A) \[\frac{3}{2}\]
B) 1
C) \[\frac{2}{3}\]
D) \[\frac{4}{9}\]
Correct Answer: B
Solution :
Since, \[\frac{2{{z}_{1}}}{3{{z}_{2}}}\]is a purely imaginary number. Let \[\frac{2{{z}_{1}}}{3{{z}_{2}}}=ib\] \[\Rightarrow \] \[\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{3}{2}ib\] Now, \[\left| \frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}+{{z}_{2}}} \right|=\left| \frac{\frac{{{z}_{1}}}{{{z}_{2}}}-1}{\frac{{{z}_{1}}}{{{z}_{2}}}+1} \right|\] \[=\left| \frac{i\frac{3}{2}b-1}{i\frac{3}{2}b+1} \right|\] \[=\frac{\sqrt{{{1}^{2}}+{{\left( \frac{3}{2}b \right)}^{2}}}}{{{1}^{2}}+{{\left( \frac{3}{2}b \right)}^{2}}}=1\]
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