Respuesta :
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
[tex]-3+ix^2y\\x^2+y+4i\\\\x^2+y=-3\\ix^2y=-4i\\\\x^2=-3-y\\x^2=-\dfrac{4}{y}\\\\-3-y=-\dfrac{4}{y}\\3y+y^2=4\\y^2+3y-4=0\\y^2-y+4y-4=0\\y(y-1)+4(y-1)=0\\(y+4)(y-1)=0\\y=-4 \vee y=1\\\\x^2=1 \vee x^2=-4\\x=-1 \vee x=1\\\\\boxed{(x,y)\in\{(-1,-4),(1,-4)\}}[/tex]
