Question

If $x=\frac{1+\sqrt{3}i}{2}$ then the value of the expression, $y=x^4-x^2+6x-4$, equals

• A. $-1+2\sqrt{3}i$
• B. $2-2\sqrt{3}i$
• C. $2+2\sqrt{3}i$
• D. $none$

Answer 1

The correct option is A $-1+2\sqrt{3}i$

$x=\frac{1+\sqrt{3}i}{2}$

$\therefore x^2={\left(\frac{1+\sqrt{3}i}{2}\right)}^2$

$\therefore x^2=\frac{1-3+2\sqrt{3}i}{4}$

$\therefore x^2=\frac{2\sqrt{3}i-2}{4}=\frac{\sqrt{3}i-1}{2}$

Now,

$x^4-x^2+6x-4=x^2(x^2-1)+6x-4$

$=\frac{\sqrt{3}i-1}{2}(\frac{\sqrt{3}i-1}{2}-1)+6\left(\frac{\sqrt{3}i+1}{2}\right)-4$

$=\left(\frac{\sqrt{3}i-1}{2}\right)\left(\frac{\sqrt{3}i-3}{2}\right)+3(\sqrt{3}i+1)-4$

$=\frac{3-3\sqrt{3}i-\sqrt{3}i+3}{4}+3\sqrt{3}i+3-4$

$=\frac{-4\sqrt{3}i}{4}+3\sqrt{3}i-1$

$=2\sqrt{3}i-1=-1+2\sqrt{3}i$

$x^4-x^2+6x-4=-1+2\sqrt{3}i$