Question

if f(z) is an analytical function of z, then prove that ( d^2/dx^2 + d^2/ dy^2) log |f(z)|= 0

Answer 1

We know that $\log |f| = \frac{1}{2} \log(f\bar{f})$. If we write $f=u+iv$, where $u, v$ are real functions, we find $\log(f\bar{f})=\log (u^2+v^2)$.

Therefore, we have

$(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}) \log|f| = \frac{1}{2} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\log(u^2+v^2)$ and let us calculate the derivatives :

$\frac{1}{2}\frac{\partial^2}{\partial x^2} \log(u^2+v^2)=\partial_x(\frac{u\partial_x u + v\partial_x v}{u^2+v^2})$ taking the second derivative gives us a long expression :

$(u^2+v^2)(u\partial_{xx}u +v\partial_{xx} v+(\partial_xu)^2+(\partial_xv)^2)-\\ -2u^2(\partial_xu)^2-2v^2(\partial_xv)^2-4uv\partial_xu\partial_xv$ all divided by $(u^2+v^2)^2$

Developping this expression gives us

$\frac{(u^2+v^2)(u\partial_{xx}u+v\partial_{xx} v) + (v\partial_x u - u\partial_x v)^2-(u\partial_x u + v\partial_x v)^2}{(u^2+v^2)^2}$

Calculating the second derivative with respect to $y$ gives us a similar expression :

$\frac{(u^2+v^2)(u\partial_{yy}u+v\partial_{yy} v) + (v\partial_y u - u\partial_y v)^2-(u\partial_y u + v\partial_y v)^2}{(u^2+v^2)^2}$

Now let us use the Cauchy-Riemann equations (as $f$ is analytic, $u, v$ should satisfy them) :

$\begin{cases} \partial_x u = \partial_y v \\ \partial_y u = -\partial_x v \end{cases}$ and therefore $(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})u = (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})v=0$

In addition, we have $(u\partial_y u + v\partial_y v)^2 = (v\partial_x u - u\partial_x v)^2$ and $(u\partial_x u+v\partial_x v)^2 = (u\partial_y v - v\partial_y u)^2$ by Cauchy-Riemann equations. Therefore, the sum

$(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}) \log|f| = \frac{1}{2} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\log(u^2+v^2)=0$