Question

Show that $(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2})\log \left|f^{\prime }\left(z\right)\right|=0$ where $f\left(z\right)$ is an analytic function.

Answer 1

To show $(\frac{{\delta }^2}{\delta x^2}+\frac{{\delta }^2}{\delta y^2})\log |f^{\prime }\left(z\right)|=0$

As we know $(\frac{{\delta }^2}{\delta x^2}+\frac{{\delta }^2}{\delta y^2})=4\frac{{\delta }^2}{\delta x\delta \overline{z}}$

Hence, $(\frac{{\delta }^2}{\delta x^2}+\frac{{\delta }^2}{\delta y^2})\left\{\log |f^{\prime }\right(z\left)|\right\}=4\frac{{\delta }^2}{\delta z\delta \overline{z}}\left\{\log |f^{\prime }\right(z\left)|\right\}$

$=\frac{4{\delta }^2}{\delta z\delta \overline{z}}\log |f^{\prime }\left(z\right){|}^2$

$=2\frac{{\delta }^2}{\delta z\delta \overline{z}}\log \left\{f^{\prime }\right(z\left)f^{\prime }\right(\overline{z}\left)\right\}$

$=2\frac{{\delta }^2}{\delta z\delta \overline{z}}\left\{\log f^{\prime }\right(z\left)\log f^{\prime }\right(\overline{z}\left)\right\}$

$=\frac{2{\delta }^2}{\delta z}[0+\frac{1}{f^{\prime }\left(\overline{z}\right)}f"(\overline{z}\left)\right]$

$=2\frac{\delta }{\delta z}\frac{f"\left(\overline{z}\right)}{f^{\prime }\left(\overline{z}\right)}$

$(\because \overline{z}$ is constant )

$=2\times 0=0$

$\therefore (\frac{{\delta }^2}{\delta x^2}+\frac{\delta }{\delta y^2})\log |f^{\prime }\left(z\right)|=0$

Hence proved