Solve for $x, y $ in this complex equation.

Question

$4x² + 3xy + (2xy - 3x²)i = 4y² - (x²/2) + (3xy - 2y²)i$

It is given that $x, y$ are real numbers. And $i = \sqrt{-1}$

Answer 1

HINT

Equating real and imaginary parts, we have that

• $4x^2+3xy=4y^2-x^2/2\implies 3xy=4y^2-9x^2/2$

• $2xy-3x^2=3xy-2y^2 \implies xy=2y^2-3x^2$

that is

• $6y^2-9x^2=4y^2-9x^2/2\implies 9x^2/2=2y^2\implies 9x^2=4y^2$

Answer 2

Equating real an imaginary parts we get the system:

$$4x^2+3xy=4y^2-\dfrac {x^2}{2}$$

$$2xy-3x^2=3xy-2y^2$$

Rearranging we get

$$9x^2-8y^2=-6xy$$

$$2y^2-3x^2=xy$$

We now elminate $y^2$: Multiply the second equation by $4$ and then add the two equations together. Assuming $x \not = 0$ (check for a solution in this case too) we get:

$$x = \dfrac {2}{3}y$$

And now it is not hard to proceed.

Make sure you check for extraneous solutions.