Question

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

Answer 1

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

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

$y=x^4-x^2+6x-4$

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

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

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