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Properties of Conjugate of a Complex Number

Let A Z 1, B ...

Question

Let $A (Z_{1}), B (Z_{2}), C (Z_{3})$ be the vertices of an equilateral triangle ABC, then the value of $a r g (\frac{Z_{2} + Z_{3} - 2 Z_{1}}{Z_{3} - Z_{2}})$ is equal to

$\frac{π}{3}$

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$\frac{π}{4}$

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$\frac{π}{2}$

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$\frac{π}{6}$

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Solution

The correct option is C
$\frac{π}{2}$

$a r g (\frac{Z_{2} + Z_{3} - 2 Z_{1}}{Z_{3} - Z_{2}}) = a r g 2 ⎧ ⎪ ⎨ ⎪ ⎩ \frac{(\frac{Z_{2} + Z_{3}}{2}) - Z_{1}}{Z_{3} - Z_{2}} ⎫ ⎪ ⎬ ⎪ ⎭$

$= a r g ⎧ ⎪ ⎨ ⎪ ⎩ \frac{(\frac{Z_{2} + Z_{3}}{2} - Z_{1})}{Z_{3} - Z_{2}} ⎫ ⎪ ⎬ ⎪ ⎭ = \frac{π}{2}$

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Similar questions

If $A (z_{1}), B (z_{2}), C (z_{3})$ be the vertices of triangle ABC in which $| A B C - ---- -$ = $\frac{π}{4}$ and $\frac{A B}{B C}$ = $\sqrt{2}$ then $z_{2}$ is equal to

Q. Statement 1: If

| z_{1} | = | z_{2} | = | z_{3} |, z_{1} + z_{2} + z_{3} = 0

and

(z_{1}), B (z_{2}), C (z_{3})

are the vertices of

Δ A B C

, then one of the values of

a r g (\frac{(z_{2} + z_{3} - 2 z_{1})}{(z_{3} - z_{2})})

\frac{π}{2}

.
Statement 2: In equilateral triangle, orthocenter coincides with centroid.

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are the vertices of the triangle ABC (in anticlockwise order). If

∠ A B C = π / 4

and

A B = \sqrt{2} (B C)

Then prove that

z_{2} = z_{3} + i (z_{1} - z_{3}) .

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are the vertices of the triangle

A B C

(in anticlockwise order). If

∠ A B C = \frac{π}{4}

and

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Q. If

z_{1}, z_{2}, z_{3}

are the vertices of a triangle in argand plane such that

| z_{1} - z_{2} | = | z_{1} - z_{3} |,

then

arg (\frac{2 z_{1} - z_{2} - z_{3}}{z_{3} - z_{2}})

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Let A(Z1),B(Z2),C(Z3) be the vertices of an equilateral triangle ABC, then the value of arg(Z2+Z3−2Z1Z3−Z2) is equal to

The correct option is C π2 arg(Z2+Z3−2Z1Z3−Z2)=arg 2⎧⎪⎨⎪⎩(Z2+Z32)−Z1Z3−Z2⎫⎪⎬⎪⎭ =arg⎧⎪⎨⎪⎩(Z2+Z32−Z1)Z3−Z2⎫⎪⎬⎪⎭=π2

Let $A (Z_{1}), B (Z_{2}), C (Z_{3})$ be the vertices of an equilateral triangle ABC, then the value of $a r g (\frac{Z_{2} + Z_{3} - 2 Z_{1}}{Z_{3} - Z_{2}})$ is equal to