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General Solution of Trigonometric Equation

If z1 and ...

Question

If $z_{1}$ and $z_{2}$ are two nonzero complex numbers such that $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ , then $a r g z_{1} - a r g z_{2}$ is equal to

A

- π

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B

- \frac{π}{2}

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C

0

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D

\frac{π}{2}

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E

π

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Solution

The correct option is C $0$
Given:-
$z_{1}$ and $z_{2}$ are two complex no. $&$

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

To find $a r g (z_{1}) + a r g (z_{2})$
Solution:-

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

Squaring both sides

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

${| z_{1} |}^{2} + {| z_{2} |}^{2} + 2 | z_{1} | | z_{2} | cos [a r g (z_{1}) - a r g (z_{2})] = {| z_{1} |}^{2} + {| z_{2} |}^{2} + 2 | z_{1} | | z_{2} |$

$2 | z_{1} | | z_{2} | cos [a r g (z_{1}) - a r g (z_{2})] = 2 | z_{1} | | z_{2} |$

$cos [a r g (z_{1}) - a r g (z_{2})] = 1$

$a r g (z_{1}) - a r g (z_{2}) = {cos}^{- 1} 1$ $a r g (z_{1}) - a r g (z_{2}) = 0$

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

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Similar questions

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 two non-zero complex numbers such that

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

a r g z_{1} + a r g z_{2} = π

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}

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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