IIT JEE , AIEEE Forum - Mathematics , Algebra

shinee

sorry 4 not typing the question properly

the q is x+3i/(2+iy) =1-i
and i solved it like this,
(x+3i)(2-iy)/(4+ysquare) =1-i
2x+3y+6i-ixy=4+ysquare-4i-iysquare
2x+3y=4+ysquare (1)
6-xy= -4-ysquare (2)
x=(4+ysquare-3y)/2 from (1)
substituting in (2)
6 -{(4+ysquare-3y)/2}y =-4-ysquare
ycube-5ysquare+4y-20=0
solving
y=(plus or minus)2i or y =5
and x=(plus or minus)3i or x=7
if we substitute y=5 and x=7, then we r getting the value as 0, but y did we neglect the imaginary answers

Ask & Discuss Questions with Community & Experts