let-and-roots-of-z-2-3z-5-0-simlify-U-n-k-0-n-k-k-and-V-n-k-0-n-1-k-1-k-

Question Number 146902 by mathmax by abdo last updated on 16/Jul/21

let α and β roots of z^{2} + 3 z + 5 = 0

simlify U_{n} = \sum_{k = 0}^{n} (α^{k} + β^{k})

and V_{n} = \sum_{k = 0}^{n} (\frac{1}{α^{k}} + \frac{1}{β^{k}})

Answered by ArielVyny last updated on 18/Jul/21

z^{2} + 3 z + 5 = 0 \to {(z + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{20}{4} = 0

{(z + \frac{3}{2})}^{2} + \frac{11}{4} = 0

(z + \frac{3}{2} - i \frac{\sqrt{11}}{2}) (z + \frac{3}{2} + i \frac{\sqrt{11}}{2}) = 0

t h e n α = - \frac{3}{2} + i \frac{\sqrt{11}}{2} a n d β = - \frac{3}{2} - i \frac{\sqrt{11}}{2}

U_{n} = \sum_{k = 0}^{n} (α^{k} + β^{k})

∣ α ∣=∣ - \frac{3}{2} + i \frac{\sqrt{11}}{2} ∣= \sqrt{{(- \frac{3}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}}

∣ α ∣= \sqrt{\frac{9}{4} + \frac{11}{4}} = \frac{\sqrt{20}}{2}

α = \frac{\sqrt{20}}{2} (- \frac{\frac{3}{2}}{\frac{\sqrt{20}}{2}} + i \frac{\frac{\sqrt{11}}{2}}{\frac{\sqrt{20}}{2}}) = \frac{\sqrt{20}}{2} (- \frac{3}{\sqrt{20}} + i \frac{\sqrt{11}}{\sqrt{20}})

α = \frac{\sqrt{20}}{2} e^{i θ} t e l q u e {\begin{cases} c o s θ = - \frac{3}{\sqrt{20}} \\ s i n θ = \sqrt{\frac{11}{20}} \end{cases}

β = \frac{\sqrt{20}}{2} e^{- i θ}

U_{n} = \sum_{k = 0}^{n} [{(\frac{\sqrt{20}}{2})}^{k} e^{i k θ} + {(\frac{\sqrt{20}}{2})}^{k} e^{- i k θ}]

U_{n} = \frac{1 - {(\frac{\sqrt{20}}{2} e^{i θ})}^{n + 1}}{1 - (\frac{\sqrt{20}}{2} e^{i θ})} + \frac{1 - {(\frac{\sqrt{20}}{2} e^{- i θ})}^{n + 1}}{1 - (\frac{\sqrt{20}}{2} e^{- i θ})}

U_{n} = \frac{1 - \frac{\sqrt{20}}{2} e^{- i θ} - {(\frac{\sqrt{20}}{2} e^{i θ})}^{n + 1} + {(\frac{\sqrt{20}}{2})}^{n + 2} e^{i θ n}}{1 - (\frac{\sqrt{20}}{2} e^{- i θ}) - (\frac{\sqrt{20}}{2} e^{i θ}) + \frac{20}{4}}

t o b e c o n t i n u e d \dots .