Question

Let $z_i,{\overline{z}}_i(i=1,2,..,5)$ are the complex roots of the equation $x^{10}+(13x-1{)}^{10}=0,$ where the bar denotes complex conjugation. Then the value of $\sum _{i=1}^5\frac{1}{z_i{\overline{z}}_i}$ is

Answer 1

Let $t=\frac{1}{x}$
$x^{10}+(13x-1{)}^{10}=0\Rightarrow 1+{(13-\frac{1}{x})}^{10}=0;x\ne 0$
$\Rightarrow 1+(13-t{)}^{10}=0\Rightarrow (13-t{)}^{10}=-1$

Using DeMoivre theorem,
$13-t=\text{cis}\left(\frac{(2k+1)\pi }{10}\right)$
Where $k=0,1,2,\cdots ,9$
$\Rightarrow t=13-\text{cis}\left(\frac{(2k+1)\pi }{10}\right)\Rightarrow \overline{t}=13-\text{cis}(-\frac{(2k+1)\pi }{10})$

Since, $\text{cis}\left(\theta \right)+\text{cis}(-\theta )=2\cos \theta$
$\therefore t\overline{t}=169-26\cos \left(\frac{(2k+1)\pi }{10}\right)+1$
$=170-26\cos \left(\frac{(2k+1)\pi }{10}\right)$

Now, the expression to find is,
$\sum _{i=1}^5{t}_i\overline{t_i}=850-26\sum _{k=0}^4\cos \frac{(2k+1)\pi }{10}$
$=850-26S$

$S=\sum _{k=0}^4\cos \frac{(2k+1)\pi }{10}\Rightarrow S=\cos \frac{\pi }{10}+\cos \frac{3\pi }{10}+\cos \frac{\pi }{2}+\cos \frac{7\pi }{10}+\cos \frac{9\pi }{10}\Rightarrow S=\cos \frac{\pi }{10}+\cos \frac{3\pi }{10}-\cos \frac{3\pi }{10}-\cos \frac{\pi }{10}\Rightarrow S=0$

$\sum _{i=1}^5{t}_i\overline{t_i}=850-26S=850\Rightarrow \sum _{i=1}^5\frac{1}{z_i{\overline{z}}_i}=850$