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Modulus of a Complex Number

If O, z1, z...

Question

If $O, z_{1}, z_{2}$ are vertices of an equilateral triangle, then

A

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

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B

(z_{1} + z_{2})^{2} = 3 z_{1} z_{2}

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C

| a r g (z_{1}) - a r g (z_{2}) | = π / 3

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D

| z_{1} + z_{2} | = 2 | z_{1} | + | z_{2} |

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Solution

The correct options are
A $| z_{1} | = | z_{2} | = | z_{1} - z_{2} |$
B $(z_{1} + z_{2})^{2} = 3 z_{1} z_{2}$
C $| a r g (z_{1}) - a r g (z_{2}) | = π / 3$
Since $0, z_{1}, z_{2}$ are the vertices of an equilateral triangle.
Hence all the sides must be equal.
Therefore
$| z_{1} - 0 | = | z_{1} |$
$= | z_{2} - 0 | = z_{2}$
$= | z_{2} - z_{1} |$
The three vertices must satisfy the relation
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$
Hence
$0^{2} + z_{1}^{2} + z_{2}^{2} = 0 (z_{1}) + z_{2} (0) + z_{3} z_{2}$
$z_{1}^{2} + z_{2}^{2} = z_{2} z_{3}$
$(z_{1} + z_{2})^{2} - 2 z_{1} z_{2} = z_{2} z_{3}$
$(z_{1} + z_{2})^{2} = 3 z_{2} z_{3}$
And each angle has to be $60^{0}$
Hence
$| a r g (z_{1}) - a r g (z_{2}) | = \frac{π}{3}$

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Similar questions

z_{1}, z_{2}

are complex numbers then the correct match List- I to List II is:

List - I	List - II
A) arg z_1z_2	1) $a r g z_{1} - a r g z_{2}$
B) $a r g z_{1}_{2}$	2) $a r g z_{1} - a r g z_{2} = \frac{π}{2}$
C) $\| z_{1} + z_{2} \| = \| z_{1} - z_{2} \|$	3) $a r g z_{1} = a r g z_{2}$
D) $\| z_{1} + z_{2} \|^{2}$	4) $a r g z_{1} + a r g z_{2}$
E) $\| z_{1} + z_{2} \| = \| z_{1} \| + \| z_{2} \|$	5) $r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})$

| z_{1} + z_{2} | = | z_{1} | + | z_{2} |

a r g z_{1} - a r g z_{2} =

| z_{1} - z_{2} | = | z_{1} | + | z_{2} |

, then prove that

a r g (z_{1}) - a r g (z_{2}) = π

z_{1}

z_{2}

are two complex numbers such that

| z_{1} | = | z_{2} |

a r g z_{1} + a r g z_{2} = π

z_{1}

z_{2}

z_{1}, z_{2}

are the complex numbers such that

| z_{1} + z_{2} | = | z_{1} | + | z_{2} |

a r g z_{1} - a r g z_{2}

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