Answer to Question #240731 in Complex Analysis for Msa

Question

Let f be a differentiable function. Establish that identity |f'(z)|^2=(Ux)^2 + (Vx)^2 = (Uy)^2 + (Vy)^2

Answer 1

Let f be a $\mathbb{C}$-differentiable function at a point $z\in\mathbb{C}$. This means that there exist the following limit: $\lim\limits_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}=f'(z)$.

In particular, if we restrict $\Delta z$ to take only real values (in this case then $\Delta z=\Delta x$), then we conclude that there exists a partial derivative $\partial f/\partial x$ at the point $z$:

$f_x(z):=\frac{\partial f}{\partial x}(z):=\lim\limits_{\Delta x\to 0}\frac{f(z+\Delta x)-f(z)}{\Delta x}=f'(z)$

In the similar way, if we restrict $\Delta z$ to take only imaginary values (in this case then $\Delta z=i\Delta y$), then we conclude that there exists a partial derivative $\partial f/\partial y$ at the point $z$:

$f_y(z):=\frac{\partial f}{\partial y}(z):=\lim\limits_{\Delta y\to 0}\frac{f(z+i\Delta y)-f(z)}{\Delta y}=i\lim\limits_{\Delta y\to 0}\frac{f(z+i\Delta y)-f(z)}{i\Delta y}=if'(z)$

Let $u(z)$ and $v(z)$ are (respectively) the real and the imaginary part of the function $f(z)$. Then

$f(z)=u(z)+iv(z)$,

$f_x(z)=u_x(z)+iv_x(z)$ and

$u_x(z)^2+v_x(z)^2=|f_x(z)|^2=|f'(z)|^2$

Similarly,

$f_y(z)=u_y(z)+iv_y(z)$ and

$u_y(z)^2+v_y(z)^2=|f_y(z)|^2=|if'(z)|^2=|f'(z)|^2$

Therefore, $|f'(z)|^2=u_x(z)^2+v_x(z)^2=u_y(z)^2+v_y(z)^2$, as required.