Question

Only For Maths Aryabatta

Let a, b, c be three distinct complex numbers such that
|a|=|b|=|c|=1 and z₁ ,z₂ are the roots of the equation az² + bz + c = 0, with |z₁| = 1.Also P and Q represent z₁ and z₂ on complex plane with ∠POQ = θ and O be the origin, then
A) b² = ac, θ = 2π/3
B) θ = 2π/3, PQ = √3
C) PQ = 2√3, b² = ac
D) θ = π/3, b² = ac

If unrelavent answer is posted for points, it will be reported.​

Answer 1

EXPLANATION.

Let a, b, c are three distinct complex number

such that

|a| = |b| = |c| = 1

Z1, Z2 are the roots of the equation

az^2 + bz + c = 0 with |z1| = 1

also p and q represent z1, z2 on complex plane

with <POQ = Ø be the origin

Solutions.

$| z_{1} z_{2} | = | \frac{c}{a} | = 1$

$| z_{1} \: + z_{2} | = | \frac{ - b}{a} | = 1$

$(z_{1} \: + z_{2})( \bar{z_{1}} + \bar{z_{2} }) | z_{1} \: + z_{2} | {}^{2} = 1$

$2 + \bar{z_{1}} z_{2} \: + z_{1} \bar{ z_{2} } = 1$

$\frac{( z_{1} \: + z_{2}) {}^{2} }{ z_{1} z_{2} } = 1$

$\frac{ {b}^{2} }{ {a}^{2} } = \frac{c}{a}$

${b}^{2} = ac$

Now,

$z_{2} = z_{1}e {}^{i \theta}$

$| z_{1} \: + z_{2} \: | = | z_{1} | \: |1 + e {}^{i \theta} |$

$2 \cos( \frac{ \theta}{2} ) | \cos( \frac{ \theta}{2} ) + i \cos( \frac{ \theta}{2} ) |$

$| z_{1} + z_{2} | = 2 \cos( \frac{ \theta}{2} ) = 1$

$\frac{ \theta}{2} = \frac{\pi}{3}$

$\theta \: \frac{2\pi}{3}$

$pq \: = | z_{2} \: - z_{1} | = | z_{1} | |e {}^{i \theta} - 1|$

$|2 \sin( \frac{ \theta}{2} ) |$

$\theta \: = \frac{2\pi}{3}$

PQ = √3

Therefore,

both Option [A] and [ B ] are correct

Answer 2

Step-by-step explanation:

[tex]EXPLANATION.

EXPLANATION.

Let a, b, c are three distinct complex number

such that

|a| = |b| = |c| = 1

Z1, Z2 are the roots of the equation

az^2 + bz + c = 0 with |z1| = 1

also p and q represent z1, z2 on complex plane

with <POQ = Ø be the origin

Solutions.

| z_{1} z_{2} | = | \frac{c}{a} | = 1∣z1z2∣=∣ac∣=1

| z_{1} \: + z_{2} | = | \frac{ - b}{a} | = 1∣z1+z2∣=∣a−b∣=1

(z_{1} \: + z_{2})( \bar{z_{1}} + \bar{z_{2} }) | z_{1} \: + z_{2} | {}^{2} = 1(z1+z2)(z1ˉ+z2ˉ)∣z1+z2∣2=1

2 + \bar{z_{1}} z_{2} \: + z_{1} \bar{ z_{2} } = 12+z1ˉz2+z1z2ˉ=1

\frac{( z_{1} \: + z_{2}) {}^{2} }{ z_{1} z_{2} } = 1z1z2(z1+z2)2=1

\frac{ {b}^{2} }{ {a}^{2} } = \frac{c}{a}a2b2=ac

{b}^{2} = acb2=ac

Now,

z_{2} = z_{1}e {}^{i \theta}z2=z1eiθ

| z_{1} \: + z_{2} \: | = | z_{1} | \: |1 + e {}^{i \theta} |∣z1+z2∣=∣z1∣∣1+eiθ∣

2 \cos( \frac{ \theta}{2} ) | \cos( \frac{ \theta}{2} ) + i \cos( \frac{ \theta}{2} ) |2cos(2θ)∣cos(2θ)+icos(2θ)∣

| z_{1} + z_{2} | = 2 \cos( \frac{ \theta}{2} ) = 1∣z1+z2∣=2cos(2θ)=1

\frac{ \theta}{2} = \frac{\pi}{3}2θ=3π

\theta \: \frac{2\pi}{3}θ32π

pq \: = | z_{2} \: - z_{1} | = | z_{1} | |e {}^{i \theta} - 1|pq=∣z2−z1∣=∣z1∣∣eiθ−1∣

|2 \sin( \frac{ \theta}{2} ) |∣2sin(2θ)∣

\theta \: = \frac{2\pi}{3}θ=32π

PQ = √3

Therefore,

both Option [A] and [ B ] are correct