Discussion Forum

1)

If $\alpha$  is a root of $z^{2}+z+1=0$  , then 

$\left(\alpha^{2014}+\frac{1}{\alpha^{2014}}\right)+\left(\alpha^{2015}+\frac{1}{\alpha^{2015}}\right)^{2}$

$+\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^{3}+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^{4}+\left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^{5}$=

Answer:

Option A

Explanation:

 $z^{2}-z+1=0$    then $z=-\omega$

So,   $| (-\omega)^{2014}+\frac{1}{(-\omega)^{2014}}|+\left[(-\omega)^{2015}+\frac{1}{(-\omega)^{2015}}\right]^{2}$

$\left[ (-\omega)^{2016}+\frac{1}{(-\omega)^{2016}}\right]^{3}+\left[(-\omega)^{2017}+\frac{1}{(-\omega)^{2017}}\right]^{4}+\left[(-\omega)^{2018}+\frac{1}{(-\omega)^{2018}}\right]^{5}$

$\left[-\omega-\frac{1}{\omega}\right]+\left[\omega^{2}+\frac{1}{\omega^{2}}\right]^{2}+8+\left[-\omega+\frac{1}{-\omega}\right]^{4}+\left[\omega^{2}+\frac{1}{\omega^{2}}\right]^{5}$

  $=[+\omega+\omega^{2}]+[\omega^{2}+\omega]^{2}+8+[-\omega-\omega^{2}]^{4}+[\omega^{2}+\omega]^{5}$

   =-1+1+8+1+1-1=8