A) \[Im\left( z \right)=0\]
B) \[\left| z \right|=\frac{\sqrt{5}}{2}\]
C) \[\left| z \right|=\frac{1}{2}\frac{\sqrt{17}}{2}\]
D) \[Re\left( z \right)=0\]
E) None of these
Correct Answer: E
Solution :
\[3\left| {{z}_{1}} \right|=\,\,4\,\left| {{z}_{2}} \right|\] \[\Rightarrow \,9{{\left| {{z}_{1}} \right|}^{2}}\,=\,16{{\left| {{z}_{2}} \right|}^{2}}\] \[9{{z}_{1}}{{\overline{z}}_{1}}\,=\,16{{z}_{2}}{{\overline{z}}_{2}}\] \[\Rightarrow \,\,\frac{3{{z}_{1}}}{2{{z}_{2}}}\,=\,\frac{8}{3}\left( \overline{\frac{{{z}_{2}}}{{{z}_{1}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,let\,\,\frac{3{{z}_{1}}}{2{{z}_{2}}}=r{{e}^{{{i}^{\theta }}}}\] \[r{{e}^{{{i}^{\theta }}}}=\,\,\frac{8}{3}\overline{\left( \frac{1}{\left( \frac{2}{3}r{{e}^{{{i}^{\theta }}}} \right)} \right)}\] \[\Rightarrow \,\,r{{e}^{{{i}^{\theta }}}}=\,\frac{4}{r}{{e}^{+i\theta }}\] \[\Rightarrow \,\,\,{{r}^{2}}\,=\,4\,\,\Rightarrow \,r=2\] \[z=\frac{3{{z}_{1}}}{2{{z}_{2}}}\,+\,\frac{2{{z}_{2}}}{3{{z}_{1}}}\,=\,r{{e}^{{{i}^{\theta }}}}+\frac{1}{r{{e}^{i\theta }}}\] \[z=2{{e}^{{{i}^{\theta }}}}\,+\,\frac{1}{2}{{e}^{-i\theta }}\] \[z=\frac{5}{2}\,\cos \,\theta +\,i\,\frac{3}{2}\,\sin \,\theta \] \[\left| z \right|=\sqrt{\frac{25}{4}+\frac{9}{4}}\,\,\,=\,\,\sqrt{\frac{17}{2}}\] \[\therefore \] No option are matched
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