Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 6M

Year: May 2014

Given, $v = x^2 - y^2 + \frac{x}{x^2+y^2}$

Differentiating with respect to r we get

$v_x = 2x - 0 + (x^2+y^2) frac{∂}{∂x} r - x \frac{∂}{∂x} (x^2+y^2 )/(x^2+y^2)^2$

$v_x = 2x + \frac{x^2+y^2 - x(2x)}{(x^2+y^2)^2}$

$v_x = 2x + \frac{(y^2 - x^2)}{(x^2+y^2)^2}$

Differentiating v with respect to y

$v_y = 0 - 2y + x \frac{-1 (2y)}{(x^2+y^2)^2}$

$v_y = - 2y - \frac{2xy}{(x^2+y^2)^2}$

Let f(z) = u + i v be analytic

Therefore, $f^1 (z) = v_y + iv_x$

$= -2y - \frac{2xy}{(x^2 + y^2)^2} + i [ 2x + \frac{y^2 - x^2}{(x^2 + y^2)^2}$

By Milne Thompsons method, put x=z & y=0

Therefore, $f^1(z) = 0 - 0 + i[2z + \frac{(-z)^2}{(z^2)^2}]$

$= i[2z - \frac{1}{z^2}]$

$f(z) = \int f^1 (z)dz$

$= \int i(zz - z^2 )dz$

$= i \bigg[\frac{2z^2}{2} - \frac{z^3}{3} \bigg]dz$

$= i(z^2 + \frac{1}{z})+k$

$u + i v = i\bigg[(x + iy)^2 + \frac{1}{(x+iy)}\bigg] + k$

$= i\bigg[x^2 + 2xiy - y^2 + \frac{x-iy}{x^2 + y^2}\bigg] + k$

$= i \bigg[x^2 - y^2 + \frac{x}{x^2 + y^2}\bigg] - 1[2xy - \frac{y}{x^2 + y^2}]$

Therefore, equating real and imaginary parts

$ u = -2xy + \frac{y}{x^2+y^2} \hspace{0.2cm} and \hspace{0.2cm} v = x^2 - y^2 + \frac{x}{x^2 + y^2}$