Show that in polar coordinates, the CR equations take the following form: \[ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{1}{r} \frac{\partial u}{\partial \theta}=-\frac{\partial v}{\partial r} .\]
Solution
Setting \[ z = re^{i\theta} = r(\cos(\theta) + i\sin(\theta) ) = x+iy \] yields \(x=r\cos(\theta), y=r\sin(\theta)\), one can identify \[ x_r = \cos(\theta)&, x_\theta = -r\sin(\theta) \\ y_r = \sin(\theta)&, y_\theta = r\cos(\theta) .\]
Now apply the chain rule: \[ u_r &= u_x x_r + u_y y_r \\ &= v_y x_r -v_x y_r && \text{CR}\\ &= v_y \cos(\theta) - v_x \sin(\theta) \\ &= {1\over r}\qty{ v_y r\cos(\theta) - v_x r\sin(\theta) } \\ &= {1\over r}\qty { v_y y_\theta + v_x x_\theta} \\ &= {1\over r} v_\theta .\] Similarly, \[ v_r &= v_x x_r + v_y y_r \\ &= v_x \cos(\theta) + v_y\sin(\theta) \\ &= -u_y\cos(\theta) + u_x\sin(\theta) && \text{CR} \\ &= {1\over r}\qty{ -u_y r\cos(\theta) + u_x r\sin(\theta)} \\ &= {1\over r}\qty{ -u_y y_\theta - u_x x_0 } \\ &= -{1\over r} u_\theta .\]
Thus \[ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} \\ .\]