Given-r-1-r-2-then-r-8-1-r-8-

Question Number 117188 by bemath last updated on 10/Oct/20

G i v e n r + \frac{1}{r} = \sqrt{2}, t h e n r^{8} + \frac{1}{r^{8}} = ?

Commented by bemath last updated on 10/Oct/20

t h a n k y o u s i r s

Answered by AbduraufKodiriy last updated on 10/Oct/20

{(r + \frac{1}{r})}^{2} = {(\sqrt{2})}^{2} \Rightarrow r^{2} + 2 + \frac{1}{r^{2}} = 2 \Rightarrow r^{2} + \frac{1}{r^{2}} = 0

r i s C o m p l e x

(r^{2} + \frac{1}{r^{2}}) = 0^{2} \Rightarrow r^{4} + 2 + \frac{1}{r^{4}} = 0 \Rightarrow r^{4} + \frac{1}{r^{4}} = - 2

{(r^{4} + \frac{1}{r^{4}})}^{2} = {(- 2)}^{2} \Rightarrow r^{8} + 2 + \frac{1}{r^{8}} = 4 \Rightarrow r^{8} + \frac{1}{r^{8}} = 2

Answered by TANMAY PANACEA last updated on 10/Oct/20

r^{2} - \sqrt{2} r + 1 = 0

r = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm \sqrt{2} i}{2} \to (\frac{1}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}}) \to (c o s \frac{π}{4} \pm i s i n \frac{π}{4})

r = e^{i \times \frac{π}{4}} o r e^{- \frac{i π}{4}}

r^{8} + \frac{1}{r^{8}}

= {(e^{i \times \frac{π}{4}})}^{8} + {(e^{- i \times \frac{π}{4}})}^{8}

= e^{i 2 π} + e^{- i 2 π} [e^{i θ} = c o s θ + i s i n θ]

= 2 c o s 2 π = 2