Question

if x,y are real and (y^2 x- 5i) and {4+i(x+y^2)} are conjugate to each other, find x and y.

Answer 1

(y²x -5i ) and { 4+ i(x² + y²)} are conjugate each other .

concept :- if any complex number in the form of a + ib then conjugate of it is
a - ib .

solution :-
y²x -5i = a - ib { like }
4 + i(x² + y²) = a + ib { like }
so, y²x = 4
y² = 4/x -----(1)

5 = x² + y² -------(2)
put eqn (1) in eqn (2)

5 = x² + 4/x

5x = x³ + 4

x³ - 5x + 4 = 0

x³ - x² + x² - x -4x + 4 =0

x²(x -1) + x (x -1) -4(x -1) = 0

(x -1)(x² + x -4) = 0

x =1 and x² +x -4 = 0

x²+ x - 4 =0
x = { -1 ±√(1 + 16) }/2
but x ≠ negative because
y² = 4/x
so, x = { -1 + √17}/2 and 1
and y² = 4/1 = 4
y= ±2

again,
y² = 4/(-1 +√17)/2
y² = 8/( -1 + √17)
y = ± 2√2/√(-1 + √17)

hence, value of x = 1 and ( -1+√17)/2
and y = ±2 and ± 2√2/√(-1+√17)