Theorem 9.1.2.
Let \(f\) be an analytic function in the domain \(D\text{,}\) and let \(z_0\) be a point in \(D\text{.}\) If \(f\,'(z_0) \ne 0\text{,}\) then \(f\) is conformal at \(z_0\text{.}\)
Section 9.1 Basic Properties of Conformal Mappings
Let \(f\) be an analytic function in the domain \(D\) and let \(z_0\) be a point in \(D\text{.}\) If \(f\,'(z_0) \ne 0\text{,}\) then we can express \(f\) in the form
\begin{equation}
f(z) = f(z_0) +f\,'(z_0)(z-z_0) +\eta(z) (z-z_0)\text{,}\tag{9.1.1}
\end{equation}
where \(\eta(z) \to 0\) as \(z \to z_0\text{.}\) If \(z\) is near \(z_0\text{,}\) then the transformation \(w=f(z)\) has the linear approximation
\begin{equation*}
S(z) =A+B(z-z_0)=Bz+A-Bz_0\text{,}
\end{equation*}
where \(A=f(z_0)\) and \(B=f\,'(z_0)\text{.}\) Because \(\eta(z) \to 0\) when \(z \to z_0\text{,}\) for points near \(z_0\) the transformation \(w=f(z)\) has an effect much like the linear mapping \(w=S(z)\text{.}\) The effect of the linear mapping \(S\) is a rotation of the plane through the angle \(\alpha =\mathrm{Arg}f\,'(z_0)\text{,}\) followed by a magnification by the factor \(|f\,'(z_0)|\text{,}\) followed by a rigid translation by the vector \(A-Bz_0\text{.}\) Consequently, the mapping \(w=S(z)\) preserves the angles at the point \(z_0\text{.}\) We now show that the mapping \(w=f(z)\) also preserves angles at \(z_0\text{.}\)
Let \(C:z(t) =x(t) +iy(t)\text{,}\) \(-1 \le t \le 1\) denote a smooth curve that passes through the point \(z(0)=z_0\text{.}\) A vector \(\mathbf{T}\) tangent to \(C\) at the point \(z_0\) is given by
\begin{equation*}
\mathbf{T}=z\,'(0)\text{,}
\end{equation*}
where the complex number \(z\,'(0)\) is expressed as a vector.
The angle of inclination of \(\mathbf{T}\) with respect to the positive \(x\)-axis is
\begin{equation*}
\beta=\mathrm{Arg}\, z\,'(0)
\end{equation*}
The image of \(C\) under the mapping \(w=f(z)\) is the curve \(K\) given by the formula \(K:w(t)=u\big(x(t),
y(t)\big) +iv\big(x(t),y(t)\big)\text{.}\) We can use the chain rule to show that a vector \(\mathbf{T}^*\) tangent to \(K\) at the point \(w_0=f(z_0)\) is given by
\begin{equation*}
\mathbf{T}^*=w\,'(0) =f\,'(z_0)z\,'(0)
\end{equation*}
The angle of inclination of \(\mathbf{T}^*\) with respect to the positive \(u\)-axis is
\begin{equation}
\gamma=\mathrm{Arg}\;f\,'(z_0) +\mathrm{Arg} z\,'(0) = \alpha+\beta\text{,}\tag{9.1.2}
\end{equation}
where \(\alpha=\mathrm{Arg}\;f\,'(z_0)\text{.}\) Therefore the effect of the transformation \(w=f(z)\) is to rotate the angle of inclination of the tangent vector \(\mathbf{T}\) at \(z_0\) through the angle \(\alpha=\mathrm{Arg}\;f\,'(z_0)\) to obtain the angle of inclination of the tangent vector \(\mathbf{T}^*\)at \(w_0\text{.}\) This situation is illustrated in Figure 9.1.1.
A mapping \(w=f(z)\) is said to be angle preserving, or conformal at \(z_0\text{,}\) if it preserves angles between oriented curves in magnitude as well as in orientation. Theorem 9.1.2 shows where a mapping by an analytic function is conformal.
Proof.
We let \(C_1\) and \(C_2\) be two smooth curves passing through \(z_0\) with tangents given by \(\mathbf{T}_1\) and \(\mathbf{T}_2\text{,}\) respectively. We let \(\beta_1\) and \(\beta_2\) denote the angles of inclination of \(\mathbf{T}_1\) and \(\mathbf{T}_2\text{,}\) respectively.
The image curves \(K_1\) and \(K_2\) that pass through the point \(w_0=f(z_0)\) have tangents denoted \(\mathbf{T}_1^*\) and \(\mathbf{T}_2^*\text{,}\) respectively. From Equation (9.1.2), the angles of inclination \(\gamma_1\) and \(\gamma_2\) of \(\mathbf{T}_1^*\) and \(\mathbf{T}_2^*\) are related to \(\beta_1\) and \(\beta_2\) by the equations
\begin{equation}
\gamma_1=\alpha+\beta_1, \text{ and } \gamma_2 = \alpha+\beta _2\text{,}\tag{9.1.3}
\end{equation}
where \(\alpha=\operatorname*{Arg}f\,'(z_0)\text{.}\) From Equations (9.1.3) we conclude that
\begin{equation*}
\gamma_2-\gamma_1=\beta_2-\beta_1\text{.}
\end{equation*}
The analytic mapping \(w=f(z)\) is conformal at the point \(z_0\text{,}\) where \(f\,'(z_0) \ne 0\) because the angle \(\gamma_2-\gamma_1\) from \(K_1\) to \(K_2\) is the same in magnitude and orientation as the angle \(\beta_2-\beta_1\) from \(C_1\) to \(C_2\text{.}\) This situation is shown in Figure 9.1.3.
Example 9.1.4.
Show that the mapping \(w=f(z) =\cos z\) is conformal at the points \(z_1=i\text{,}\) \(z_2=1\text{,}\) and \(z_3=\pi+i\text{,}\) and determine the angle of rotation given by \(\alpha=\mathrm{Arg}f\,'(z)\) at the given points.
Solution.
Because \(f\,'(z) =-\sin z\text{,}\) we conclude that the mapping \(w=\cos z\) is conformal at all points except \(z=n\pi\text{,}\) where \(n\) is an integer. Calculation reveals that
\begin{align*}
f\,'(i) \amp =-\sin(i) =-i\sinh1,\\
f\,'(1) \amp = -\sin1, \text{ and }\\
f\,'(\pi+i) \amp =-\sin (\pi + i) = i\sinh 1\text{.}
\end{align*}
Therefore the angles of rotation are given, respectively, by
\begin{align*}
\alpha_1 \amp = \mathrm{Arg}\big(f\,'(i)\big) = -\frac{\pi}{2},\\
\alpha_2 \amp =\mathrm{Arg}\big(f\,'(1)\big) =\pi, \text{ and }\\
\alpha_3 \amp =\mathrm{Arg}\big(f\,'(\pi+i)\big) =\frac {\pi}{2}\text{.}
\end{align*}
Let \(f\) be a nonconstant analytic function. If \(f\,'(z_0)=0\text{,}\) then \(z_0\) is called a critical point of \(f\text{,}\) and the mapping \(w=f(z)\) is not conformal at \(z_0\text{.}\) Theorem 9.1.5 shows what happens at a critical point.
Theorem 9.1.5.
Let \(f\) be analytic at \(z_0\text{.}\) If \(f\,'(z_0) = 0,\ldots,f^{(k-1)}(z_0)=0\) and \(f^{(k)}(z_0) \ne 0\text{,}\) then the mapping \(w=f(z)\) magnifies angles at the vertex \(z_0\) by a factor \(k\text{.}\)
Proof.
Since \(f\) is analytic at \(z_0\text{,}\) it has a Taylor series expansion. Because \(a_n=\displaystyle\frac{f^{(n) }(z_0) }{n!}=0\text{,}\) for \(n=1,\ 2,...,\ k-1\text{,}\) the series representation for \(f\) is
\begin{equation}
f(z) = f(z_0) + a_k(z-z_0)^k + a_{k+1}(z-z_0)^{k+1} + \cdots\text{.}\tag{9.1.4}
\end{equation}
From Equation (9.1.4) we conclude that
\begin{equation}
f(z) - f(z_0) = (z-z_0)^kg(z)\text{,}\tag{9.1.5}
\end{equation}
where \(g\) is analytic at \(z_0\) and g\((z_0) = a_k \ne 0\text{.}\) Consequently, if \(w=f(z)\) and \(w_0=f(z_0)\text{,}\) then using Equation (9.1.5) we obtain
\begin{equation}
\arg(w-w_0) =\mathrm{Arg}[f(z)-f(z_0)] = k\mathrm{Arg}(z-z_0) + \mathrm{Arg}[g(z)]\text{.}\tag{9.1.6}
\end{equation}
If \(C\) is a smooth curve that passes through \(z_0\) and \(z \to z_0\) along \(C\text{,}\) then \(w\to w_0\) along the image curve \(K\text{.}\) The angle of inclination of the tangents \(\mathbf{T}\) to \(C\) and \(\mathbf{T}^*\) to \(K\text{,}\) respectively, are then given by the following limits:
\begin{equation}
\beta = \lim_{z \to z_0} \mathrm{Arg}(z-z_0) \text{ and } \gamma = \lim\_{w\to w_0}\ \mathrm{Arg}(w-w_0)\text{.}\tag{9.1.7}
\end{equation}
\begin{equation}
\gamma = \lim_{z \to z_0} (k\mathrm{Arg}(z-z_0) + \mathrm{Arg}[g(z)]) = k\beta+\delta\text{,}\tag{9.1.8}
\end{equation}
where \(\delta = \mathrm{Arg}[g(z_0)] = \mathrm{Arg}(a_k)\text{.}\)
If \(C_1\) and \(C_2\) are two smooth curves that pass through \(z_0\text{,}\) and \(K_1\) and \(K_2\) are their images, then from Equation (9.1.8) it follows that
\begin{equation*}
\Delta\gamma = \gamma_2 - \gamma_1 = k(\beta_2-\beta_1) = k\Delta\beta
\end{equation*}
That is, the angle \(\Delta\gamma\) from \(K_1\) to \(K_2\) is \(k\) times as large as the angle \(\Delta\beta\) from \(C_1\) to \(C_2\text{.}\) Therefore angles at the vertex \(z_0\) are magnified by the factor \(k\text{.}\) This situation is shown in Figure 9.1.6.
Example 9.1.7.
Show that the mapping \(w=f(z)=z^2\) maps the unit square \(S=\{x+iy: 0\lt x\lt 1, \; 0\lt y\lt 1\}\) onto the region in the upper half-plane \(\mathrm{Im}(w)>0\text{,}\) which lies under the parabolas
\begin{equation*}
u=1-\frac{1}{4}v^2, \text{ and } u=-1+\frac{1}{4}v^2\text{,}
\end{equation*}
as shown in Figure 9.1.8.
Solution.
The derivative is \(f\,'(z)=2z\text{,}\) and we conclude that the mapping \(w=z^2\) is conformal for all \(z \ne 0\text{.}\) Note that the right angles at the vertices \(z_1=1\text{,}\) \(z_2=1+i\text{,}\) and \(z_3=i\) are mapped onto right angles at the vertices \(w_1=1, \, w_2=2i\text{,}\) and \(w_3=-1\text{,}\) respectively. At the point \(z_0=0\text{,}\) we have \(f\,'(0) =0\) and \(f\,''(0) \ne 0\text{.}\) Hence angles at the vertex \(z_0=0\) are magnified by the factor \(k=2\text{.}\) In particular, the right angle at \(z_0=0\) is mapped onto the straight angle at \(w_0=0\text{.}\)
Another property of a conformal mapping \(w=f(z)\) is obtained by considering the modulus of \(f\,'(z_0)\) . If \(z_1\) is near \(z_0\text{,}\) we can use Equation (9.1.1) and neglect the term \(\eta(z_1)(z_1-z_0)\text{.}\) We then have the approximation
\begin{equation}
w_1-w_0 = f(z_1)-f(z_0) \approx f\,'(z_0)(z_1-z_0)\text{.}\tag{9.1.9}
\end{equation}
From Equation (9.1.9), the distance \(|w_1-w_0|\) between the images of the points \(z_1\) and \(z_0\) is given approximately by \(|f\,'(z_0)| \, |z_1-z_0|\text{.}\) Therefore we say that the transformation \(w=f(z)\) changes small distances near \(z_0\) by the scale factor \(|f\,'(z_0)|\text{.}\) For example, the scale factor of the transformation \(w=f(z) =z^2\) near the point \(z_0=1+i\) is \(|f\,'(1+i)| = |2(1+i)| = 2\sqrt{2}\text{.}\)
We also need to say a few things about the inverse transformation \(z=g(w)\) of a conformal mapping \(w=f(z)\) near a point \(z_0\) , where \(f\,'(z_0) \ne 0\text{.}\) A complete justification of the following assertions relies on theorems studied in advanced calculus. We express the mapping \(w=f(z)\) in the coordinate form
1
See, for instance, R. Creighton Buck, Advanced Calculus, 3rd ed. (New York, McGraw-Hill), pp. 358—361, 1978.
\begin{equation}
u = u(x,y), \text{ and } v=v(x,y)\text{.}\tag{9.1.10}
\end{equation}
The mapping in Equations (9.1.10) represents a transformation from the \(xy\) plane into the \(uv\) plane, and the Jacobian determinant, \(J(x,y)\text{,}\) is defined by
\begin{equation}
J(x,y) = \begin{vmatrix} u_x(x,y) \amp u_y(x,y)\\ v_x(x,y) \amp v_y(x,y). \end{vmatrix}\tag{9.1.11}
\end{equation}
The transformation in Equations (9.1.10) has a local inverse, provided \(J(x,y) \ne 0\text{.}\) Expanding Equation (9.1.11) and using the Cauchy-Riemann equations, we obtain
\begin{align}
J(x_0,y_0) \amp = u_x(x_0,y_0) v_y(x_0,y_0) -v_x(x_0,y_0) u_y(x_0,y_0)\notag\\
\amp = u_x^2(x_0,y_0) +v_x^2(x_0,y_0)\tag{9.1.12}\\
\amp = |f\,'(z_0)|^2 \ne 0\text{.}\notag
\end{align}
Consequently, Equations (9.1.11) and (9.1.12) imply that a local inverse \(z=g(w)\) exists in a neighborhood of the point \(w_0\text{.}\) The derivative of \(g\) at \(w_0\) is given by the familiar expression
\begin{align*}
g'(w_0) \amp = \lim_{w\to w_0} \frac{g(w)-g(w_0)}{w-w_0}\\
\amp = \lim_{z \to z_0}\ \frac{z-z_0}{f(z)-f(z_0)}\\
\amp = \frac{1}{f\,'(z_0)}=\frac{1}{f\,'\big(g(w_0)\big)}\text{.}
\end{align*}
Exercises Exercises
1.
State where the following mappings are conformal.
(a)
\(w=\exp z\text{.}\)
Solution.
All complex numbers \(z\text{.}\)
(b)
\(w=\sin z\text{.}\)
(c)
\(w=z^2+2z\text{.}\)
Solution.
All complex numbers \(z\) except \(z=0\) and \(z=-2\text{.}\)
(d)
\(w=\exp(z^2+1)\text{.}\)
(e)
\(w=\displaystyle\frac{1}{z}\text{.}\)
Solution.
All complex numbers \(z\) except \(z=0\text{.}\)
(f)
\(w=\displaystyle\frac{z+1}{z-1}\text{.}\)
2.
For Exercises 2-5, find the angle of rotation \(\alpha = \mathrm{Arg}\big(f\,'(z)\big)\) and the scale factor \(|f\,'(z)|\) of the mapping \(w=f(z)\) at the indicated points.
(a)
\(w=\displaystyle\frac{1}{z}\) at the points \(1\text{,}\) \(1+i\text{,}\) and \(i\text{.}\)
(b)
\(w=\ln r+i\theta\text{,}\) where \(-\frac{\pi}{2}\lt \theta\lt \frac{3\pi}{2}\) at the points \(1\text{,}\) \(1+i\text{,}\) \(i\text{,}\) and \(-1\text{.}\)
Solution.
\(f\,'(1)=1, \; \alpha = \mathrm{Arg}\big(f\,'(1)\big) = 0, \; |f\,'(1)|=1\text{;}\) \(f\,'(1)=\frac{1}{2}-\frac{i}{2}, \; \alpha = \mathrm{Arg}\big(f\,'(1+i)\big) = -\frac{\pi}{4}, \; |f\,'(1+i)=\frac{\sqrt{2}}{2}\text{;}\) \(f\,'(1)=-i, \; \alpha = \mathrm{Arg}\big(f\,'(i)\big) = -\frac{\pi}{2}, \; |f\,'(1)| = 1\text{.}\)
(c)
\(w=r^\frac{1}{2}\cos\frac{\theta}{2}+ir^\frac{1}{2}\sin\frac{\theta}{2}\text{,}\) where \(-\pi\lt \theta\lt \pi\text{,}\) at the points \(i, \, 1, \, -i\text{,}\) and \(3+4i\text{.}\)
(d)
\(w=\sin z\) at the points \(\frac{\pi}{2}+i, \, 0\text{,}\) and \(-\frac{\pi}{2}+i\text{.}\)
Solution.
\(f\,'(\frac{\pi}{2}+i)=-i\sinh 1, \; \alpha - \mathrm{Arg}\big(f\,'(-\frac{\pi}{2}+i)\big) = -\frac{\pi}{2}, \; f\,'(\frac{\pi}{2}+i) = \sinh 1\text{;}\) \(f\,'(-\frac{\pi}{2}+i) = i\sinh 1, \; \alpha = \mathrm{Arg}\big(f\,'(-\frac{\pi}{2}+i)\big) = \frac{\pi}{2}, \; f\,'(-\frac{\pi}{2}+i) = \sinh 1\text{;}\) \(f\,'(0)=1, \; \alpha = \mathrm{Arg}\big(f\,'(0)\big) = 0, \; |f\,'(0)| = 1\text{.}\)
(e)
Consider the mapping \(w=z^2\text{.}\) If \(a\ne 0\) and \(b\ne 0\text{,}\) show that the lines \(x=a\) and \(y=b\) are mapped onto orthogonal parabolas.
(f)
Consider the mapping \(w=z^\frac{1}{2}\text{,}\) where \(z^\frac{1}{2}\) denotes the principal branch of the square root function. If \(a>0\) and \(b>0\text{,}\) show that the lines \(x=a\) and \(y=b\) are mapped onto orthogonal curves.
Solution.
\(|f\,'(a+ib)| = \frac{1}{|2\sqrt{a+ib}|} = \frac{1}{2(a^2+b^2)^\frac{1}{4}} \ne 0\text{,}\) hence \(f(z)\) is conformal at \(z = a+ib\text{.}\) The lines \(z_1(t) = a+(b+t)i\) and, \(z_2(t) = (a+t)+ib\) intersect orthogonally at the point \(z_1(0) = z_2(0) = a + ib\text{,}\) therefore, their image curves will intersect orthogonally at the point \(a + ib\text{.}\)
(g)
Consider the mapping \(w=\exp z\text{.}\) Show that the lines \(x=a\) and \(y=b\) are mapped onto orthogonal curves.
(h)
For \(w=\sin z\) show that the line segment \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}, \; y=0\text{,}\) and the vertical line \(x=a\text{,}\) where \(|a| \lt \frac{\pi}{2}\text{,}\) are mapped onto orthogonal curves.
Solution.
\(|f\,'(a+ib)| = |\cos(a+ib)| = \sqrt{\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b} \ne 0\text{,}\) hence \(f(z)\) is conformal at \(z=a+ib\text{.}\) The lines \(z_1(t)=a+ti\) and \(z_2(t)=a+t\) intersect orthogonally at the point \(z_1(0) = z_2(0) = a\text{.}\) Therefore, their image curves will intersect orthogonally at the point \(\sin(a + ib)\text{.}\)
(i)
Consider the mapping \(w=\mathrm{Log} \, z\text{,}\) where \(\mathrm{Log} \, z\) denotes the principal branch of the logarithm function. Show that the positive \(x\)-axis and the vertical line \(x=1\) are mapped onto orthogonal curves.
(j)
If \(f\) is analytic at \(z_0\) and \(f\,'(z_0) \ne 0\text{,}\) show that the function \(g(z) =\overline{f(z)}\) preserves the magnitude, but reverses the sense, of angles at \(z_0\text{.}\)
Solution.
First show that the mapping \(W = \bar{Z}\) preserves the magnitude, but reverses the sense of angles at \(Z_0\text{.}\) Then consider the mapping \(w=f(z)\) as a composition.
(k)
If \(w=f(z)\) is a mapping, where \(f(z)\) is not analytic, then what behavior would you expect regarding the angles between curves?
