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Show that the function $v = e^x (xsiny + ycosy)$ satisfies Laplace equation. Find its corresponding analytic function & its harmonic conjugate.
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Given imaginary part $v = e^x (xsiny + ycosy)$

$\therefore v_{x} = \frac{\partial V}{\partial x} = e^x (sin y) + (xsin y + ycosy)e^x \\ \therefore v^{2}_{x} = \frac{\partial^2 V}{\partial x^2} = e^x (sin y) + e^x (sin y) + (xsiny + ycosy)e^x \\ v^{2}_{x} = e^x (-xsiny + y(-cosy) - siny - siny) \\ v_{y} = \frac{\partial V}{\partial y} = e^x (-xsiny + y(-csoy) - siny -siny) \\ v_{y} = \frac{\partial V}{\partial y} = e^x (-xsiny - ycosy) - 2siny) \\ v^{2}_{y} = e^x(-xsiny + y(-cosy) - siny - siny) \\ = e^x(-xsiny - ycosy - 2siny)$

$v^{2}_{y} = -2e^x (siny) - e^x (xsiny + ycosy)$

$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0$

V satisfies Laplace equation

Now, let

$\psi_{1}(x,y) = v_{y} = e^x (xcosy + y(-siny) + cosy) \\ \psi_{2}(x,y) = v_{x} = e^x siny + (xsiny + ycosy)e^x$

Now put x = z, y = 0

$\psi_{1}(z, 0) = e^z (z + 1) \\ \psi_{2} (z,0) = 0$

Now, by Milne Thompson method,

$f^1 (z) = \psi_{1}(z,0) + i\psi_{2}(z,0) \\ = e^z (z+1) + i(0) \\ f^1 (z) = e^z(z + 1)$

Integrating above equation,

$F(z) = \int e^z (z + 1)dz \\ = (z+1)e^z + c$

(using Leibnitz rule)

Analytic function,

$F(z) = (z + 1)e^z + c$

Now put z = x + iy

$\therefore u + iv = (x + iy)e^{(x + iy)} + c \\ = (x + iy)e^x * e^{iy} + c \\ = e^x (x + iy) (cosy + isiny) + c$

$U + iV = e^x (xcosy - ysiny) + i(xsiny + ycosy) + c$

The above equation of a is corresponding harmonic conjugate

Analytic function $f(z) = ze^z$

Corresponding harmonic conjugate $w = e^x(xcosy - ysiny)$

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