If $x=\left(\frac{1+i}{2}\right)$ , (where $i=\sqrt{-1}$ ) then the expression $2x^4-2x^2+x+3$  equals:

$\therefore x=\left(\frac{1+i}{2}\right)$

$\Rightarrow$        $2x-1=i\Rightarrow 4x^2+1-4x=-1$

$\Rightarrow$            $2x^2-2x+1=0$

Since $2x^4-2x^2+x+3$

$=\left(2x^2-2x+1\right)\left(x^2+x\right)+\left(3-x^2\right)$

$=0+3-{\left(\frac{1+i}{2}\right)}^2=3-\left(i/2\right)$