A) 3/2
B) 1
C) 2/3
D) 4/9
Correct Answer: A
Solution :
As given, let \[\frac{2{{z}_{1}}}{3{{z}_{2}}}=iy\]or \[\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{3}{2}iy\], so that \[\left| \frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}+{{z}_{2}}} \right|=\left| \frac{\frac{{{z}_{1}}}{{{z}_{2}}}-1}{\frac{{{z}_{1}}}{{{z}_{2}}}+1} \right|=\left| \frac{\frac{3}{2}iy-1}{\frac{3}{2}iy+1} \right|=\left| \frac{1-\frac{3}{2}iy}{1+\frac{3}{2}iy} \right|=1\] \[\left\{ \because \,\,\,\,|z|\,=\,|\overline{z}| \right\}\]
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