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Geometrical Representation of Argument and Modulus

If |z1 + z2...

Question

If $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ , then show that $a r g (z_{1}) = a r g (z_{2})$ .

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Solution

$| z_{1} + z_{2} | =$ $| z_{1} | + | z_{2} |$
or ${| z_{1} + z_{2} |}_{1}^{2} = {| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2} + 2 | z_{1} | | z_{2} |$
or ${| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2} + 2 R e (z_{1} {¯ ¯ ¯ z}_{2}) = {| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2} + 2 | z_{1} | | z_{2} |$
or $2 R e (z_{1} {¯ ¯ ¯ z}_{2}) = 2 | z_{1} | | z_{2} |$
$⟹ c o s (θ_{1} - θ_{2}) = 1$
$⟹ θ_{1} - θ_{2} = 0$
$⟹ a r g (z_{1}) = a r g (z_{2})$
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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} | = | 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}

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

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