An algebra problem by abhi jith

Let $z_1 = a + bi, z_2 = c + di$ be two distinct complex numbers. If $z + \bar{z} = 2|z - 1|,$ then:

$z_1 +\bar{z_1} = 2|z_1 - 1| \Rightarrow (a + bi) + (a - bi) = 2\sqrt{(a-1)^{2} + b^{2}} \Rightarrow 2a = 2\sqrt{(a-1)^{2} + b^{2}},$

or $a^{2} = (a-1)^{2} + b^{2} \Rightarrow a^{2} = a^{2} - 2a + 1 + b^{2} \Rightarrow a = \frac{b^{2} + 1}{2}$ . Likewise, $c = \frac{d^{2} + 1}{2}$ for $z_2$ , which produces $z_1 = \frac{b^{2} + 1}{2} + bi, z_2 = \frac{d^{2} + 1}{2} + di.$

If $arg(z_1 - z_2) = \frac{\pi}{4},$ then we have: $z_1 - z_2 = \frac{b^{2} - d^{2}}{2} + (b - d)i$ , and $\frac{\pi}{4} = arctan(\frac{b - d}{\frac{b^{2} - d^{2}}{2}}) = arctan(\frac{2\cdot(b - d)}{(b+d)(b-d)}) = arctan(\frac{2}{b+d}) \Rightarrow b + d = 2.$ , which now produces:

$z_1 = \frac{b^{2} + 1}{2} + bi; z_2 = \frac{(2-b)^{2} + 1}{2} + (2 - b)i.$

Upon observation,

$Re(z_1 + z_2) = b^{2} - 2b + 3$ , which only equals 2 for $b = 1 \Rightarrow$ choice A is incorrect,

$Im(z_1 + z_2) = 2 \Rightarrow$ choice B is correct.

Both $z_1,z_2$ lie on the parabola $Im(z)^{2} = 2Re(z) - 1 \Rightarrow$ choice C is correct.

In order for $z_1, z_2$ to not exist, we require $Re(z_1, z_2) = Im(z_1, z_2) = 0$ for some real value of $b$ . This is impossible since $Re(z_1), Re(z_2) > 0$ for all $b \in \mathbb{R} \Rightarrow$ choice D is incorrect.

Hence, choices B and C are the correct combination.