AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

since x2+y+4i and −3+x2yi are conjugate. So,
x2+y+4i=−3−x2yi
(x2+y)+4i=(−3)−x2yi
Compare both the sides.
x2+y‌‌=−3...(1)
4‌‌=−x2y
y‌‌=‌

4

−x2

y‌‌=−‌

4

x2

....(2)
Substitute the value of y in equation (1) and (2)

x2−‌

4

x2

=−3
x4+3x2−4=0
(x2+4)(x2−1=0)
x=±1
Substitute the value of x in equation (2)
y=‌

−4

(1)2

y=−4
The value is calculated as,
(|x2|+|y2|)‌‌=|x2|+|y2|+2|x||y|
‌‌=(1)2+(−4)2+2(1)(−4)
‌‌=1+16+8
‌‌=25