If alpha , beta are the roots of the equatio | vedclass

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Question

$\alpha+\beta=2$

$\alpha . \beta=4$

by solving,

$\alpha=2[\cos \pi / 3+i \sin \pi / 3]$

$\beta=2[\cos \pi / 3-i \sin \pi / 3]$

$\mathrm

Vedclass · Accepted Answer

${2^{n + 1}}\cos \left( {\frac{{n\pi }}{3}} \right)$

Vedclass · Answer

$\alpha+\beta=2$

$\alpha . \beta=4$

by solving,

$\alpha=2[\cos \pi / 3+i \sin \pi / 3]$

$\beta=2[\cos \pi / 3-i \sin \pi / 3]$

$\mathrm{SO},$

$2^{n}\left[\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}\right]+2^{n}\left[\cos \frac{n \pi}{3}-i \sin \frac{n \pi}{3}\right]$

$=2^{n}\left[\cos \frac{n \pi}{3}\right]$

$=2^{n+1} \cdot \cos \frac{n \pi}{3}$

If $\alpha , \beta $ are the roots of the equation $x^2 - 2x + 4 = 0$ , then the value of $\alpha ^n +\beta ^n$ is

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