$\displaystyle \ \ Since\ f(z)=u+iv\ is\ analytic \\[2ex] \displaystyle By\ cauchys\ Reimanns\ equations, \\[3ex] \displaystyle \frac{\partial{}u}{\partial{}x}=\frac{\partial{}v}{\partial{}y}….\left(1\right)\ and\ \frac{\partial{}u}{\partial{}y}=-\frac{\partial{}v}{\partial{}x}….\left(2\right) \\[2ex] $

$\displaystyle Differentiate\ (1)\ w.r.t.\ `x’\ and\ (2)\ w.r.t.\ `y’ \\[2ex] \displaystyle \frac{{\partial{}}^2u}{\partial{}x^2}=\frac{{\partial{}}^2v}{\partial{}x\partial{}y}\ and\ \frac{{\partial{}}^2u}{\partial{}y^2}=-\frac{{\partial{}}^2v}{\partial{}y\partial{}x} \\[2ex] $

$\displaystyle We\ assume,\\[2ex] \displaystyle \frac{{\partial{}}^2v}{\partial{}x\partial{}y}=\frac{{\partial{}}^2v}{\partial{}y\partial{}x}\\[3ex] and\ on\ adding\ above\ two\ equations,\ we\ get, \\[2ex] \displaystyle \ \frac{{\partial{}}^2u}{\partial{}x^2}+\frac{{\partial{}}^2u}{\partial{}y^2}=\frac{{\partial{}}^2v}{\partial{}x\partial{}y}-\frac{{\partial{}}^2v}{\partial{}y\partial{}x}=0 \\[2ex] \displaystyle \therefore{}'u^{'}is\ harmonic. \\[2ex] $

$\displaystyle Similarly,\ differentiate\ (1)\ w.r.t.\ ‘y’\ and\ ‘2’\ w.r.t.\ ‘x’ \\[2ex] \displaystyle \frac{{\partial{}}^2u}{\partial{}y\partial{}x}=\frac{{\partial{}}^2v}{\partial{}y^2}\ and-\frac{{\partial{}}^2u}{\partial{}x\partial{}y}=\frac{{\partial{}}^2v}{\partial{}x^2} \\[2ex] $

$\displaystyle We\ assume,\ \\[2ex] \displaystyle \frac{{\partial{}}^2u}{\partial{}x\partial{}y}=\frac{{\partial{}}^2u}{\partial{}y\partial{}x}\\[2ex] and\ on\ adding\ above\ two\ equatons,\\[2ex]we\ get, \\[2ex] \displaystyle \frac{{\partial{}}^2v}{\partial{}y^2}+\frac{{\partial{}}^2v}{\partial{}x^2}=\frac{{\partial{}}^2u}{\partial{}y\partial{}x}-\frac{{\partial{}}^2u}{\partial{}x\partial{}y}=0 \\[3ex] \displaystyle \therefore{}'v^{'}is\ harmonic. \\[2ex] $