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

## Answers

Answered by

76

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

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

Answered by

53

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.

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