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 \(\left| \frac{2z_1 + 7z_2}{2z_1 - 7z_2} \right|.\)
Let z_1 = a + bi and z_2 = c + di. Then z_2/z_1 = pure imaginary gives you d/a = -c/b. Let d/a = k and c/b = -k.
Then d = ka and c = -bk. A little bit of algebra then gives you
\(\left| \frac{2z_1 + 7z_2}{2z_1 - 7z_2} \right| = \frac{2}{7}\)
+118667
\(Let \qquad ki=\frac{z_2}{z_1} \qquad k\in Z\\ \displaystyle \left| \frac{2z_1+7z_2}{2z_1-7z_2}\right|\\ =\displaystyle \left| \frac{\frac{2z_1+7z_2}{z_1}}{\frac{2z_1-7z_2}{z_1}}\right|\\ =\displaystyle \left| \frac{2+7ki}{2-7ki}\right|\\ =\displaystyle \left| \frac{(2+7ki)^2}{4+49k^2}\right|\\ =\displaystyle \left| \frac{(4-49k^2)+28ki}{4+49k^2}\right|\\ =\displaystyle \frac{\sqrt{(4-49k^2)^2+(28k)^2}}{4+49k^2}\\ =\displaystyle \frac{\sqrt{ 16+2401k^4-392k^2 +784k^2}}{4+49k^2}\\ =\displaystyle \frac{\sqrt{ 2401k^4+392k^2+16 }}{4+49k^2}\\ =\displaystyle \frac{ 49k^2+4 }{4+49k^2}\\ =1 \)
You need to check that.
Let \qquad ki=\frac{z_2}{z_1} \qquad k\in Z\\
\displaystyle \left| \frac{2z_1+7z_2}{2z_1-7z_2}\right|\\
=\displaystyle \left| \frac{\frac{2z_1+7z_2}{z_1}}{\frac{2z_1-7z_2}{z_1}}\right|\\
=\displaystyle \left| \frac{2+7ki}{2-7ki}\right|\\
=\displaystyle \left| \frac{(2+7ki)^2}{4+49k^2}\right|\\
=\displaystyle \left| \frac{(4-49k^2)+28ki}{4+49k^2}\right|\\
=\displaystyle \frac{\sqrt{(4-49k^2)^2+(28k)^2}}{4+49k^2}\\
=\displaystyle \frac{\sqrt{ 16+2401k^4-392k^2 +784k^2}}{4+49k^2}\\
=\displaystyle \frac{\sqrt{ 2401k^4+392k^2+16 }}{4+49k^2}\\
=\displaystyle \frac{ 49k^2+4 }{4+49k^2}\\
=1
