A) \[{{z}_{2}}={{\overline{z}}_{1}}\]
B) \[{{z}_{2}}=\frac{1}{{{z}_{1}}}\]
C) \[arg\,({{z}_{1}})=\]arg \[({{z}_{2}})\]
D) \[|{{z}_{1}}|\,=\,|{{z}_{2}}|\]
Correct Answer: C
Solution :
\[|{{z}_{1}}+{{z}_{2}}|\,=\,|{{z}_{1}}|+|{{z}_{2}}|\] \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}\,=\,|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}+2|{{z}_{1}}||{{z}_{2}}|\] Þ \[|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}+2\operatorname{Re}|{{z}_{1}}{{\bar{z}}_{2}}|\] \[=\,|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}+2|{{z}_{1}}||{{z}_{2}}|\] Þ \[2\operatorname{Re}|{{z}_{1}}{{\bar{z}}_{2}}|=2|{{z}_{1}}||{{z}_{2}}|\] Þ \[2|{{z}_{1}}||{{\bar{z}}_{2}}|\cos ({{\theta }_{1}}-{{\theta }_{2}})=2|{{z}_{1}}||{{z}_{2}}|\] Þ \[\cos ({{\theta }_{1}}-{{\theta }_{2}})=1\]or \[{{\theta }_{1}}-{{\theta }_{2}}=0\] \ \[arg({{z}_{1}})=arg({{z}_{2}})\]
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