Question

The equation $Z^{10}+(13 Z-1)^{10}=0$ has 5 pairs of complex roots $a_1, b_1, a_2, b_2, a_3, b_3, a_4, b_4$, $a _5, b _5$. Each pair $a _{ i }, b _{ i }$ are complex conjugate. Find $\sum \frac{1}{ a _{ i } b _{ i }}$.

Answer 1

$Z^{10}+Z^{10}\left(13-\frac{1}{Z}\right)^{10}=0$
$\therefore \left(13-\frac{1}{Z}\right)^{10}=-1=\cos \pi+i \sin \pi $
$13-\frac{1}{Z}=(\cos (2 m +1) \pi+i \sin 2 m \pi+\pi)^{1 / 10}$
$= e ^{i \frac{(2 m +1) \pi}{10}} $
$\therefore \frac{1}{ Z }=13- e ^{i \frac{(2 m +1) \pi}{10}}$
substituting $m =0,1,2, \ldots \ldots .9 we$ get

Let $\frac{1}{ Z _1}=\frac{1}{ a _1}$ and $\frac{1}{ Z _{10}}=\frac{1}{ b _1}$ and so on
$\therefore \frac{1}{ a _{ i } b _{ i }} =169-13\left[ e ^{i \frac{\pi}{10}}+ e ^{-i \frac{\pi}{10}}\right]+1 $
$=169-13\left[ e ^{\frac{3+\pi}{10}}+ e ^{-\frac{3+\pi}{10}}\right]+1$
$\therefore \frac{1}{ a _{ i } b _{ i }} =170-26 R Re ^{i \frac{\pi}{10}}$
and $\frac{1}{ a _2 b _2}=170-26 Re ^{i \frac{3 \pi}{10}}$ etc
$\therefore \frac{1}{ a _{ i } b _{ i }} =850-26\left[\cos \frac{\pi}{10}+\cos \frac{3 \pi}{10}+\cos \frac{5 \pi}{10}+\cos \frac{3 \pi}{10}+\cos \frac{9 \pi}{10}\right] $
$ =850-26\left[\cos 18^{\circ}+\cos 54^{\circ}+\cos 90^{\circ}+\cos 126^{\circ}+\cos 162^{\circ}\right]$
$ =850$