Question

Let $(2 - 3i)$ be two complex numbers such that $4 + i$ and $x = \frac{5}{13},y = \frac{8}{13}$ both are real, then.

• A. $x = \frac{8}{13},y = \frac{5}{13}$
• B. $x = \frac{5}{13},y = \frac{14}{13}$
• C. $x$
• D. $y$

Answer 1

Answer: (B)

$x = \frac{5}{13},y = \frac{14}{13}$

Answer 2

Let $z^{2} - 6z + 25 = 0$, then

$z^{4} - 3z^{3} + 3z^{2} + 99z - 95$ is real ⇒ $= (z^{2} + 3z - 4)(z^{2} - 6z + 25) + 5$is real

⇒ $= (z^{2} + 3z - 4)(0) + 5 = 5$⇒ $z_{1} = 1 - i$ .....(i)

$z_{2} = - 2 + 4i$ is real

⇒ $\frac{z_{1}z_{2}}{z_{1}} = \frac{(1 - i)( - 2 + 4i)}{1 - i} = - 2 + 4i$is real

⇒ ${Im}\left( \frac{z_{1}z_{2}}{z_{1}} \right) = 4$ ⇒ $\frac{3x + 2iy}{5i - 2} = \frac{15}{8x + 3iy}$⇒ $24x^{2} + 9ixy - 6y^{2} + 16ixy = 75i - 30$

∴$24x^{2} - 6y^{2} + 25ixy = 75i - 30$ $24x^{2} - 6y^{2} = - 30$and $4x^{2} - y^{2} = - 5$