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Let \(z_1\) and \(z_2\) be complex numbers such that \(\frac{z_2}{z_1}\) is pure imaginary and \(2z_1 \neq 7z_2.\) Compute \(\left| \frac{2z_1 + 7z_2}{2z_1 - 7z_2} \right|.\)
 

 Dec 27, 2018
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\(\dfrac{2z_1+7z_2}{2z_1-7z_2} = \\ \dfrac{2+7\frac{z_2}{z_1}}{2-7\frac{z_2}{z_1}}\\ \text{let }w = 2+ 7\frac{z_2}{z_1}\\ \dfrac{2+7\frac{z_2}{z_1}}{2-7\frac{z_2}{z_1}} = \dfrac{w}{w^*}\)

 

\(\text{You should know, or you can prove it as an exercise, that }\\ \forall z \in \mathbb{C},~z\neq 0,~\left|\dfrac{z}{z^*}\right| = 1 \text{ so,}\\ \left|\dfrac{2z_1+7z_2}{2z_1-7z_2}\right| = 1\)

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 Dec 27, 2018
edited by Rom  Dec 28, 2018

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