1-i-3-2-2020-1-i-3-2-2020-A-A-4-

Question Number 81966 by naka3546 last updated on 17/Feb/20

{(\frac{1 + i \sqrt{3}}{2})}^{2020} + {(\frac{1 - i \sqrt{3}}{2})}^{2020} = A

A^{4} = ?

Answered by TANMAY PANACEA last updated on 17/Feb/20

\frac{1}{2} = c o s \frac{π}{3} \frac{\sqrt{3}}{2} = s i n \frac{π}{3}

{(c o s \frac{π}{3} + i s i n \frac{π}{3})}^{2020} + {(c o s \frac{π}{3} - i s i n \frac{π}{3})}^{2020}

{(e^{i \times \frac{π}{3}})}^{2020} + {(e^{- i \times \frac{π}{3}})}^{2020}

e^{i θ} + e^{- i θ} = 2 c o s θ w h e r e [θ = \frac{2020 π}{3}]

A = 2 c o s θ

A^{4} = 16 {(c o s θ)}^{4}

n o w c o s (\frac{2020 π}{3})

c o s (2020 \times 60)

= c o s (121200)

= c o s (336 \times 360 + 240)

= - c o s 60

= \frac{- 1}{2}

s o r e q u i r e d a n s w e r i s 16 \times {(\frac{- 1}{2})}^{4} = 1

Commented by naka3546 last updated on 17/Feb/20

θ = \frac{4 π}{3} \Rightarrow A = 2 \cos θ = - 1