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Geometrical Representation of Argument and Modulus

If for comple...

Question

If for complex number $z_{1}$ and $z_{2}, a r g (z_{1}) - a r g (z_{2}) = 0$ , then $∣ z_{1} - z_{2} ∣$ is equal to

A

∣ z_{1} ∣ + ∣ z_{2} ∣

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B

∣ z_{1} ∣ - ∣ z_{2} ∣

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C

∣ | z_{1} ∣ - ∣ z_{2} | ∣

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D

0

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Solution

The correct option is D $∣ | z_{1} ∣ - ∣ z_{2} | ∣$
Let $a r g (z_{1}) = a r g (z_{2}) = θ$
Let $z_{1} = r_{1} c i s (θ) = r_{2} c i s (θ)$
$z_{1} - z_{2} = (r_{1} - r_{2}) c i s (θ)$
$| z_{1} - z_{2} | = | r_{1} - r_{2} |$
$| z_{1} - z_{2} | = | | z_{1} | - | z_{2} | |$

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Q. If for complex numbers

z_{1}

z_{2}, a r g (z_{1}) - a r g (z_{2}) = 0

| z_{1} - z_{2} |

z_{1}

z_{2}

are two nonzero complex numbers such that

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

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

z_{1} & z_{2}

are two non-zero complex numbers such that

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

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

z_{1}

z_{2}

are two non-zero 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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