A) \[-\pi \]
B) \[-\frac{\pi }{2}\]
C) 0
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Since, \[|{{z}_{1}}+{{z}_{2}}|=|{{z}_{1}}|+|{{z}_{2}}|\] On squaring both sides, we get \[|{{z}_{1}}{{|}^{2}}|{{z}_{2}}{{|}^{2}}+2|{{z}_{1}}|\,|{{z}_{2}}|\cos (\arg {{z}_{1}}-\arg {{z}_{2}})\] \[=|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}|+2|{{z}_{1}}|\,\,|{{z}_{2}}|\] \[\Rightarrow \]\[2|{{z}_{1}}|{{z}_{2}}|\cos (\arg {{z}_{1}}-\arg {{z}_{2}})=2|{{z}_{1}}|\,\,|{{z}_{2}}|\] \[\Rightarrow \]\[\cos (\arg {{z}_{1}}-\arg {{z}_{2}})=1\] \[\Rightarrow \]\[\arg ({{z}_{1}})-\arg ({{z}_{2}})=0\]\[\Rightarrow \]\[\Rightarrow \]
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