Question

Evaluate the mentioned below:
(iv)$x^6+x^4+x^2+1,whenx=(1+i)/\sqrt{2}$

Answer 1

Given that-
$x=(1+i)/\sqrt{2}$
$\Rightarrow \sqrt{2}x=1+i$
$\Rightarrow (\sqrt{2}x{)}^2=(1+i{)}^{2}squaringonbothside$
$\Rightarrow 2x^2=1+i^2+2i$
$\Rightarrow x^2=i\dots \left(1\right)$
$\Rightarrow x^6+x^4+x^2+1$
$=(x^2{)}^3+(x^2{)}^2+x^2+1$
$=(i{)}^3+(i{)}^2+i+1fromeqn\left(1\right)$
$=-i-1+i+1$
=0
Therefore, $x^6+x^4+x^2+1=0$