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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 $$\lambda ∈ R$$
$${z_1\over z_2}={ 3 \lambda \over2}i$$