question involving analycity of $f=u+iv$

Question

Let $f=u+iv:\mathbb C\to\mathbb C$ be analytic. Then is it true that $\dfrac{\delta^2 v}{\delta x^2}+\dfrac{\delta^2 v}{\delta y^2}=0?$

Answer 1

Let us consider the Cauchy Riemann conditions

$\frac {\partial u} {\partial x}$ = $\frac {\partial v} {\partial y}$

and

$\frac {\partial u} {\partial y}$=-$\frac {\partial v} {\partial x}$

So you are asking if $\dfrac{\delta^2 v}{\delta x^2}+\dfrac{\delta^2 v}{\delta y^2}=0$

Lets find $\dfrac{\delta^2 v}{\delta x^2}$

$\frac \partial {\partial x}$($\frac {\partial v} {\partial x}) $= $\frac \partial {\partial x}$(-$\frac {\partial u} {\partial y}$) = -$\frac {\partial u} {\partial x \partial y}$

Nowlets find $\dfrac{\delta^2 v}{\delta y^2}$

$\frac \partial {\partial y}$($\frac {\partial v} {\partial y}) $= $\frac \partial {\partial y}$($\frac {\partial u} {\partial x}$) = $\frac {\partial u} {\partial y \partial x}$

So we end up with

$\frac {\partial u} {\partial y \partial x}$-$\frac {\partial u} {\partial x \partial y}$=0 (mixed derivatives are equal)

So your statement is correct

Answer 2

Hint: See Cauchy–Riemann equations.