If $z_{1} and z_{2}$ are two non-zero complex numbers such that $|z_{1} + z_{2}| = |z_{1}| + |z_{2}|$ , then ${Argz}_{1} - {Argz}_{2} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- π$

$π$

$\frac{π}{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$|z_{1} + z_{2}| = |z_{1}| + |z_{2}| Let z_{1} = x_{1} + {iy}_{1} z_{2} = x_{2} + {iy}_{2} \Rightarrow | z_{1} | = \sqrt{{x_{1}}^{2} + {y_{1}}^{2}}, | z_{2} | = \sqrt{{x_{2}}^{2} + {y_{2}}^{2}} a n d | z_{1} + z_{2} | = | (x_{1} + x_{2}) + i (y_{1} + y_{2}) | = \sqrt{{(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2}} ∴ \sqrt{{(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2}} = \sqrt{{x_{1}}^{2} + {y_{1}}^{2}} + \sqrt{{x_{2}}^{2} + {y_{2}}^{2}} \Rightarrow {(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2} = {x_{1}}^{2} + {y_{1}}^{2} + {x_{2}}^{2} + {y_{2}}^{2} + 2 \sqrt{({x_{1}}^{2} + {y_{1}}^{2}) ({x_{2}}^{2} + {y_{2}}^{2})} \Rightarrow 2 x_{1} x_{2} + 2 y_{1} y_{2} = 2 \sqrt{{x_{1}}^{2} {x_{2}}^{2} + {x_{1}}^{2} {y_{2}}^{2} + {y_{1}}^{2} {x_{2}}^{2} + {y_{1}}^{2} {y_{2}}^{2}} \Rightarrow {(x_{1} x_{2} + y_{1} y_{2})}^{2} = {x_{1}}^{2} {x_{2}}^{2} + {y_{1}}^{2} {y_{2}}^{2} + {y_{1}}^{2} {x_{2}}^{2} + {x_{1}}^{2} {y_{2}}^{2} \Rightarrow {x_{1}}^{2} {y_{2}}^{2} + {x_{2}}^{2} {y_{1}}^{2} - 2 x_{1} x_{2} y_{1} y_{2} = 0 {\Rightarrow (x_{1} y_{2} - x_{2} y_{1})}^{2} = 0 x_{1} y_{2} = x_{2} y_{1} ∴ \frac{y_{2}}{x_{2}} = \frac{y_{1}}{x_{1}} \Rightarrow Arg z_{2} = Arg z_{1} ∴ A r g z_{2} - A r g z_{1} = 0$