Question

If −3+ix2y and x2+y+4i are conjugate of each other then (x,y)= ?

Answer 1

Let X be original complex no and X¹ be its conjugate
if a complex number is z=A+iB. and its conjugate would be z¹= A-iB


Z= -3 + i(x²y)

Z'= -3-i(x²y). ---------- (1)

on comparing (1) with the given conjugate Z¹= (x²+y) + 4i

x²+y = -3 --------- (2) and

x²y= -4 -----------(3)

From (2)
x²= -(3+y)

Substituting value of x² in (3)
-(3+Y)Y= -4
y²+3y-4=0
on solving we get
y= 1, -4

at y= 1, x=Not defined as 3+y will be positive and any square cannot be negative at any point

at y= -4. X= +1, -1


so (X,Y) = (1, -4) and (-1, -4) .......ANSWER

Answer 2

Thank you for asking this question, here is your answer:

Let z1=−3+ix2y and z2=x2+y+4i

z1=−3−ix2y  and z2¯¯¯¯¯=x2+y−4i

z1=z2 and z2=z1

−3−ix2y=x2+y+4i  and x2+y−4i=−3+ix2y

x2+y=−3 and x2y=−4

−4/y +y=−3

y2+3y−4=0

y=−4 or y=1

wheny=−4,x=±1

wheny=1,x2=−4 (this is not possible)

(x,y)=(±1,−4)

If there is any confusion please leave a comment below.