question_answer

If\[{{z}_{1}}\]and\[{{z}_{2}}\]are two non-zero complex numbers such that\[|{{z}_{1}}+{{z}_{2}}|=|{{z}_{1}}|+|{{z}_{2}}|\]then \[\arg ({{z}_{1}})-\arg ({{z}_{2}})\]is equal to

A) \[-\frac{\pi }{2}\]             

B)        0                             

C)        \[-\pi \]               

D)        \[\frac{\pi }{2}\]

Correct Answer: B

Solution :

Let and Now, given that                                                                 Again, squaring, we get Alternate Solution Since;        On squaring both sides, we get \[|{{z}_{1}}+{{z}_{2}}|\,=|{{z}_{1}}|\,+|{{z}_{2}}|\] \[\Rightarrow \]                \[|{{z}_{1}}{{|}^{2}}\,+|{{z}_{2}}{{|}^{2}}+2\operatorname{Re}\,({{z}_{1}}{{\overline{z}}_{2}})\]                                 \[=|{{z}_{1}}{{|}^{2}}\,+|{{z}_{2}}{{|}^{2}}+2|{{z}_{1}}|\,\,\,|{{z}_{2}}|\]           \[\operatorname{Re}\,({{z}_{1}}{{\overline{z}}_{2}})\,=|{{z}_{1}}|\,\,|{{z}_{2}}|\]           \[|{{z}_{1}}|\,\,|{{z}_{2}}|\,\,\,\cos ({{\theta }_{1}}-{{\theta }_{2}})\,=|{{z}_{1}}|\,\ |{{z}_{2}}|\]           \[{{\theta }_{1}}-{{\theta }_{2}}=0\]           \[\arg ({{z}_{1}})\,-\arg ({{z}_{2}})\,=0\]