• Any function which satisfy the relation$\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0 $ is said to satisfies Laplace equation.
• Given u=sin x cosh y+2 cos x sinh y+ x2-y2+4xy
• Differentiating u with respect to x we have

$\dfrac{\partial u}{\partial x}=\cos x \cosh y-2\sin x \sinh y +2x+4y$

• Differentiating it again with respect to x we have

$\dfrac{\partial^2u}{\partial x^2}=-\sin x \cos hy-2 \cos x \sin hy+2$.......1

• Differentiating u with respect to y we have

$\dfrac{\partial u}{\partial y}=\sin x \sin hy+2 \cos x \cos hy-2y+4x$

• Differentiating it again with respect to y we have

$\dfrac{\partial^2 u}{\partial y^2}=\sin x \cos hy+2 \cos x \sin hy-2$..........2

• Adding 1 and 2 we get$\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0 $

Thus u=sin x cosh y+2 cos x sinh y+ x2-y2+4xy satisfies Laplace's equation.

• Given u=sin x cosh y+2 cos x sinh y+ x2-y2+4xy.
• Differentiating u with respect to x we have

$\psi1(x,y)=\dfrac{\partial u}{\partial x}=\cos x \cos hy-2\sin x \sinh y +2x+4y$

• Differentiating u with respect to y we have

$\psi2(x,y)=\dfrac{\partial u}{\partial y}=\sin x \sin hy+2\cos x \cos hy-2y+4x$

• By Milne thompson method
• f'(*z)=$\psi1(z,0)-i \psi2(z,0)$.......(Substituting x=z,y=0 in$\psi1$ and$\psi2$ )
• f'(z)=cosz +2z-i (2cosz +4z)

$\therefore $f(z)=$\int(\cos z+2z)\mathrm{d}z-i \int(2\cos z+4z)\mathrm{d}z$

$\therefore f(z)=\sin z+z^2-i (2\sin z+2z^2)$