For what values of x and y are the nos 3+ix2y and x2+y+4i conjugate complexes?

Complex conjugate of 3+i.x2y is 3-i.x2y where x and y are real.
It is given that complex conjugate of 3+i.x2y is x2+y+i.4
therefore,
3-i.x2y=x2+y+i.4
comparing the real and imaginary parts, we have:

x2+y=3 ........(1) x2y=-4 ..........(2)y=-4x2now, from 1, we getx2-4x2=3x4-3x2-4=0x4-4x2+x2-4=0x2(x2-4)+1(x2-4)=0(x2-4)(x2+1)=0x2-4=0 or x2+1=0x2=4 or x2=-1Rejected

x=±2 since x is a real number

therefore 2 and -2 are two values of x and 

y=-4x2=-44=-1

hope this helps you

  • 5
Since they are conjugate complexes , 3 = x2+ y and x2y = - 4
now, solving them yields x =± 2 and y = - 1
  • 3
plssolve in detail.
  • 0
i get the answer. thankxx
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