A) \[{{z}_{2}}=1\]
B) \[{{z}_{2}}=\frac{1}{{{z}_{1}}}\]
C) \[arg\left( {{z}_{1}} \right)=arg\left( {{z}_{2}} \right)\]
D) \[\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|\]
Correct Answer: C
Solution :
[c] \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}={{(\left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|)}^{2}}\] \[{{\left| {{Z}_{1}} \right|}^{2}}+{{\left| {{Z}_{2}} \right|}^{2}}+2\left| {{Z}_{1}} \right|\left| {{Z}_{2}} \right|COS\left( {{\theta }_{1}}-{{\theta }_{1}} \right)={{\left| {{Z}_{1}} \right|}^{2}}+{{\left| {{Z}_{2}} \right|}^{2}}2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\]\[cos\left( {{\theta }_{1}}-{{\theta }_{2}} \right)=1\] \[{{\theta }_{1}}-{{\theta }_{2}}=0{}^\circ \] \[Arg({{z}_{1}})=Arg({{z}_{2}})\]
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