Let z1and z2 be complex numbers such that z2z1 is pure imaginary and 2z1≠7z2. Compute |2z1+7z22z1−7z2|.
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
|2z1+7z22z1−7z2|=27
+118703
Letki=z2z1k∈Z|2z1+7z22z1−7z2|=|2z1+7z2z12z1−7z2z1|=|2+7ki2−7ki|=|(2+7ki)24+49k2|=|(4−49k2)+28ki4+49k2|=√(4−49k2)2+(28k)24+49k2=√16+2401k4−392k2+784k24+49k2=√2401k4+392k2+164+49k2=49k2+44+49k2=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
