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

Marks: 6M

Year: Dec 2015

Let $f (z)=u+iv(ax^2+bx^2y^2+cy^4+dx^2-2y^2) + i(4x^3y-exy^3+4xy)$

Comparing both sides we get,

$u =ax^2+bx^2y^2+cy^4+dx^2 - 2y^2 \text{and} \hspace{0.2cm} v = 4x^3y-exy^3+4xy$

Partially differentiating w. r. t x we get

$ux=4ax^3+2bxy^2+0+2dx=0\hspace{0.2cm} \text{and} \hspace{0.2cm} vx=12x^2y+ey3+4y$

Partially differentiating w. r. t x we get

$ uy=0+2bx^2y+4cy^3+0-4y \hspace{0.2cm} and \hspace{0.2cm} vy=4x^3-3exy^2+4x $

Since f (z) is analytic, by Cauchy Reimann’s equations,

ux=vy and uy= - vx

$∴ 4ax^3+2bxy^2 + 2dx = 4x^3-3exy^2 + 4x \hspace{0.2cm} \text{and} \hspace{0.2cm} 2b2^y + 4xy^3 - 4y = -(12x^2y - ey^3 + 4y)$

Comparing coefficient of x and y, we get

4a=4

∴a=1

2d=4

∴d=2

2b=-3e_____(1)

2b=-12

∴b=-6______(2)

From (1) & (2) , 2 (-6) = -3e

∴ e=4

4c=e

∴4c=4

∴c=1

Ans: a=1 ; b=-6 ; c=1 ; d=2 ; e=4