Mathematics - Complex Number Question with Solution | TestHub

 $\left(2+\alpha \right){}^4=a+b\alpha$
${\left(4+{\alpha }^2+4\alpha \right)}^2=a+b\alpha$

Now using, $\alpha =\frac{-1+i\sqrt{3}}{2}\Rightarrow {\left(2\alpha +1\right)}^2={\left(i\sqrt{3}\right)}^2$

$\Rightarrow {\alpha }^2=-\left(1+\alpha \right)$

So, $a+b\alpha ={\left(4-\alpha -1+4\alpha \right)}^2$

$a+b\alpha ={\left(3+3\alpha \right)}^2=9\left(1+2\alpha +{\alpha }^2\right)$
$a+b\alpha =9\left(1+2\alpha +\left(-1-\alpha \right)\right)=9\alpha$

$\Rightarrow a=0, b=9\Rightarrow a+b=9$