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

Marks: 8M

Year: Dec 2015

By Cauchy Reimann’s equations,

ux=vy and uy= - vx………………(1)

Given , $u + v = \frac{2 sin 2x}{e^{2y} + e^{-2y} - 2 cos 2x}\\ \hspace{1cm} = \frac{2 sin 2x}{2 cosh 2y - 2 cos 2x} \\ \hspace{1cm} = \frac{sin 2x}{cosh 2y - cos 2x} $

Let D = cosh2 y - cos 2 x Different partially W.R.T(x)

$\therefore ux + vx = \frac{(cosh 2 y - cos 2 x) cos 2 x . 2 - sin 2 x (0 + sin 2x - 2)}{D^2}\\ \hspace{1.7cm} = \frac{2(cosh 2y cos 2 x - cos^2 2 x - sin^2 - 2x)}{D^2} \\ \hspace{1.7cm} = \frac{2(cosh 2 y cos 2x -1)}{D^2} .......... ..... (2)$

Similarly differentiating 'u + v' partially w.r.t 'y'

$\therefore uy + vy = \frac{(cos h 2y - cos 2x).0 - sin 2x . (sinh 2y.2 - 0)}{D^2}\\ \therefore vy + uy = \frac{- 2 sin 2x sinh 2y}{D^2}\\ \therefore ux + vx = \frac{- 2 sin 2x sinh 2y}{D^2} (From 1) ....... (3)$

Adding (2) & (3)

$\therefore 2ux = \frac{2(cosh 2 y cos 2 x - 1)}{D^2} + \frac{- 2 sin 2 x sinh 2y}{D^2}\\ \therefore u_x = \frac{cosh 2y cos 2x - 1 - sin 2x sinh 2y}{D^2}$

Similarly, substract (3) from (2)

$\therefore 2vx = \frac{2 (cos 2y cos 2x - 1)}{D^2} + \frac{2 sin 2x sin h 2y}{D^2} \\ \therefore vx = \frac{cosh 2y cos 2x - 1 + sin 2x sinh. 2y}{D^2}\\ \therefore f^1 (z) = ux + ivx \\ \frac{2 (cos 2y cos 2x - 1 - sin 2x sinh 2y)}{(cos 2y - cos 2x)^2} + i \frac{cosh 2y cos 2x - 1 + sin 2x sinh2y}{(cosh2y - cos2x)^2}$

By Milne Thompson’s method, put x=2 & y=0

$\therefore f^1(z) = \frac{1.cos2z - 1 - 0}{(1 - cos2z)^2} + i. \frac{1.cos2z - 1+ 0}{(1 - cos2z)^2}\\ = \frac{-1(1-cos2z)}{(1 - cos2z)^2}+i\frac{-1(1 - cosz)}{(1 - cos 2z)^2}\\ = \frac{-1}{1 - cos2z}(1 + i)\\ = \frac{-1}{2 sin^2 z}(1 + i)\\ \frac{1 + i}{2} cosec^2z\\ \therefore f(z) = \int f^1(z) dz\\ = - \int \frac{1 + i}{2} cosec^2 z dz\\ \therefore F(z) = \frac{1 + i}{2} cot z + k$