Modulus squared of derivative of analytic function

Question

In my lecturers notes it is mentioned:

$(u_{x})^2 + (v_{x})^2 = |\frac{dw}{dz}|^2,$ where $w = u(x, y) + iv(x, y), \quad z = x + iy.$

But this isn't immediately obvious to me, if anyone could offer some clarity that would be great.

Answer 1

The Wirtinger derivatives are defined as $$ \left\{ \begin{array}{l} \partial_z = \frac{1}{2}(\partial_x - i\partial_y) \\ \partial_\bar{z} = \frac{1}{2}(\partial_x + i\partial_y) \\ \end{array} \right. $$ Given the fact that $w$ is analytic, that is $\partial_\bar{z}w = 0$, which translates into the Cauchy-Riemann equations $u_x = v_y$ and $u_y = - v_x$, you can compute $|\partial_zw|^2$ and prove the result straightforwardly: $$ \partial_zw = \frac{1}{2}(\partial_x-i\partial_y)(u+iv) = \frac{1}{2}(u_x+iv_x-iu_y+v_y) = u_x+iv_x \\ \Longrightarrow |\partial_zw|^2 = (u_x)^2 + (v_x)^2 $$

Answer 2

You may like this. Let $w=f(z)$. Then $\frac{dw}{dz}=f'(z)=u_x+iv_x$ $$ \left|\frac{dw}{dz}\right|^2=\left|f'(z)\right|^2=|u_x+iv_x|^2=u_x^2+v_x^2.$$