If-n-1-k-2-k-3-1-k-3-1-n-Solve-for-complex-numbees-z-4-3z-3-z-2-3z-1-0-

Question Number 188224 by Shrinava last updated on 26/Feb/23

If Ω = \underset{n = 1}{\sum^{\infty}} {(\underset{k = 2}{\prod^{\infty}} \frac{k^{3} - 1}{k^{3} + 1})}^{n}

Solve for complex numbees :

z^{4} + {3 z}^{3} + Ω z^{2} + 3 z + 1 = 0

Answered by aleks041103 last updated on 27/Feb/23

\frac{k^{3} - 1}{k^{3} + 1} = \frac{(k - 1) (k^{2} + k + 1)}{(k + 1) (k^{2} - k + 1)}

\underset{k = 2}{\prod^{m}} \frac{k^{3} - 1}{k^{3} + 1} = \underset{k = 2}{\prod^{m}} \frac{k - 1}{k + 1} \underset{k = 2}{\prod^{m}} \frac{k^{2} + k + 1}{k^{2} - k + 1}

\underset{k = 2}{\prod^{m}} \frac{k^{2} + k + 1}{k^{2} - k + 1} = \frac{\underset{k = 2}{\prod^{m}} k^{2} + k + 1}{\underset{k = 2}{\prod^{m}} k^{2} - k + 1} =

= \frac{\underset{k = 2}{\prod^{m}} k^{2} + k + 1}{\underset{k = 2}{\prod^{m}} {(k - 1)}^{2} + (k - 1) + 1} =

= \frac{\underset{k = 2}{\prod^{m}} k^{2} + k + 1}{\underset{k = 1}{\prod^{m - 1}} k^{2} + k + 1} = \frac{m^{2} + m + 1}{3}

\underset{k = 2}{\prod^{m}} \frac{k - 1}{k + 1} = \underset{k = 2}{\prod^{m}} \frac{k - 1}{k} \underset{k = 2}{\prod^{m}} \frac{k}{k + 1} =

= \frac{1}{m} \frac{2}{m + 1} = \frac{2}{m (m + 1)}

\underset{k = 2}{\prod^{m}} \frac{k^{3} - 1}{k^{3} + 1} = \frac{2 (m^{2} + m + 1)}{3 m (m + 1)}

\Rightarrow \underset{k = 2}{\prod^{\infty}} \frac{k^{3} - 1}{k^{3} + 1} = \underset{m \to \infty}{l i m} \frac{2 (m^{2} + m + 1)}{3 m (m + 1)} = \frac{2}{3}

\Rightarrow Ω = \underset{n = 1}{\sum^{\infty}} {(\frac{2}{3})}^{n} = \frac{2}{3} \underset{n = 0}{\sum^{\infty}} {(\frac{2}{3})}^{n} =

= \frac{2}{3} \frac{1}{1 - \frac{2}{3}} = 2

z^{4} + 3 z^{3} + Ω z^{2} + 3 z + 1 = 0

z^{4} + 3 z^{3} + 2 z^{2} + 3 z + 1 = 0

z = 0 \Rightarrow n o t s o l .

z^{2} + 3 z + 2 + 3 \frac{1}{z} + \frac{1}{z^{2}} = 0

w = z + \frac{1}{z} \Rightarrow w^{2} = 2 + z^{2} + \frac{1}{z^{2}} \Rightarrow z^{2} + \frac{1}{z^{2}} = w^{2} - 2

w^{2} - 2 + 3 w + 2 = 0

\Rightarrow w^{2} + 3 w = 0 \Rightarrow w = 0; - 3

\Rightarrow z + \frac{1}{z} = 0 \Rightarrow z^{2} + 1 = 0 \Rightarrow z_{1, 2} = \pm i

z + \frac{1}{z} = - 3 \Rightarrow z^{2} + 3 z + 1 = 0 \Rightarrow z_{3, 4} = \frac{- 3 \pm \sqrt{5}}{2}

\Rightarrow z \in {i, - i, \frac{\sqrt{5} - 3}{2}, - \frac{3 + \sqrt{5}}{2}}; Ω = 2

Commented by Shrinava last updated on 28/Feb/23

perfect dear professor thank you