Question

The real values of $\frac{(1 + i)x - 2i}{3 + i}$and$+ \frac{(2 - 3i)y + i}{3 - i} = i$for which the equation is $x = - 1,y = 3$ $x = 3,y = - 1$= $x = 0,y = 1$ is satisfied, are.

• A. $x = 1,y = 0$
• B. $z_{1}$
• C. $z_{2}$
• D. None of these

Answer 1

Answer: (C)

$z_{2}$

Answer 2

Equation $(z) = 0$

⇒ $\frac{8\sin\theta}{1 + 4\sin^{2}\theta}$

Equating real and imaginary parts, we get

$\sin\theta = 0$ ......(i)

$\therefore$ ......(ii)

From (i) and (ii), we get $\theta = n\pi$

Aliter : $n = 0$ .