f_A\) is continuous there.
As the theorem below illustrates, this more expanded setting gives rise to a tidy conceptual understanding of Möbius transformations. Before we get there, we need to generalize the notion of a circle to reflect the geometry of the Riemann sphere.

Definition 1.27.9. Generalized circle.

We call any line or circle in \(\C\) a generalized circle. Furthermore we say that a line \(\ell\) is a generalized circle containing \(\infty\).

Remark 1.27.10. Generalized circle.

Let \(S^2=\{(x,y,z)\in \R^3\colon x^2+y^2+z^2=1\}\text{,}\) and let \(\phi\colon S^2\rightarrow \C*=\C\cup \{\infty\}\) be the stereographic projection. (See Definition 1.4.19.) By a circle in \(S^2\) we mean any latitude line with respect to a chosen pole \(P=(a,b,c)\in S^2\text{.}\) (Such a latitude line can be described as the intersection of \(S^2\) with a plane of the form \(ax+by+cz=d\) for some \(d\in \R\text{.}\)) It can be shown that the image under \(\phi\) of any circle in \(C\subseteq S^2\) is either a circle or line, and that \(\phi(C)\) is a line if and only if \(C\) contains \(N=(0,0,1)\text{.}\)
In other words, generalized circles in \(\C\) correspond precisely to circles in \(S^2\) under the stereographic projection; and in particular, lines in \(\C\) correspond to circles containing \(N=(0,0,1)\) (which corresponds to \(\infty\) under \(\phi\)). This explains where this terminology comes from.
According to \((6)\) of Theorem 1.27.11, once we know the value of a Möbius transformation \(f\) at three distinct inputs \(z_1,z_2,z_3\text{,}\) we can determine \(f(z)\) for any \(z\in \C^*\text{.}\) This can often be done in a pleasing geometric manner using the following properties also enjoyed by \(f\text{.}\)
  1. \(f\) maps generalized circles to generalized circles.
  2. \(f\) is continuous. As such given any connected region \(R\text{,}\) \(f(R)\) is connected. Furthermore because \(f\) has a continuous inverse, it follows that \(f\) maps the boundary of \(R\) to the boundary of \(f(R)\text{.}\)
  3. \(f\) is conformal. Thus if two paths intersect at a point \(z_0\) with oriented angle \(\theta\text{,}\) then their images under \(f\) intersection at \(f(z_0)\) with the same angle.
The next example illustrates how to get as much as possible out of these properties.

Example 1.27.12. Möbius transformation.

Let \(g\colon \C^*\rightarrow \C^*\) be the unique Möbius transformation satisfying
\begin{align*} g(0) \amp = -1 \amp g(i)\amp =0 \amp g(\infty)\amp = 1\text{.} \end{align*}
  1. Find a complex matrix \(A\) such that \(g=f_A\text{.}\)
  2. Explicitly express \(g\) as a composition of dilations, translations, and inversions.
  3. Find a complex matrix \(B\) such that \(g^{-1}=f_B\text{.}\)
  4. For each region (or curve) below, determine its image under \(g\text{.}\)
    1. The imaginary axis.
    2. The right half-plane \(\{z\in \C\colon \Re z > 0\}\)
    3. The real axis.
    4. The upper half-plane \(\{z\in\C\colon \Im z> 0\}\)
    5. The line \(\Im z=1\text{.}\)
Solution.
  1. Writing \(g(z)=(az+b)/(cz+d)\text{,}\) since \(g(0)=-1\) and \(g(\infty)=1\text{,}\) we see that \(b=-d\) and \(a=c\text{,}\) Thus \(g(z)=(az+b)/(az-b)\text{.}\) Next since \(a(-b)-ab=-2ab\ne 0\text{,}\) we must have \(a\ne 0\) and \(b\ne 0\text{.}\) Lastly, since \(g(i)=0\text{,}\) we have \(ai=-b\text{,}\) or \(a=ib\text{.}\) Thus we can choose
    \begin{equation} A=\begin{bmatrix} ib\amp b\\ ib\amp -b \end{bmatrix}\tag{1.100} \end{equation}
    for any \(b\ne 0\in \C\text{.}\) In particular, we can set \(b=-i\text{,}\) in which case
    \begin{equation*} A=\begin{bmatrix} 1\amp -i\\ 1\amp i \end{bmatrix}\text{.} \end{equation*}
    Note that although there are infinitely many different matrices \(A\) satisfying \(g=f_A\) (one for each nonzero \(b\)), they all give rise to the same function \(g\text{.}\) Indeed, given any \(A\) as in (1.100) we have
    \begin{equation*} g(z)=\frac{ibz+b}{ibz-b}=\frac{ib}{ib}\frac{z-i}{z+1}=\frac{z-i}{z+i}\text{.} \end{equation*}
    My particular choice of \(b=-i\) yields a particularly simple matrix that makes life easier in the next two parts.
  2. Using your favorite algorithm for writing an invertible matrix as a product of elementary matrices, we have
    \begin{equation*} \begin{bmatrix} 1\amp i\\ 1\amp -i \end{bmatrix} =\begin{bmatrix} 1\amp 0\\ 1\amp 1 \end{bmatrix} \begin{bmatrix} 1\amp 0\\ 0\amp -2i \end{bmatrix} \begin{bmatrix} 1\amp i \\ 0\amp 1 \end{bmatrix}\text{.} \end{equation*}
    Writing this equality as \(A=BCD\text{,}\) we have
    \begin{equation*} g=f_A=f_B\circ f_C\circ f_D\text{,} \end{equation*}
    where
    \begin{align*} f_B(z) \amp =\frac{z}{z+1}=\frac{1}{1+1/z}\\ f_C(z) \amp = \frac{z}{-2i}=\frac{i}{2}z\\ f_D(z) \amp =z+i \text{.} \end{align*}
    We see that \(f_C\) is a dilation and \(f_D\) is a translation. Furthermore we can write \(f_B=p\circ q\circ p\text{,}\) where \(p(z)=1/z\) (inversion), and \(q(w)=w+1\) (translation). Thus
    \begin{equation*} g=p\circ q\circ p\circ f_C\circ f_D \end{equation*}
    is our desired breakdown of \(g\) into dilations, translations, and inversions.
  3. We have \(g^{-1}=f_B\text{,}\) where \(B\) is any scalar multiple of
    \begin{equation*} A^{-1}=\frac{i}{2}\begin{bmatrix} -i\amp -i\\ -1\amp 1 \end{bmatrix}\text{.} \end{equation*}
    We take
    \begin{equation*} B=2A=\begin{bmatrix} 1\amp 1\\ -i\amp i \end{bmatrix} \end{equation*}
    Thus
    \begin{equation*} g^{-1}(z)=\frac{z+1}{-iz+i}=i\cdot\frac{z+1}{z-1}\text{.} \end{equation*}
  4. It is possible to answer this question without ever making using of our explicit formula for \(g\text{,}\) using instead only the the fact that \(g\) is biholomorphic, conformal, maps generalized circles to generalized circles, and maps \(0,1,\infty\) to \(-1,0,1\text{,}\) respectively. For the sake of brevity, however, we will “cheat” here and there, and evaluate \(g\) at some other points. For example, it helps us to observe that \(g\) has a pole at \(z_0=-i\) (and thus \(g(-i)=\infty\)).
    Let \(\ell\) be the imaginary axis. We know \(g(\ell\cup\{\infty\})\) is a generalized circle containing \(g(0)=-1\text{,}\) \(g(i)=0\) and \(g(\infty)=1\text{.}\) There is no circle containing these three points, so \(g(\ell\cup \{\infty\})\) must be the real axis. Since \(g(\infty)=1\text{,}\) we conclude that \(g(\ell)=\R-\{1\}\text{.}\)
    Let’s nail down some names for various open half-planes:
    \begin{align*} N\amp = \{z\in \C\colon \Im z> 0\} \\ S \amp =\{z\in \C \colon \Im z< 0\}\\ E \amp =\{z\in \C\colon \Re z< 0\}\\ W\amp = \{z\in \C\colon \Re z> 0\} \text{.} \end{align*}
    As you can tell, I am thinking in terms of north, south, east, and west. Note that each of these regions is open and connected. Since \(g(\ell)=\R-\{0\}\) and \(g\) is bijective, we have \(g(E)\subseteq N\cup S\text{.}\) Since \(g(E)\) is connected and and since \(N\cap S=\emptyset\text{,}\) we must have \(g(E)\subseteq N\) or \(g(E)\subseteq S\text{.}\) Since \(g(1)=(1-i)/(1+i)=-i\in S\text{,}\) we see that \(g(E)\subseteq S\text{.}\) Applying the inverse \(g^{-1}\) to this inclusion, we see further that \(E\subseteq g^{-1}(S)\text{.}\) Since \(g^{-1}(S)\subseteq W\) or \(g^{-1}(S)\subseteq E\) (using similar reasoning as above), the inclusion \(E\subseteq g^{-1}(S)\) implies \(g^{-1}(S)\subseteq E\text{.}\) We conclude that \(S\subseteq g(E)\) and thus that \(g(E)=S\text{.}\)
    Since \(g(0)=-1\text{,}\) \(g(1)=-i\text{,}\) and \(g(\infty)=1\text{,}\) we see that \(g(\R\cup \{\infty\})\) is a generalized circe containing \(-1,-i, 1\text{.}\) The unit circle, \(S^1=\{z\in \C\colon \abs{z}=1\}\) is the only such generalized circle. Thus \(g(\R\cup \{\infty\})=S^1\) and \(g(\R)=S^1-\{1\}\text{.}\)
    Continue to denote the upper half-plane as \(N\text{.}\) Since \(g(\R)=S^1-\{1\}\) and \(g(N)\) is connected, we must have \(g(N)\) lying entirely within the unit circle or entirely without it. Since \(g(i)=0\text{,}\) we conclude that \(g(N)\subseteq B_1(0)\text{.}\) Now making use of \(g^{-1}\) we can argue further (similarly as above) that \(g^{-1}(B_1(0))\subseteq N \text{,}\) and hence \(B_1(0)\subseteq g(N)\text{.}\) We conclude that \(g(N)=B_1(0)\text{.}\)
    Since the line \(\ell'\) defined as \(\Im z=1\) lies in the upper half-plane, we see that \(g(\ell' \cup \{\infty\})\) must be a generalized circle \(C\) lying within \(g(N)=B_1(0)\cup \{1\}\) and containing \(g(i)=0\) and \(g(\infty)=1\text{.}\) Since \(C\) is bounded, it must be an honest to goodness circle. We could get one more point on \(C\) by computing \(g(1+i)\text{,}\) but instead we argue using conformality. The paths \(\gamma(t)=(t+1)i\) and \(\psi(t)=t+i\) intersect at \(i\) where their tangents form an oriented angle of \(\pi/2\) (from \(\phi\) to \(\gamma\)). Since \(g\) is conformal, the paths \(g\circ \gamma\) and \(g\circ \psi\) intersect at \(g(i)=0\) at an oriented angle of \(\pi/2\text{.}\) Since \(g\circ \gamma\) points tangentially along the real line (from above), and \(g\circ\psi\) points tangentially along the circle \(C\) at the point \(0\text{,}\) we see that the tangent line to \(C\) at \(0\) is vertical! A geometric argument then shows we must have \(C=\{z\in \C\colon \abs{z-1/2}=1/2\}\text{.}\)
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