Cauchy-Riemann Equations

Reading

Section 3.1

Problems

Practice problems:

(Pages 41-42) 1. 2, 5, 6, 7, 8, 9, 10

Be ready to present propositions 3.6 and 3.7, problems 8, 9, 10

Topics to know

1. If \(f\) is complex-differentiable, then \(f_x\) and \(f_y\) both exist and relate to complex derivative \(f'\):
   • For real \(h\): \(\displaystyle \lim_{h\to 0}\frac{f(z+h) - f(z)}{h} = f_x\).
   • For imaginary \(h=it\), \(\displaystyle \lim_{h\to 0}\frac{f(z+h) - f(z)}{h} = \frac{f_y}{i}\).
   • So \(f_y = if_x\). Equivalently \(u_x = v_y\) and \(u_y = -v_x\).
2. Converse is true if the partial derivatives are continuous: If \(f_x\) and \(f_y\) exist in a neighborhood of \(z\) and \(f_y = i f_x\). Then \(f\) is complex-differentiable and \(f'(z) = f_x = -i f_y\).
   • Rewrite of mean value theorem: \(\displaystyle \frac{f(x+h) - f(x)}{h} = f'(x+\theta h)\) where \(0<\theta < 1\) is some specific but unknown number. (\(x+\theta h\) is a way to specify an element between \(x\) and \(x+h\))
   • Suppose \(h = h_1 + i h_2\), z = x + i y.
   • \(\displaystyle \frac{f(z+h) - f(z)}{h} = \frac{u(z+h) - u(z)}{h} + i \frac{v(z+h) - v(z)}{h}\).
   • \(u(z+h) - u(z) = u(x+h_1, y+h_2) - u(x+h_1, y) + u(x+h_1, y) - u(x, y)\) which equals \(h_2 u_y(x+h_1, y+\theta_2 h_2) + h_1 u_x(x+\theta_1 h_1, y)\) where \(\theta_1,\theta_2\in (0,1)\).
   • That becomes \(h_2 u_y(z_2) + h_1 u_x(z_1)\) where \(z_2 = (x+h_1) + i(y+\theta_2 h_2)\), \(z_1 = x + \theta_1 h_1 + i y\). In both cases \(z_1,z_2\to z\) as \(h\to 0\).
   • Similarly \(v(z+h) - v(z) = h_2 v_y(z_4) + h_1 v_x(z_3)\) where \(z_3,z_4\to z\) as \(h\to 0\).
   • Therefore \(\displaystyle \frac{f(z+h) - f(z)}{h} = \frac{h_2}{h} \bigl[u_y(z_2) + iv_y(z_4)\bigr] + \frac{h_1}{h} \bigl[u_x(z_1) + iv_x(z_3)\bigr]\)
   • Since \(f_y = if_x\), we can write \(\displaystyle f_x(z) = \frac{h_2}{h} f_y(z) + \frac{h_1}{h}f_x(z)\)
   • Take the difference: \(\frac{f(z+h) - f(z)}{h} - f_x(z) = \frac{h_2}{h} \bigl[(u_y(z_2) - u_y(z)) + i(v_y(z_1) - v_y(z))\bigr] + \frac{h_1}{h} \bigl[(u_y(z_4) - u_y(z)) + i(v_y(z_3) - v_y(z))\bigr]\)
   • Because of continuity, each of the terms goes to \(0\). The factors in front are bound by \(1\). Therefore the whole thing goes to \(0\).
   • Geometrically, instead of trying to go straight from \(z+h\) to \(z\), we instead move vertically then horizontally.
3. \(f\) is called analytic at \(z\) if it is complex-differentiable everywhere in a neighborhood of \(z\). It is called analytic on a set \(S\) if it is complex-differentiable everywhere in an open set containing \(S\).