Let-a-and-b-be-complex-numbers-representing-the-points-A-and-B-respectively-in-the-complex-plane-If-a-b-b-a-1-and-O-is-the-origin-Then-OAB-is-

Question Number 139055 by EnterUsername last updated on 21/Apr/21

Let a and b be complex numbers representing the points

A and B respectively in the complex plane .

If \frac{a}{b} + \frac{b}{a} = 1 and O is the origin . Then Δ OAB is ?

Answered by MJS_new last updated on 22/Apr/21

\frac{a}{b} + \frac{b}{a} = 1 \Rightarrow b = a (\frac{1}{2} \pm \frac{\sqrt{3}}{2} i)

a = p + q i \Rightarrow b = \frac{p \pm \sqrt{3} q}{2} + \frac{q \mp \sqrt{3} p}{2} i

A = (\begin{matrix} p \\ q \end{matrix}) \land B = (\begin{matrix} \frac{p \pm \sqrt{3} q}{2} \\ \frac{q \mp \sqrt{3} p}{2} \end{matrix})

A O = (\begin{matrix} - p \\ - q \end{matrix}) \land A B = (\begin{matrix} - \frac{p \pm \sqrt{3} q}{2} \\ - \frac{q \mp \sqrt{3} p}{2} \end{matrix})

\cos α = \frac{∣ A O ∙ A B ∣}{∣ A O ∣ \times ∣ A B ∣} = \frac{1}{2} \Rightarrow α = \frac{π}{3}

Commented by EnterUsername last updated on 22/Apr/21

A = (\begin{matrix} p \\ q \end{matrix}) \land B . I {d o n}^{'} t u n d e r s t a n d

Commented by mr W last updated on 22/Apr/21

i t m e a n s

A = (\begin{matrix} p \\ q \end{matrix}) a n d B = (\begin{matrix} \frac{p \pm \sqrt{3} q}{2} \\ \frac{q \mp \sqrt{3} p}{2} \end{matrix})

\land m e a n s a n d

\lor m e a n s o r

Commented by EnterUsername last updated on 23/Apr/21

O k t h a n k s . I m i s t a k e n e d i t f o r “ t h e v e c t o r p e r p e n d i c u l a r t o \dots^{″}

Commented by MJS_new last updated on 23/Apr/21

we use \overset{⇀}{a} ⊥ \overset{⇀}{b} for this

Answered by mr W last updated on 22/Apr/21

s a y a = {A e}^{α i}

s a y b = {B e}^{β i}

l e t k = \frac{A}{B}, θ = α - β

\frac{A}{B} e^{(α - β) i} + \frac{B}{A} e^{- (α - β) i} = 1

{k e}^{θ i} + \frac{1}{k} e^{- θ i} = 1

(k + \frac{1}{k}) \cos θ + (k - \frac{1}{k}) \sin θ i = 1

\Rightarrow (k + \frac{1}{k}) \cos θ = 1

\Rightarrow (k - \frac{1}{k}) \sin θ = 0 \Rightarrow {\begin{cases} k - \frac{1}{k} = 0 \Rightarrow k = 1 \\ \sin θ = 0 \Rightarrow θ = n π \end{cases}

w i t h k = 1 :

\frac{A}{B} = 1

\cos θ = \frac{1}{2} \Rightarrow θ = 2 n π \pm \frac{π}{3}

\Rightarrow Δ A O B i s e q u i l a t e r a l

w i t h \sin θ = 0 :

\cos θ = \pm 1

k + \frac{1}{k} = \pm 1 \Rightarrow n o s o l u t i o n f o r k > 0

t h e r e f o r e o n l y p o s s i b i l i t y i s

Δ A O B i s e q u i l a t e r a l t r i a n g l e .

Commented by EnterUsername last updated on 22/Apr/21

T h a n k s S i r