Points z1 z2 are adjacent vertices of a regular octagon. The vertex z3 adjacent to z2(z3 ≠ z1) is represented by :

Question

Points z1 z2 are adjacent vertices of a regular octagon. The vertex z3 adjacent to z2(z3 ≠ z1) is represented by : (A) z2+(1/√2)(1 ± i)(z1+z2) (B) z2+(1/√2)(1 ± i)(z1-z2) (C) z2+(1/√2)(1 ± i)(z2-z1) (D) z1+(1/√2)(1 ± i)(z1+z2)

Tardigrade Admin · Accepted Answer

(z3-z2/z2-z1)= e ± i (π/4) ∵ |z1-z2|=|z2-z3| <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/8fa15a3f8ac4de24bd883ec1be49912e-.png" /> ⇒ z3=z2+((1+i/√2))(z2-z1)

Q. Points $z_{1} & z_{2}$ are adjacent vertices of a regular octagon. The vertex $z_{3}$ adjacent to $z_{2} (z_{3} \neq = z_{1})$ is represented by :

$z_{2} + \frac{1}{2} (1 \pm i) (z_{1} + z_{2})$

$z_{2} + \frac{1}{2} (1 \pm i) (z_{1} - z_{2})$

$z_{2} + \frac{1}{2} (1 \pm i) (z_{2} - z_{1})$

$z_{1} + \frac{1}{2} (1 \pm i) (z_{1} + z_{2})$

$\frac{z _{3} - z _{2}}{z _{2} - z _{1}} = e^{\pm i \frac{π}{4}}$
$∵ ∣ z_{1} - z_{2} ∣ = ∣ z_{2} - z_{3} ∣$

$\Rightarrow z_{3} = z_{2} + (\frac{1 + i}{2}) (z_{2} - z_{1})$

Q. Points z1​&z2​ are adjacent vertices of a regular octagon. The vertex z3​ adjacent to z2​(z3​=z1​) is represented by :

z2​+2​1​(1±i)(z1​+z2​)

z2​+2​1​(1±i)(z1​−z2​)

z2​+2​1​(1±i)(z2​−z1​)

z1​+2​1​(1±i)(z1​+z2​)