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Differentiability of a function at a point \(z_0\) means that, locally near \(z_0\), the increment \(\Delta f_{z_0}\) is approximately equal to the product of \(\Delta z\) and a constant:
\[
\Delta f_{z_0} = \Delta f_{z_0}(\Delta z) = f(z_0 + \Delta z) - f(z_0) \approx A \cdot \Delta z
\]
with \(A = f'(z_0)\). Note that multiplying the complex argument \(\Delta z\) by the complex constant \(A\) geometrically means a rotation by \(\operatorname {arg}A\) and a scaling by \(\left|A\right|\). That is, if a smooth curve \(\gamma\) passes through the point \(z_0\), with slope \(\varphi\) at \(z_0\) (the angle between the tangent vector to \(\gamma\) at \(z_0\) and the axis \(\operatorname {Re}z\) equal to \(\varphi\)), then the angle of the image curve \(f(\gamma )\) at the point \(f(z_0)\) will equal \(\varphi + \operatorname {arg}A\). The angle-preservation property: if \(f\) is differentiable at \(z_0\) and the smooth curves \(\gamma_1\), \(\gamma_2\) intersect at \(z_0\) at some angle, then \(f(\gamma_1)\) and \(f(\gamma_2)\) will intersect at \(f(z_0)\) at the same angle.
What is the geometric meaning of \(\left|A\right| = \left|f'(z_0)\right|\)? If we draw a circle of small radius \(\rho\) around \(z_0\), its image will approximately correspond to a circle of radius \(\left|A\right| \cdot \rho\) around \(f(z_0)\). The smaller \(\rho\) is, the more the resulting figure will resemble a circle. \(\left|f'(z_0)\right|\) is also called the coefficient of linear dilation at the point \(z_0\).
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ, two panels: on the left, horizontal and vertical lines through z0 = 0.8 + 0.5i; on the right, their images under w = z^2 (the parabolas (t^2-1/4, t) and (16/25 - t^2, 8t/5)), intersecting at w0 = z0^2 = 0.39 + 0.8i at the same right angle. Interactive: dragging the point z0, choosing the function (z^2, e^z, 1/z), a slider for the angle between the original curves; showing the tangent lines and labeling arg fβ(z0) and |fβ(z0)|.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/320/010_conformal_angles_fig.tex
Find the set of all points \(z_{0}\) at which the coefficient of linear dilation for the following maps equals one:
- \(w = \frac{1+i z}{1-i z}\).
Find the set of all points \(z_{0}\) at which the angle of rotation for the following maps equals zero:
\(w = z^{2}-2 z\);
\(w = \frac{i}{z}\);
Lengths of curves
If a regular function \(f\) maps a curve \(\gamma = \gamma (t)\) bijectively (injectively) onto a curve \(f \circ \gamma\), their lengths can be computed by the formulas
\[
\left|\gamma \right| = \int _{\gamma }\left|dz\right|, \quad \left|f \circ \gamma \right| = \int _{f \circ \gamma }\left|dw\right| = \int _{\gamma } \left|f'(z)\right| \; \left|dz\right|
\]
Find the lengths of the images of the following curves \(\Gamma\) under the given maps:
\(\Gamma = \left\{ z: z = i t+1,0 \leqslant t \leqslant 1\right\} , w = z^{2}\);
\(\Gamma = \left\{ z: z = (1+i) t, 0 \leqslant t \leqslant 2 \pi \right\} , w = e^{z}\);
\(f(z) = z^2\) is an entire function. Let us check injectivity: suppose \(f(it_1 + 1) = f(it_2 + 1)\). I.e.
\[
1 - t_1^2 + 2it_1 = 1 - t_2^2 + 2it_2
\]
In particular, the imaginary parts are equal: \(2t_1 = 2t_2\). Hence, \(t_1 = t_2\). We have proved injectivity.
We have
\[
\begin{align} \left|f \circ \Gamma \right| & = \int _{\Gamma }\left|f'(z)\right| \; \left|dz\right| = \int _{\Gamma } \left|2z\right| \; \left|dz\right| = \int _0^1 \left|2(it + 1)\right| \cdot \left|i\right| \; dt = \\ & = 2 \int _0^1 \sqrt{1 + t^2}\, dt = 2 \cdot \left[\frac{t\sqrt{1+t^2}}{2} + \frac{1}{2}\ln \left(t + \sqrt{1+t^2}\right)\right]_{0}^{1} = \sqrt{2} + \ln \left(1 + \sqrt{2}\right) \end{align}
\]
We have
\[
\begin{align} \left|f \circ \Gamma \right| & = \int _{\Gamma }\left|f'(z)\right| \; \left|dz\right| = \int _{\Gamma } \left|e^{z}\right| \; \left|dz\right| = \int _0^{2\pi } \left|e^{(1+i)t}\right| \cdot \left|1 + i\right| \; dt = \\ & = \sqrt{2}\int _0^{2\pi }e^{t} dt = \sqrt{2}(e^{2\pi } - 1) \end{align}
\]
\(\sqrt{2}+\ln (1+\sqrt{2})\);
\(\sqrt{2}\left(e^{2 \pi }-1\right)\);
Find the lengths of the images of the following curves \(\Gamma\) under the given maps:
\(\Gamma = \left\{ z: z = i t, 0 \leqslant t \leqslant 2 \pi \right\} , w = e^{z}\);
\(\Gamma = \left\{ z: z = (1+i) t, 0 \leqslant t \leqslant 1\right\} , w = z^{m}(m = 1,2, \ldots )\).
Dilation of areas
\(f\) is called univalent if it is injective on its domain of definition. That is, it does not glue points together. Let \(f: D \to f(D) =: G\) be holomorphic and univalent. How do we compute the area of \(G\)? We have
\[
S(G) = \iint _G dudv \enclose {circle}{=}
\]
Note that \(f: (x,y) \mapsto (u,v)\),
\[
\mathbf{J}_f = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} = \begin{pmatrix} A & -B \\ B & A \end{pmatrix}, \qquad \operatorname {det}\mathbf{J}_f = A^2 + B^2 = \left|f'\right|^2
\]
where \(u_x = v_y = A\), \(v_x = -u_y = B\). Then
\[
\enclose {circle}{=} \iint _{D} \left|\operatorname {det}\mathbf{J}_f\right| \; dxdy = \iint _{D} \left|f'(z)\right|^2 \; dx dy
\]
Let \(f(z) = z^2\), \[
D = \left\{ z \in \mathbb {C} \; : \; 2 < \left|z\right| < 3, \; \left|\arg z\right| < \frac{\pi }{4}\right\}
\] \(G := f(D)\). Find \(S(G)\).
We must first check whether \(f\) is univalent on \(D\). Suppose \(z_1^2 = z_2^2\). Then \(z_1 = z_2\) or \(z_1 = -z_2\). There are no antipodal points in the domain. I.e. \(z_1 = z_2\), so the function is univalent.
We have
\[
\begin{align} S(G) & = \iint _{D} \left|f'(z)\right|^2 \; dxdy = \iint _{D} 4\left|z\right|^2 \; dxdy = 4 \cdot \int _{2}^3 \left(\int _{-\pi /4}^{\pi /4} r^2 \; d\varphi \right) \cdot r \cdot dr =\\ & = 4 \cdot \frac{\pi }{2} \cdot \int _2^3 r^3 \; dr = 2\pi \cdot \frac{3^4 - 2^4}{4} = \frac{65}{2}\pi \end{align}
\]
\(S(G) = 2\pi \cdot \frac{3^4 - 2^4}{4} = \frac{65}{2}\pi\).
Find the areas of the images of the domains \(D\) under the given maps:
\(D = \left\{ z: 0 < \operatorname {Re} z < 1,\; |\operatorname {Im} z| < \pi \right\} , w = e^{z}\);
\(D = \left\{ z: -1 < \operatorname {Re} z < 0,\; 1 < \operatorname {Im} z < 2\right\}\), \(w = \frac{z-1}{z+1}\).
The function \(e^{z}\) is injective on the open strip \(\left|\operatorname {Im}z\right| < \pi\) (width exactly \(2\pi\), boundary excluded). We have \(\left|\left(e^{z}\right)'\right|^{2} = \left|e^{z}\right|^{2} = e^{2x}\), hence
\[
S = \iint _{D} e^{2x} \, dxdy = 2\pi \int _{0}^{1}e^{2x}\, dx = 2\pi \cdot \frac{e^{2}-1}{2} = \pi \left(e^{2} - 1\right)
\]
A fractional-linear (MΓΆbius) map is injective (globally on \(\bar{\mathbb {C}}\)), so there is nothing to check. We have \(w' = \frac{2}{(z+1)^{2}}\), i.e., denoting \(u := x + 1 \in (0, 1)\):
\[
S = \iint _{D}\frac{4}{\left|z+1\right|^{4}}\, dxdy = \int _{1}^{2}\left(\int _{0}^{1} \frac{4\, du}{\left(u^{2}+y^{2}\right)^{2}}\right) dy = \int _{1}^{2}\left[\frac{2}{y^{2}\left(1+y^{2}\right)} + \frac{2}{y^{3}}\operatorname {arctg}\frac{1}{y}\right]dy
\]
The first term is integrated using the decomposition \(\frac{1}{y^2(1+y^2)} = \frac{1}{y^2} - \frac{1}{1+y^2}\):
\[
2\int _{1}^{2}\left(\frac{1}{y^{2}} - \frac{1}{1+y^{2}}\right)dy = 2\left[-\frac{1}{y} - \operatorname {arctg}y\right]_{1}^{2} = 1 + \frac{\pi }{2} - 2\operatorname {arctg}2
\]
The second, by parts, using \(\left(\operatorname {arctg}\frac{1}{y}\right)' = -\frac{1}{1+y^{2}}\):
\[
2\int _{1}^{2}\frac{\operatorname {arctg}\frac{1}{y}}{y^{3}}\, dy = \left[-\frac{\operatorname {arctg}\frac{1}{y}}{y^{2}} + \frac{1}{y} + \operatorname {arctg}y\right]_{1}^{2} = \operatorname {arctg}2 - \frac{1}{2} - \frac{1}{4}\operatorname {arctg}\frac{1}{2}
\]
Adding and using \(\operatorname {arctg}2 + \operatorname {arctg}\frac{1}{2} = \frac{\pi }{2}\), we obtain
\[
S = \frac{1}{2} + \frac{\pi }{2} - \operatorname {arctg}2 - \frac{1}{4}\operatorname {arctg}\frac{1}{2} = \frac{1}{2}+\frac{3}{4} \operatorname {arctg} \frac{1}{2}
\]
\(\pi \left(e^{2}-1\right)\);
\(\frac{1}{2}+\frac{3}{4} \operatorname {arctg} \frac{1}{2}\).
Find the areas of the images of the domains \(D\) under the given maps:
- \(D = \left\{ z: 2 < |z| < 3,|\arg z| < \frac{\pi }{4}\right\} , w = z^{2} ;\)