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