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 ΔOAB is ?


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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

	$O k t h a n k s . I m i s t a k e n e d i t f o r ‘ ‘ t h e v e c t o r p e r p e n d i c u l a r t o . . .^{″}$
	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
	$say a=Ae^(αi) say b=Be^(βi) let k=(A/B), θ=α−β (A/B)e^((α−β)i) +(B/A)e^(−(α−β)i) =1 ke^(θi) +(1/k)e^(−θi) =1 (k+(1/k))cos θ+(k−(1/k))sin θ i=1 ⇒(k+(1/k))cos θ=1 ⇒(k−(1/k))sin θ=0 ⇒ { ((k−(1/k)=0 ⇒k=1)),((sin θ=0 ⇒θ=nπ)) :} with k=1: (A/B)=1 cos θ=(1/2) ⇒θ=2nπ±(π/3) ⇒ΔAOB is equilateral with sin θ=0: cos θ=±1 k+(1/k)=±1 ⇒no solution for k>0 therefore only possibility is ΔAOB is equilateral triangle.$
	$s a y a = {A e}^{α i}$ $s a y b = {B e}^{β i}$ $l e t k = \frac{A}{B}, θ = α - β$ $\frac{A}{B} e^{(α - β) i} + \frac{B}{A} e^{- (α - β) i} = 1$ ${k e}^{θ i} + \frac{1}{k} e^{- θ i} = 1$ $(k + \frac{1}{k}) \cos θ + (k - \frac{1}{k}) \sin θ i = 1$ $\Rightarrow (k + \frac{1}{k}) \cos θ = 1$ $\Rightarrow (k - \frac{1}{k}) \sin θ = 0 \Rightarrow {\begin{cases} k - \frac{1}{k} = 0 \Rightarrow k = 1 \\ \sin θ = 0 \Rightarrow θ = n π \end{cases}$ $w i t h k = 1 :$ $\frac{A}{B} = 1$ $\cos θ = \frac{1}{2} \Rightarrow θ = 2 n π \pm \frac{π}{3}$ $\Rightarrow Δ A O B i s e q u i l a t e r a l$ $w i t h \sin θ = 0 :$ $\cos θ = \pm 1$ $k + \frac{1}{k} = \pm 1 \Rightarrow n o s o l u t i o n f o r k > 0$ $t h e r e f o r e o n l y p o s s i b i l i t y i s$ $Δ A O B i s e q u i l a t e r a l t r i a n g l e .$
	Commented by EnterUsername last updated on 22/Apr/21

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