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Properties of Conjugate of a Complex Number

If z1, z2...

Question

If $z_{1}$ , $z_{2}$ are conjugate complex numbers and arg $(z_{1} z_{2}) =$

A

a r g z_{1} + a r g z_{2} = 0

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B

a r g z_{1} + a r g z_{2} + n π

,

n ϵ {- 1, 0, 1}

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C

a r g z_{1} + a r g z_{2} + n π

,

n ϵ N

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D

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

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Solution

The correct option is A $a r g z_{1} + a r g z_{2} = 0$
$z_{1}$ and $z_{2}$ are conjugates of each other.
Hence, $z_{1} z_{2} = | z_{1} |^{2}$
$z_{1} z_{2}$ is real and positive.
Hence, arg $(z_{1} z_{2}) = 0$
Also, arg $z_{1}$ +arg $z_{2} = 0$
Hence, option A is correct.

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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) $r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})$

Q. Let two pairs of non-zero conjugate complex numbers

(z_{1}, z_{2})

(z_{3}, z_{4})

then prove value of

a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}})

a r g (z_{1}) - a r g (z_{4}) + a r g (z_{2}) - a r g (z_{3}) = 0

z_{1}, z_{2}

z_{3}, z_{4}

are two pairs of conjugate complex numbers, then

a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}})

z_{1}, z_{2}

z_{3}, z_{4}

are 2 pairs of complex conjugate numbers, then

a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}})

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

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

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