A) \[\operatorname{Re}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=0\]
B) \[\operatorname{Im}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=0\]
C) \[\operatorname{Re}({{z}_{1}}{{z}_{2}})=0\]
D) \[\operatorname{Im}({{z}_{1}}{{z}_{2}})=0\]
Correct Answer: A
Solution :
We have \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}=|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}\] Þ \[|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}+2|{{z}_{1}}||{{z}_{2}}|\cos ({{\theta }_{1}}-{{\theta }_{2}})\]\[=|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}\] Where \[{{\theta }_{1}}=arg({{z}_{1}}),{{\theta }_{2}}=arg({{z}_{2}})\] Þ \[\cos ({{\theta }_{1}}-{{\theta }_{2}})=0\,\,\,\,\Rightarrow {{\theta }_{1}}-{{\theta }_{2}}=\frac{\pi }{2}\] Þ \[arg\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=\frac{\pi }{2}\Rightarrow \operatorname{Re}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=\frac{|{{z}_{1}}|}{|{{z}_{2}}|}\cos \left( \frac{\pi }{2} \right)=0\] Note: Also \[\operatorname{Re}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=0\Rightarrow \operatorname{Re}({{z}_{1}}\overline{{{z}_{2}}})=0\] Þ \[{{z}_{1}}\overline{{{z}_{2}}}\] is purely imaginary.
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