Let $z \in \mathbb{C}$ be such that $\frac{z^{2}+3i}{z-2+i}=2+3i$. Then the sum of squares of all possible values of $z^{2}$ is
- (1) $19-2i$
- (2) $-19-2i$
- (3) $19+2i$
- (4) $-19+2i$
Solution:
$$\frac{z^{2}+3i}{z-2+i} = 2+3i$$
$$z^{2}+3i = (z-2+i)(2+3i)$$
$$z^{2}+3i = z(2+3i) – 2(2+3i) + i(2+3i)$$
$$z^{2}+3i = z(2+3i) – 4 – 6i + 2i – 3$$
$$z^{2}+3i = z(2+3i) – 7 – 4i$$
$$z^{2} – z(2+3i) + 7 + 7i = 0$$
Let the roots (possible values of $z$) be $z_1$ and $z_2$.
Sum of roots: $z_1 + z_2 = -\frac{b}{a}$
$$z_1 + z_2 = 2+3i$$
Product of roots: $z_1 z_2 = \frac{c}{a}$
$$z_1 z_2 = 7+7i$$
$$z_1^2 + z_2^2 = (z_1+z_2)^2 – 2z_1 z_2$$
$$z_1^2 + z_2^2 = (2+3i)^2 – 2(7+7i)$$
$$z_1^2 + z_2^2 = (4 + 12i + 9i^2) – (14 + 14i)$$
$$z_1^2 + z_2^2 = (4 + 12i – 9) – 14 – 14i$$
$$z_1^2 + z_2^2 = -5 + 12i – 14 – 14i$$
$$z_1^2 + z_2^2 = -19 – 2i$$
Ans. (2)