The complex number associated with the vertices A - B - C of - ABC are e i - - - respectively -where - - are the complex cube roots of unity and Re e i -Re- then the complex number of the point where angle bisector of A meets the circumcircle of the triangle- is A- e i -B- -C- e - i -D- -

Clearly, ∠ DOB = ∠ COD = A ⇒ z = ω e^iA and ω = ze^iA (Applying rotation about O) ⇒ z^2 = ωω =1 ⇒ z = 1 (∵ A and D are on opposite sides of BC) ...

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Byju's Answer

Standard XII

Mathematics

Rotation Concept

The complex n...

Question

The complex number associated with the vertices $A, B, C$ of $Δ A B C$ are $e^{i θ}, ω, ¯ ¯ ¯ ω$ , respectively [where $ω, ¯ ¯ ¯ ω$ are the complex cube roots of unity and $R e (e^{i θ}) > R e (ω)$ , then the complex number of the point where angle bisector of $A$ meets the circumcircle of the triangle, is

e^{i θ}

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e^{- i θ}

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ω ¯ ¯ ¯ ω

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ω + ¯ ¯ ¯ ω

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Solution

The correct option is D $ω + ¯ ¯ ¯ ω$

Clearly,
$∠ D O B = ∠ C O D = A$
$\Rightarrow z = ω e^{i A}$ and $¯ ¯ ¯ ω = z e^{i A}$ (Applying rotation about $O$ )
$\Rightarrow z^{2} = ω ¯ ¯ ¯ ω = 1$
$\Rightarrow z = - 1$
( $∵ A$ and $D$ are on opposite sides of $B C$ )

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The complex number associated with the vertices A,B,C of ΔABC are eiθ,ω,¯¯¯ω, respectively [where ω,¯¯¯ω are the complex cube roots of unity and Re(eiθ)>Re(ω), then the complex number of the point where angle bisector of A meets the circumcircle of the triangle, is

The correct option is D ω+¯¯¯ω Clearly, ∠DOB=∠COD=A ⇒z=ωeiA and ¯¯¯ω=zeiA (Applying rotation about O) ⇒z2=ω¯¯¯ω=1 ⇒z=−1 (∵A and D are on opposite sides of BC)

The correct option is D $ω + ¯ ¯ ¯ ω$

Clearly,
$∠ D O B = ∠ C O D = A$
$\Rightarrow z = ω e^{i A}$ and $¯ ¯ ¯ ω = z e^{i A}$ (Applying rotation about $O$ )
$\Rightarrow z^{2} = ω ¯ ¯ ¯ ω = 1$
$\Rightarrow z = - 1$
( $∵ A$ and $D$ are on opposite sides of $B C$ )