Question

If $a$ and $b$ are real numbers such that $(2+\alpha {)}^4=a+b\alpha$, where $\alpha =\frac{-1+i\sqrt{3}}{2},$ then $a+b$ is equal to

• A. $33$
• B. $57$
• C. $9$
• D. $24$

Answer 1

The correct option is C $9$

Given: $(2+\alpha {)}^4=a+b\alpha$
$\Rightarrow {(2+\frac{(-1+i\sqrt{3})}{2})}^4=a+b\alpha$
$\Rightarrow {(\frac{3}{2}+\frac{i\sqrt{3}}{2})}^4=a+b\alpha$
$\Rightarrow 9{(\frac{\sqrt{3}}{2}+\frac{i}{2})}^4=a+b\alpha$
$\Rightarrow 9(e^{i\pi /6}{)}^4=a+b(\frac{-1}{2}+\frac{i\sqrt{3}}{2})$
$\Rightarrow 9\left(e^{i2\pi /3}\right)=(a-\frac{b}{2})+i\left(\frac{b\sqrt{3}}{2}\right)$
$\Rightarrow -\frac{9}{2}+\frac{9\sqrt{3}}{2}i=(a-\frac{b}{2})+i\left(\frac{b\sqrt{3}}{2}\right)$
Compare real and imaginary parts, we get
$\frac{b\sqrt{3}}{2}=\frac{9\sqrt{3}}{2}$
$\Rightarrow b=9$
and $a-\frac{b}{2}=-\frac{9}{2}$
$\Rightarrow a=0$
$\therefore a+b=9$