A) \[\frac{37}{33}\]
B) \[2\]
C) 1
D) \[3\]
E) \[\frac{33}{37}\]
Correct Answer: C
Solution :
Since, \[\frac{5{{z}_{2}}}{11{{z}_{1}}}=iy\] (say) or \[\frac{{{z}_{2}}}{{{z}_{1}}}=\frac{11}{5}iy\] Now, \[\left| \frac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|=\frac{\left| 2+3\frac{{{z}_{2}}}{{{z}_{1}}} \right|}{\left| 2-3\frac{{{z}_{2}}}{{{z}_{1}}} \right|}\] \[=\frac{\left| 2+\frac{33}{5}iy \right|}{\left| 2-\frac{33}{5}iy \right|}=1\]
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