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Polar Representation of a Complex Number

| z 1 + z 2 |...

Question

$| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ is possible if

A

z_{2} =_{1}

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B

z_{2} = \frac{1}{z_{1}}

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C

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

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D

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

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Solution

The correct option is C $a r g z_{1} = a r g z_{2}$
Given $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$

Squaring, ${| z_{1} + z_{2} |}^{2} = {(| z_{1} | + | z_{2} |)}^{2}$

$\Rightarrow {| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2} + 2 R e | z_{1} | | z_{2} | = {| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2} + 2 | z_{1} | | z_{2} |$

$\Rightarrow 2 R e | z_{1} | ∣ ∣_{2} ∣ ∣ = 2 | z_{1} | | z_{2} |$

$\Rightarrow | z_{1} | ∣ ∣_{2} ∣ ∣ cos (θ_{1} - θ_{2}) = | z_{1} | | z_{2} |$

$\Rightarrow cos (θ_{1} - θ_{2}) = 1 \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2}$

$\Rightarrow a r g z_{1} = a r g z_{2}$

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Similar questions

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

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

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} |

, then show that

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

, then prove that

a r g (z_{1}) - a r g (z_{2}) = π

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