If $\alpha$ and $\beta$ are the roots of x²–2x+4 = 0 then ${\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=$

x²–2x+4 = 0$\Rightarrow x=\frac{2\pm \sqrt{4-16}}{2}=\frac{2\pm i\sqrt{12}}{2}=1\pm i\sqrt{3}$

$\begin{array}{l}\mathrm{\alpha }=1+\mathrm{i}\sqrt{3},\mathrm{\beta }=1-\mathrm{i}\sqrt{3}\\ {\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=(1+\mathrm{i}\sqrt{3}{)}^n+(1-\mathrm{i}\sqrt{3}{)}^{\mathrm{n}}\\ {\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=2^n\left[{\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)}^n+{\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)}^n\right]\\ {\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=2^{\mathrm{n}}\cdot \left(cos\frac{\mathrm{n\pi }}{3}+i\sin \frac{n\pi }{3}\right)+2^{\mathrm{n}}\left(\cos \frac{\mathrm{n\pi }}{3}-i\sin \frac{n\pi }{3}\right)\\ {\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=2^{\mathrm{n}}\left(2\cos \frac{\mathrm{n\pi }}{3}\right)\\ {\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=2^{\mathrm{n}+1}\cos \frac{\mathrm{n\pi }}{3}\end{array}$