First slide

Theory of equations

Question

Let $α, β be the roots of x^{2} - 2 xcos ϕ + 1 = 0$ then the equation whose roots are $α^{n} and β^{n}$ is

Easy

A

$x^{2} - 2 xcos nϕ - 1 = 0$
B

$x^{2} - 2 xcos nϕ + 1 = 0$
C

$x^{2} - 2 xsin nϕ + 1 = 0$
D

$x^{2} + 2 xsin nϕ - 1 = 0$

Solution

The given equation is $x^{2} - 2 xcos ϕ + 1 = 0$
$∴ x = \frac{2 \cos ϕ \pm \sqrt{4 \cos^{2} ϕ - 4}}{2} = \cos ϕ \pm isin ϕ$
$Let α = \cos ϕ + i \sin ϕ, then β = \cos ϕ - i \sin ϕ ∴ α^{n} + β^{n} = (\cos ϕ + i \sin ϕ)^{n} + (\cos ϕ - i \sin ϕ)^{n} = 2 \cos n ϕ and \begin{aligned} α^{n} β^{n} = (\cos n ϕ + i \sin n ϕ) (\cos n ϕ - i \sin n ϕ) \\ = \cos^{2} n ϕ + \sin^{2} n ϕ = 1 \end{aligned}$
$∴$ Required equation is
$x^{2} - 2 xcos nϕ + 1 = 0$

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