Question

The equation $z^{10}+(13z-1{)}^{10}=0$ has $5$ pairs of complex roots $a_1,b_1,a_2,b_2,a_3,b_3,a_4,b_4,a_5,b_5$. If each pair $a_i,b_i$ are complex conjugates, then

• A. $\sum _{i=1}^5(\frac{1}{a_i}+\frac{1}{b_i})=130$
• B. $\sum _{i=1}^5(\frac{1}{a_i}+\frac{1}{b_i})=260$
• C. $\sum _{i=1}^5\left(\frac{1}{a_i{b}_i}\right)=1700$
• D. $\sum _{i=1}^5\left(\frac{1}{a_i{b}_i}\right)=850$

Answer 1

Answer: (A), (D)

The correct options are
A $\sum _{i=1}^5(\frac{1}{a_i}+\frac{1}{b_i})=130$
D $\sum _{i=1}^5\left(\frac{1}{a_i{b}_i}\right)=850$

$z^{10}+(13z-1{)}^{10}=0\Rightarrow z^{10}[1+{(13-\frac{1}{z})}^{10}]=0\Rightarrow {(13-\frac{1}{z})}^{10}=-1\Rightarrow 13-\frac{1}{z}=[\cos (2m+1)\pi +i\sin (2m+1)\pi {]}^{1/10}\Rightarrow 13-\frac{1}{z}=e^{i(2m+1)\pi /10}\therefore \frac{1}{z}=13-e^{i(2m+1)\pi /10}$

For the complex roots, substituting $m=0,1,2,\dots ,9$
$\frac{1}{a_1}=\frac{1}{z_1}=13-e^{i\left(1\pi \right)/10}\frac{1}{b_1}=\frac{1}{z_{10}}=13-e^{i\left(19\pi \right)/10}=13-e^{-i\left(1\pi \right)/10}\frac{1}{a_2}=\frac{1}{z_2}=13-e^{i\left(3\pi \right)/10}\frac{1}{b_2}=\frac{1}{z_9}=13-e^{i\left(17\pi \right)/10}=13-e^{-i\left(3\pi \right)/10}⋮\frac{1}{a_5}=\frac{1}{z_5}=13-e^{i\left(5\pi \right)/10}\frac{1}{b_5}=\frac{1}{z_6}=13-e^{i\left(15\pi \right)/10}=13-e^{-i\left(5\pi \right)/10}$

Now, $\sum _{i=1}^5(\frac{1}{a_i}+\frac{1}{b_i})$
$=13\times 10-2Re[e^{i\left(1\pi \right)/10}+e^{i\left(3\pi \right)/10}+e^{i\left(5\pi \right)/10}+e^{i\left(7\pi \right)/10}+e^{i\left(9\pi \right)/10}]=130-2[\cos \frac{\pi }{10}+\cos \frac{3\pi }{10}+\cos \frac{5\pi }{10}+\cos \frac{7\pi }{10}+\cos \frac{9\pi }{10}]=130-2[\cos \frac{\pi }{10}+\cos \frac{3\pi }{10}+\cos \frac{\pi }{2}-\cos \frac{3\pi }{10}-\cos \frac{\pi }{10}]=130$

Also, $\frac{1}{a_1{b}_1}=(13-e^{i\left(1\pi \right)/10})(13-e^{-i\left(1\pi \right)/10})$
$\Rightarrow \frac{1}{a_1{b}_1}=169-13[e^{i\left(1\pi \right)/10}+e^{-i\left(1\pi \right)/10}]+1\Rightarrow \frac{1}{a_1{b}_1}=170-26\left[\cos \frac{\pi }{10}\right]$

Now, $\sum _{i=1}^5\left(\frac{1}{a_i{b}_i}\right)$
$=170\times 5-26[\cos \frac{\pi }{10}+\cos \frac{3\pi }{10}+\cos \frac{5\pi }{10}+\cos \frac{7\pi }{10}+\cos \frac{9\pi }{10}]=850$