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Prove that real and imaginary parts of an analytic function f(z) = u + iv are harmonic function.
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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.

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