Question

If α is the non real 5th root of unity and Z1 and Z2 are any two non zero complex numbers then find the value of complex expression:

$E = \frac{P}{Q} ,\ where\\\\ P= \Sigma_{p=0}^4 \ |\ Z_1 + \alpha^p\ Z_2\ |^2\\\\ Q = ( | Z_1 |^2 + | Z_2 |^2 )\\\\Answer=5$

Answer 1

This seems to involve a lot of trigonometric steps..

$\alpha=5^{th}\ root\ of\ 1=e^{\frac{i2\pi}{5}}\\Let\ Z_1=a e^{iA},\ \ \ Z_2=be^{iB} \\|Z_1|^2=a^2,\ \ |Z_2|^2=b^2,\ \ \ =\ \textgreater \ Q=a^2+b^2\\\\ P=| ae^{iA}+be^{iB}|^2+|ae^{iA}+be^{iB} \alpha |^2+|ae^{iA}+be^{iB}\alpha^2|^2+\\.\ \ \ \ \ |ae^{iA}+be^{iB}\alpha^3|^2+|ae^{iA}+be^{iB}\alpha^4|^2\\\\P=| ae^{iA}+be^{iB}|^2+|ae^{iA}+be^{iB} e^{\frac{i2\pi}{5}} |^2\\.\ \ \ \ +| ae^{iA}+be^{iB}e^{\frac{i4\pi}{5}}|^2+|ae^{iA}+be^{iB} e^{\frac{i6\pi}{5}} |^2\\. \ \ \ \ \ +|ae^{iA}+be^{iB} e^{\frac{i8\pi}{5}} |^2$

$=(a CosA+b CosB)^2+(a SinA+b SinB)^2\\ + [a CosA+b Cos(B+\frac{2\pi}{5})]^2+[ a SinA+b Sin(B+\frac{2\pi}{5})]^2\\ +[aCosA+b Cos(B+\frac{4\pi}{5})]^2+[ a SinA+b Sin(B+\frac{4\pi}{5})]^2\\ +[aCosA+b Cos(B+\frac{6\pi}{5})]^2+[ a SinA+b Sin(B+\frac{6\pi}{5})]^2\\ +[aCosA+b Cos(B+\frac{6\pi}{5})]^2+[ a SinA+b Sin(B+\frac{8\pi}{5})]^2\\\\=5(a^2+b^2)+2ab*\\ . \ \ [ Cos(A-B)+Cos(A-B-\frac{2\pi}{5})+ Cos(A-B-\frac{4\pi}{5})\\ . \ \ \ \ \ +Cos(A-B-\frac{6\pi}{5})+Cos(A-B-\frac{8\pi}{5}]$

$=5*(a^2+b^2)+2ab* [Cos(A-B)+2 Cos(A-B-\pi)*Cos(\frac{3\pi}{5})\\ . \ \ \ +2Cos(A-B-\pi)*Cos(\pi/5)] \\\\P=5(a^2+b^2)+2ab* [Cos(A-B)-2Cos(A-B)*(Cos\frac{3\pi}{5}+Cos\frac{\pi}{5})]\\\\we\ know\ Cos\frac{3\pi}{5}+Cos\frac{\pi}{5}=\frac{1}{2}\\\\Hence,\ P=5*(a^2+b^2)\\\\E=\frac{P}{Q}=5$