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 \[|{{z}_{1}}+{{z}_{2}}|\,=|{{z}_{1}}|\,+|{{z}_{2}}|\] \[\Rightarrow \] \[|{{z}_{1}}{{|}^{2}}\,+|{{z}_{2}}{{|}^{2}}+2\operatorname{Re}\,({{z}_{1}}{{\overline{z}}_{2}})\] \[=|{{z}_{1}}{{|}^{2}}\,+|{{z}_{2}}{{|}^{2}}+2|{{z}_{1}}|\,\,\,|{{z}_{2}}|\] \[\operatorname{Re}\,({{z}_{1}}{{\overline{z}}_{2}})\,=|{{z}_{1}}|\,\,|{{z}_{2}}|\] \[|{{z}_{1}}|\,\,|{{z}_{2}}|\,\,\,\cos ({{\theta }_{1}}-{{\theta }_{2}})\,=|{{z}_{1}}|\,\ |{{z}_{2}}|\] \[{{\theta }_{1}}-{{\theta }_{2}}=0\] \[\arg ({{z}_{1}})\,-\arg ({{z}_{2}})\,=0\]You need to login to perform this action.
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