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Harmonic Function:
The function $\Phi(x, y)$ which has continuous partial derivatives of first and second order and which satisfied Laplace's equation in two dimensions is called a harmonic function.
i.e. $\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$ or $\overline{V}^2 \phi = 0$
Given f(z) = u + iv is an analytic function
By Cauchy Riemann equations.
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} $
Now
$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial x} \bigg(\frac{\partial u}{\partial x} \bigg) + \frac{\partial}{\partial y} \bigg(\frac{\partial u}{\partial y}\bigg)$$
$$= \frac{\partial}{\partial x} \bigg(\frac{\partial v}{\partial y} \bigg) + \frac{\partial}{\partial y} \bigg(\frac{- \partial v}{\partial x} \bigg)$$
$$= \frac{\partial^2 v}{\partial x \partial y} - \frac{\partial^2 v}{\partial x}{\partial y}$$
$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$$
u satisfies Laplace's equation
$\therefore$ u is harmonic.
Similarly,
$$\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial x} \bigg(\frac{\partial v}{\partial x} \bigg) + \frac{\partial}{\partial y} \bigg(\frac{\partial v}{\partial y}\bigg)$$
$$= \frac{\partial}{\partial x} \bigg(\frac{- \partial u}{\partial y}\bigg) + \frac{\partial}{\partial y} \bigg(\frac{\partial u}{\partial x} \bigg)$$
$$= - \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial x \partial y}$$
$$\therefore \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$$
$\therefore$ v satisfies Laplace's equation
$\therefore$ v is harmonic.