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Properties of Modulus

lf z1 and ...

Question

lf $z_{1}$ and $z_{2}$ are two complex numbers such that $| z_{1} | = | z_{2} | + | z_{1} - z_{2} |$ , then $arg (z_{1}) - arg (z_{2})$

A

0

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B

\frac{π}{2}

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C

\frac{- π}{2}

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D

π

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Solution

The correct option is B $0$
Let $A$ denote the complex number $z_{1}$ and $B$ denote the complex number $z_{2}$ on the Argand Plane.
$| z_{1} | = O A$ , where $O$ is the origin.
$| z_{2} | = O B$
and $| z_{1} - z_{2} | = A B$ .
Since, $O A = O B + A B$ , $O, A$ and $B$ are collinear with $B$ lying between $O$ and $B$ .
Hence, $a r g (z_{1}) = a r g (z_{2})$

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