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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 (a/b)+(b/a)=1 and O is the origin. Then ΔOAB is ?

$$\mathrm{Let}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{representing}\:\mathrm{the}\:\mathrm{points} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{respectively}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}. \\ $$$$\mathrm{If}\:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\mathrm{1}\:\mathrm{and}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{Then}\:\Delta\mathrm{OAB}\:\mathrm{is}\:? \\ $$

Answered by MJS_new last updated on 22/Apr/21

(a/b)+(b/a)=1 ⇒ b=a((1/2)±((√3)/2)i) a=p+qi ⇒ b=((p±(√3)q)/2)+((q∓(√3)p)/2)i A= ((p),(q) ) ∧B= ((((p±(√3)q)/2)),(((q∓(√3)p)/2)) ) AO= (((−p)),((−q)) ) ∧ AB= (((−((p±(√3)q)/2))),((−((q∓(√3)p)/2))) ) cos α =((∣AO•AB∣)/(∣AO∣×∣AB∣))=(1/2) ⇒ α=(π/3)

$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1}\:\Rightarrow\:{b}={a}\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${a}={p}+{q}\mathrm{i}\:\Rightarrow\:{b}=\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}+\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}\mathrm{i} \\ $$$${A}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\wedge{B}=\begin{pmatrix}{\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}}\\{\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}}\end{pmatrix} \\ $$$${AO}=\begin{pmatrix}{−{p}}\\{−{q}}\end{pmatrix}\:\wedge\:{AB}=\begin{pmatrix}{−\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}}\\{−\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{cos}\:\alpha\:=\frac{\mid{AO}\bullet{AB}\mid}{\mid{AO}\mid×\mid{AB}\mid}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\alpha=\frac{\pi}{\mathrm{3}} \\ $$

Commented by EnterUsername last updated on 22/Apr/21

A= ((p),(q) ) ∧B. I don′t understand

$${A}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\wedge{B}.\:{I}\:{don}'{t}\:{understand} \\ $$

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

it means A= ((p),(q) ) and B= ((((p±(√3)q)/2)),(((q∓(√3)p)/2)) ) ∧ means and ∨ means or

$${it}\:{means} \\ $$$${A}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\:{and}\:{B}=\begin{pmatrix}{\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}}\\{\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\wedge\:{means}\:{and} \\ $$$$\vee\:{means}\:{or} \\ $$

Commented by EnterUsername last updated on 23/Apr/21

Ok thanks. I mistakened it for “the vector perpendicular to...”

$${Ok}\:{thanks}.\:{I}\:{mistakened}\:{it}\:{for}\:``{the}\:{vector}\:{perpendicular}\:{to}...'' \\ $$

Commented by MJS_new last updated on 23/Apr/21

we use a^⇀ ⊥b^⇀ for this

$$\mathrm{we}\:\mathrm{use}\:\overset{\rightharpoonup} {{a}}\bot\overset{\rightharpoonup} {{b}}\:\mathrm{for}\:\mathrm{this} \\ $$

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.

$${say}\:{a}={Ae}^{\alpha{i}} \\ $$$${say}\:{b}={Be}^{\beta{i}} \\ $$$${let}\:{k}=\frac{{A}}{{B}},\:\theta=\alpha−\beta \\ $$$$\frac{{A}}{{B}}{e}^{\left(\alpha−\beta\right){i}} +\frac{{B}}{{A}}{e}^{−\left(\alpha−\beta\right){i}} =\mathrm{1} \\ $$$${ke}^{\theta{i}} +\frac{\mathrm{1}}{{k}}{e}^{−\theta{i}} =\mathrm{1} \\ $$$$\left({k}+\frac{\mathrm{1}}{{k}}\right)\mathrm{cos}\:\theta+\left({k}−\frac{\mathrm{1}}{{k}}\right)\mathrm{sin}\:\theta\:{i}=\mathrm{1} \\ $$$$\Rightarrow\left({k}+\frac{\mathrm{1}}{{k}}\right)\mathrm{cos}\:\theta=\mathrm{1} \\ $$$$\Rightarrow\left({k}−\frac{\mathrm{1}}{{k}}\right)\mathrm{sin}\:\theta=\mathrm{0}\:\Rightarrow\begin{cases}{{k}−\frac{\mathrm{1}}{{k}}=\mathrm{0}\:\Rightarrow{k}=\mathrm{1}}\\{\mathrm{sin}\:\theta=\mathrm{0}\:\Rightarrow\theta={n}\pi}\end{cases} \\ $$$$ \\ $$$${with}\:{k}=\mathrm{1}: \\ $$$$\frac{{A}}{{B}}=\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\theta=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\Delta{AOB}\:{is}\:{equilateral} \\ $$$$ \\ $$$${with}\:\mathrm{sin}\:\theta=\mathrm{0}: \\ $$$$\mathrm{cos}\:\theta=\pm\mathrm{1} \\ $$$${k}+\frac{\mathrm{1}}{{k}}=\pm\mathrm{1}\:\Rightarrow{no}\:{solution}\:{for}\:{k}>\mathrm{0} \\ $$$$ \\ $$$${therefore}\:{only}\:{possibility}\:{is} \\ $$$$\Delta{AOB}\:{is}\:{equilateral}\:{triangle}. \\ $$

Commented by EnterUsername last updated on 22/Apr/21

Thanks Sir

$${Thanks}\:{Sir} \\ $$

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