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If z1, z2 are complex numbers such that,\(\frac{2z_1}{3z_1}\) is purely imaginary number, then finds  \(|\frac{z_1-z_2}{z_1+z_2}|\) 

 

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since , \(\frac{2z_1}{3z_1}\)  is purely imaginary 

\(\frac{2z_1}{3z_1} \) ==\(\lambda\)  for some \(\lambda ∈ R\)  

\({z_1\over z_2}={ 3 \lambda \over2}i\) 

Now , 

 

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