Question

Find the value of x and y for which (3 + 4i) x² – (4 – 3i) y = 3x – 4y + 6i, where x, y ∈ R.
1. x = 0 and y = 2
2. \(x = 1\;and\;y = \frac{2}{3}\)
3. \(x = 0,\;1\;and\;y = 2,\;\frac{2}{3}\)
4. \(x = 0,\;1\;and\;y = 1,\;\frac{1}{3}\)

Answer 1

Correct Answer - Option 3 : \(x = 0,\;1\;and\;y = 2,\;\frac{2}{3}\)

CONCEPT:

Two complex number z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ are equal if and only if their real and imaginary parts are equal individually.

I.e. z₁ = z₂

⇒ Re(z₁) = Re(z₂) and I_m (z₁) = I_m(z₂)

CALCULATION:

Given (3 + 4i) x² – (4 – 3i) y = 3x – 4y + 6i

⇒ 3x² – 4y = 3x – 4y

⇒ 3x² - 3x = 0

∴ x² - x = 0

⇒ x = 0, 1

Now we will equate the imaginary part.

∴ 4x² + 3y = 6

So, when x = 0; y = 2 and when x = 1, y = 2 / 3