Question

Find the original function without finding the corresponding conjugate
u(x,y)=x/x^2+y^2

Answer 1

For find tho original function we use the Cauchy-Riemann condition:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \\$

$u = \frac{x}{x^2 + y^2}$

$\frac{\partial u}{\partial x} = \frac{y^2 - x^2}{(x^2+y^2)^2}=\frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y} = -\frac{2yx}{(x^2+y^2)^2} =-\frac{\partial v}{\partial x}$

$v = \int\frac{y^2 - x^2}{(x^2+y^2)^2}dy + \phi(x) = -\frac{y}{x^2+y^2} + \phi(x)$

$\frac{\partial v}{\partial x} = \frac{2yx}{(x^2+y^2)^2} + \phi\prime(x) = \frac{2yx}{(x^2+y^2)^2}$

$\phi\prime(x) = 0 ,\phi(x) = C$

So: $v(x,y) = -\frac{y}{x^2+y^2} + C$ , and $w(u,v) = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$