A) \[\frac{2}{3}\]
B) \[\frac{3}{2}\]
C) \[\frac{4}{9}\]
D) \[1\]
E) \[\frac{9}{4}\]
Correct Answer: D
Solution :
\[\because \]\[\frac{2{{z}_{1}}}{3{{z}_{2}}}\]is purely imaginary \[\Rightarrow \] \[\frac{2{{z}_{1}}}{3{{z}_{2}}}=ki\]where \[k\in R\] \[\Rightarrow \] \[\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{3}{2}ki\] Now, \[\left| \frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}+{{z}_{2}}} \right|=\left| \frac{\frac{{{z}_{1}}}{{{z}_{2}}}-1}{\frac{{{z}_{1}}}{{{z}_{2}}}+1} \right|=\left| \frac{3ki-2}{3ki+2} \right|\] \[=1\]
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