Complex Analysis

Lecture 28

Harmonic functions

Precise definition

\(u: \Omega\subset \R^n \ra \R^n\) harmonic says:

  1. \(u\) and partials of order 1 and 2 are continuous;
  2. \(\Delta u = 0\).

Scale factor

Last remarks on conformal mappings

\(f: z \mapsto w\) conformal (i. e. analytic, \(f'(z_0) \neq 0\)).

For \(z\) near \(z_0, z\neq z_0\): \[\frac{|f(z)-f(z_0)|}{|z-z_0|} \approx |\,f'(z_0)|.\]

\(\implies |f(z)-f(z_0)| \approx |\,f'(z_0)| \cdot |z-z_0|\)

Scale factor

Last remarks on conformal mappings

\(\implies |\,f'(z_0)| \) represents the scaling factor or dilatation factor. That is

  • stretching, if \(\,\left|\,f'(z_0)\right| \gt 1\);
  • shrinking, if \(\,\left|\,f'(z_0)\right| \lt 1\).

Scale factor

Example: \(f(z) = z^2\) at \(z_0=1+i\), (\(z=x+iy, \) \(w = u+iv\))

\(u = x^2-y^2\),   \(v= 2xy\)

Scaling factor = \(|\,f'(z_0)|\) \(= 2 |z_0|\) \(=2 \sqrt{2}\).

Poisson's integral formula

Let \(C_0\) be a positively oriented circle of radius \(r_0\), centred at the origin.

Recall Cauchy: if \(f\) is analytic in and on \(C_0\), then for \(z_0 \in \text{Int} \,C_0\) \[f(z) = \frac{1}{2\pi i}\int_{C_0} \frac{f\left( \zeta \right)}{\zeta - z}d \zeta \qquad (1) \]

Recall for \(z = r e^{i\theta}\), \(r\gt 0\) the inverse point to \(z\) relative to the circle \(C_0\) is \(r^* e^{i \theta}\) with \(r^*\) such that \(r^* \cdot r = r^2_0\).

Poisson's integral formula

Recall \((z^*)^*=z\). Note also:

\(\qquad z^* = r^* e^{i \theta}\) \(= \dfrac{r_0^2}{r} e^{i \theta}\) \(= \dfrac{r_0^2}{re^{-i\theta}}\) \(= \dfrac{r_0^2}{\conj{z}}\) \(=\dfrac{\zeta \conj{\zeta}}{\conj{z}}\) \(\;(2)\;\)

for any \(\zeta\in C_0\). Now fix \(z \in \text{Int}\, C_0, \; z_0\neq 0\) and note \[\ds \int_{C_0} \frac{f\left( \zeta \right)}{\zeta - z}d \zeta = 0 \quad(\text{Cauchy}),\] since \(\zeta\mapsto \dfrac{f(\zeta)}{\zeta -z^*}\) is analytic in and on \(C_0\) (\(z^*\in \text{Ext}\, C_0\)).

Poisson's integral formula

Combine this with \((1)\) implies

\(f(z) = \dfrac{1}{2 \pi i} \ds \int_{C_0} \left( \frac{1}{\zeta -z} - \frac{1}{\zeta - z^*} \right) f(\zeta) d \zeta\) \(\quad (3)\)

\(\quad I = \ds \left( \frac{\zeta}{\zeta -z} - \frac{\zeta}{\zeta - z^*} \right) \frac{1}{\zeta} \) \(= \ds \left( \frac{\zeta}{\zeta -z} - \frac{1}{1 - \frac{\conj{\zeta}}{\conj{z}}} \right) \frac{1}{\zeta}\)

\(\;\;\quad = \ds \left( \frac{\zeta}{\zeta -z} - \frac{\conj{z}}{\conj{z} - \conj{\zeta}} \right) \frac{1}{\zeta}\) \(= \ds \frac{\zeta \conj{\zeta} - z \conj{z}}{|\zeta - z|^2} \cdot \frac{1}{\zeta}\) \(\quad (4)\)

Poisson's integral formula

Now for \(z = r e^{i \theta} \) put \(\zeta = r_0 e^{i \phi}\), \(0\leq \phi \leq 2 \pi \).

Then \(d\zeta = r_0i e^{i \phi}d\phi\) and \[I = \frac{r_0^2 - r^2}{|\zeta - z|^2} \cdot \frac{1}{r_0 e^{i \phi}} \qquad (5) \]

Poisson's integral formula

Using the cosine rule: \[|\zeta - z|^2 = r_0^2+r^2-2r_0r \cos \left(\phi - \theta\right)\quad (5)'\]

Poisson's integral formula

\((3)\) implies \[f\left(re^{i \theta}\right)= \frac{1}{2\pi i} \int_{0}^{2 \pi}\frac{r_0^2-r^2}{|\zeta - z|^2}\cdot \frac{f\left(r_0e^{i\phi}\right)r_0 i e^{i\phi}d \phi}{r_0 e^{i\phi}}\quad \quad\]

\(\qquad \quad= \ds \frac{r_0^2-r^2}{2 \pi } \int_{0}^{2 \pi} \frac{f\left(r_0e^{i\phi}\right) d\phi}{r_0^2-2r_0r \cos \left(\phi - \theta\right) +r^2} \). 😀