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 \(\left| \frac{2z_1 + 7z_2}{2z_1 - 7z_2} \right|.\)

mathtoo Dec 27, 2018

#1**+1 **

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

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

.Rom Dec 27, 2018