Show that \(f = u+iv\) with \(u, v\in C^1({\mathbf{R}})\) satisfying the Cauchy-Riemann equations on \(\Omega\), then \(f\) is holomorphic on \(\Omega\) with \begin{align*} f'(z) = {\frac{\partial f}{\partial x}\,} = {1\over i} {\frac{\partial f}{\partial y}\,} = {1\over 2}\qty{u_x + iv_x} .\end{align*} Conversely, show that if \(f\) is holomorphic, then \(f\) satisfies the Cauchy-Riemann equations.
Holomorphic \(\implies\) CR:
Suppose \(f'(z_0)\) exists for all \(z_0\in {\mathbf{C}}\), so the following limit exists: \begin{align*} f'(z_0) \coloneqq\lim_{h\to 0, h\in {\mathbf{C}}} {f(z_0 + h) - f(z_0) \over h} .\end{align*} Approach along \(\left\{{t + 0i {~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}\): \begin{align*} f'(z_0) = f'(x_0, y_0) = \lim_{t\to 0, t\in {\mathbf{R}}} {f(x_0 + t, y_0) - f(x_0, y_0) \over t} \coloneqq f_x(x_0, y_0) .\end{align*} Approach along \(\left\{{0 + ti {~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}\): \begin{align*} f'(z_0) = f'(x_0, y_0) = \lim_{t\to 0, t\in {\mathbf{R}}} {f(x_0, y_0 + t) - f(x_0, y_0) \over it} \coloneqq{1\over i} f_y(x_0, y_0) .\end{align*} Thus \begin{align*} if_x = f_y \implies i(u_x + i v_x) = u_y + i v_y \\ \implies -v_x + iu_x = u_y + iv_y \\ \implies u_x = v_y,\, u_y = -v_x .\end{align*}
CR \(\implies\) holomorphic: A straightforward but messy calculation, not likely to be useful for quals!
Show that in polar coordinates, the CR equations take the following form: \begin{align*} \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{1}{r} \frac{\partial u}{\partial \theta}=-\frac{\partial v}{\partial r} .\end{align*}
Setting \begin{align*} z = re^{i\theta} = r(\cos(\theta) + i\sin(\theta) ) = x+iy \end{align*} yields \(x=r\cos(\theta), y=r\sin(\theta)\), one can identify \begin{align*} x_r = \cos(\theta)&, x_\theta = -r\sin(\theta) \\ y_r = \sin(\theta)&, y_\theta = r\cos(\theta) .\end{align*}
Now apply the chain rule: \begin{align*} u_r &= u_x x_r + u_y y_r \\ &= v_y x_r -v_x y_r && \text{CR}\\ &= v_y \cos(\theta) - v_x \sin(\theta) \\ &= {1\over r}\qty{ v_y r\cos(\theta) - v_x r\sin(\theta) } \\ &= {1\over r}\qty { v_y y_\theta + v_x x_\theta} \\ &= {1\over r} v_\theta .\end{align*} Similarly, \begin{align*} v_r &= v_x x_r + v_y y_r \\ &= v_x \cos(\theta) + v_y\sin(\theta) \\ &= -u_y\cos(\theta) + u_x\sin(\theta) && \text{CR} \\ &= {1\over r}\qty{ -u_y r\cos(\theta) + u_x r\sin(\theta)} \\ &= {1\over r}\qty{ -u_y y_\theta - u_x x_0 } \\ &= -{1\over r} u_\theta .\end{align*}
Thus \begin{align*} \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} \\ .\end{align*}