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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

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

\(\frac{z_1}{z_2}\) = \(\frac{3λ}{2}\)i

Now,

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