Question

The equation $Z^{10}+(13Z−1{)}^{10}=0$ 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 $∑\frac{1}{a_i{b}_i}$.

Answer 1

Answer: 850

$Z^{10}+Z^{10}{\left(13−\frac{1}{Z}\right)}^{10}=0$
$∴{\left(13−\frac{1}{Z}\right)}^{10}=−1=\cos π+i\sin π$
$13−\frac{1}{Z}=(\cos (2m+1)π+i\sin 2mπ+π{)}^{1/10}$
$=e^{i\frac{(2m+1)Ï€}{10}}$
$∴\frac{1}{Z}=13−e^{i\frac{(2m+1)π}{10}}$
substituting $m=0,1,2,…….9we$ get

Let $\frac{1}{Z_1}=\frac{1}{a_1}$ and $\frac{1}{Z_{10}}=\frac{1}{b_1}$ and so on
$∴\frac{1}{a_i{b}_i}=169−13\left[e^{i\frac{π}{10}}+e^{−i\frac{π}{10}}\right]+1$
$=169−13\left[e^{\frac{3+π}{10}}+e^{−\frac{3+π}{10}}\right]+1$
$∴\frac{1}{a_i{b}_i}=170−26RR{e}^{i\frac{π}{10}}$
and $\frac{1}{a_2{b}_2}=170−26R{e}^{i\frac{3π}{10}}$ etc
$∴\frac{1}{a_i{b}_i}=850−26\left[\cos \frac{π}{10}+\cos \frac{3π}{10}+\cos \frac{5π}{10}+\cos \frac{3π}{10}+\cos \frac{9π}{10}\right]$
$=850−26\left[\cos 18^{∘}+\cos 54^{∘}+\cos 90^{∘}+\cos 126^{∘}+\cos 162^{∘}\right]$
$=850$