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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Answer: \(\left|\dfrac{z_1-z_2}{z_1+z_2}\right| = 1\)
(The condition in the question should read: \(\dfrac{2z_1}{3z_2}\) is purely imaginary, with \(z_2 \ne 0\).)
Since \(\dfrac{2z_1}{3z_2}\) is purely imaginary, let \(\dfrac{2z_1}{3z_2} = \lambda i\), where \(\lambda \in \mathbb{R}\).
\(\therefore \dfrac{z_1}{z_2} = \dfrac{3\lambda}{2}\, i = k i\), where \(k = \dfrac{3\lambda}{2}\) is real.
Divide numerator and denominator by \(z_2\):
\(\left|\dfrac{z_1-z_2}{z_1+z_2}\right| = \left|\dfrac{\frac{z_1}{z_2}-1}{\frac{z_1}{z_2}+1}\right| = \left|\dfrac{ki-1}{ki+1}\right| = \dfrac{|-1+ki|}{|1+ki|}\)
\(= \dfrac{\sqrt{1+k^2}}{\sqrt{1+k^2}} = 1\)
Answer: \(\left|\dfrac{z_1-z_2}{z_1+z_2}\right| = 1\)
(The condition should read \(\dfrac{2z_1}{3z_2}\) is purely imaginary; \(\dfrac{2z_1}{3z_1} = \dfrac{2}{3}\) is real, so that must be a typo.)
Let \(\dfrac{2z_1}{3z_2} = i\lambda\), where \(\lambda\) is real and \(\lambda \ne 0\). Then
\[\frac{z_1}{z_2} = \frac{3i\lambda}{2} = ik,\quad k = \frac{3\lambda}{2}\in\mathbb{R}\]
Divide numerator and denominator by \(z_2\):
\[\frac{z_1-z_2}{z_1+z_2} = \frac{ik-1}{ik+1}\]
Now \(|ik-1| = \sqrt{k^2+1}\) and \(|ik+1| = \sqrt{k^2+1}\), so
\[\left|\frac{z_1-z_2}{z_1+z_2}\right| = \frac{\sqrt{k^2+1}}{\sqrt{k^2+1}} = 1\]
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