Let $a$, $b$, $c$ be distinct complex numbers with $|a| = |b| = |c| = 1$ and $z_1$, $z_2$ be the roots of the equation $az^2 + bz + c = 0$ with $|z_1| = 1$. Also $P$ and $Q$ are the points representing the complex numbers $z_1$ and $z_2$ respectively in the complex plane with $\angle POQ = \theta$ (where $O$ being the origin) then which of the following is/are <b>correct</b>?
Step-by-Step Solution
Step 1: Since $|a|=|b|=|c|=1$ and $z_1, z_2$ are roots of $az^2+bz+c=0$, by Vieta's formulas:
$$z_1 + z_2 = -\frac{b}{a}, \quad z_1 z_2 = \frac{c}{a}.$$
Step 2: Since $|a|=|c|=1$, we have $|z_1 z_2| = \left|\dfrac{c}{a}\right| = 1$, so $|z_1||z_2|=1$. Given $|z_1|=1$, it follows $|z_2|=1$.
Step 3: Since $|z_1|=|z_2|=1$, both roots lie on the unit circle. Also $|z_1+z_2| = \left|\dfrac{b}{a}\right| = \dfrac{|b|}{|a|} = 1$. So option (d) $|z_1+z_2|=1$ is correct.
Step 4: Since $|z_1|=|z_2|=1$ and $|z_1+z_2|=1$, the triangle formed by $O$, $P$, $Q$ is equilateral (all sides equal to 1 is not immediate; use the law of cosines). With $|OP|=|OQ|=1$ and $|z_1+z_2|=1$:
$$|z_1-z_2|^2 = |z_1|^2 + |z_2|^2 - 2\text{Re}(z_1\overline{z_2}) = 2 - 2\cos\theta.$$
Also $|z_1+z_2|^2 = 2 + 2\cos\theta = 1$, so $\cos\theta = -\dfrac{1}{2}$, giving $\theta = \dfrac{2\pi}{3}$. So option (b) is correct.
Step 5: Compute $PQ = |z_1 - z_2|$:
$$|z_1-z_2|^2 = 2 - 2\cos\theta = 2 - 2\left(-\frac{1}{2}\right) = 3 \Rightarrow PQ = \sqrt{3}.$$
So option (c) is correct.
Step 6: Check option (a). $b^2 = ac$ would mean the discriminant $b^2 - 4ac = ac - 4ac = -3ac \neq 0$ in general, so the roots are not necessarily equal. Since $a$, $b$, $c$ are distinct, $b^2 = ac$ is not necessarily true. Option (a) is not always correct.
Correct Answer: 2, 3, 4