If Ω = Σ_(n=1) ^∞ (Π_(k=2) ^∞ ((k^3 − 1)/(k^3 + 1)))^n Solve for complex numbees: z^4 + 3z^3 + Ωz^2 + 3z + 1 = 0


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
	$((k^3 −1)/(k^3 +1))=(((k−1)(k^2 +k+1))/((k+1)(k^2 −k+1))) Π_(k=2) ^m ((k^3 −1)/(k^3 +1))=Π_(k=2) ^m ((k−1)/(k+1))Π_(k=2) ^m ((k^2 +k+1)/(k^2 −k+1)) Π_(k=2) ^m ((k^2 +k+1)/(k^2 −k+1)) = ((Π_(k=2) ^m k^2 +k+1)/(Π_(k=2) ^m k^2 −k+1)) = =((Π_(k=2) ^m k^2 +k+1)/(Π_(k=2) ^m (k−1)^2 +(k−1)+1))= =((Π_(k=2) ^m k^2 +k+1)/(Π_(k=1) ^(m−1) k^2 +k+1))=((m^2 +m+1)/3) Π_(k=2) ^m ((k−1)/(k+1))=Π_(k=2) ^m ((k−1)/k)Π_(k=2) ^m (k/(k+1))= =(1/m) (2/(m+1))=(2/(m(m+1))) Π_(k=2) ^m ((k^3 −1)/(k^3 +1))=((2(m^2 +m+1))/(3m(m+1))) ⇒Π_(k=2) ^∞ ((k^3 −1)/(k^3 +1))=lim_(m→∞) ((2(m^2 +m+1))/(3m(m+1)))=(2/3) ⇒Ω=Σ_(n=1) ^∞ ((2/3))^n =(2/3)Σ_(n=0) ^∞ ((2/3))^n = =(2/3) (1/(1−(2/3)))=2 z^4 +3z^3 +Ωz^2 +3z+1=0 z^4 +3z^3 +2z^2 +3z+1=0 z=0⇒not sol. z^2 +3z+2+3(1/z)+(1/z^2 )=0 w=z+(1/z)⇒w^2 =2+z^2 +(1/z^2 )⇒z^2 +(1/z^2 )=w^2 −2 w^2 −2+3w+2=0 ⇒w^2 +3w=0⇒w=0;−3 ⇒z+(1/z)=0⇒z^2 +1=0⇒z_(1,2) =±i z+(1/z)=−3⇒z^2 +3z+1=0⇒z_(3,4) =((−3±(√5))/2) ⇒z∈{i,−i,(((√5)−3)/2),−((3+(√5))/2)} ; Ω=2$
	$\frac{k^{3} - 1}{k^{3} + 1} = \frac{(k - 1) (k^{2} + k + 1)}{(k + 1) (k^{2} - k + 1)}$ $\underset{k = 2}{\prod^{m}} \frac{k^{3} - 1}{k^{3} + 1} = \underset{k = 2}{\prod^{m}} \frac{k - 1}{k + 1} \underset{k = 2}{\prod^{m}} \frac{k^{2} + k + 1}{k^{2} - k + 1}$ $\underset{k = 2}{\prod^{m}} \frac{k^{2} + k + 1}{k^{2} - k + 1} = \frac{\underset{k = 2}{\prod^{m}} k^{2} + k + 1}{\underset{k = 2}{\prod^{m}} k^{2} - k + 1} =$ $= \frac{\underset{k = 2}{\prod^{m}} k^{2} + k + 1}{\underset{k = 2}{\prod^{m}} {(k - 1)}^{2} + (k - 1) + 1} =$ $= \frac{\underset{k = 2}{\prod^{m}} k^{2} + k + 1}{\underset{k = 1}{\prod^{m - 1}} k^{2} + k + 1} = \frac{m^{2} + m + 1}{3}$ $\underset{k = 2}{\prod^{m}} \frac{k - 1}{k + 1} = \underset{k = 2}{\prod^{m}} \frac{k - 1}{k} \underset{k = 2}{\prod^{m}} \frac{k}{k + 1} =$ $= \frac{1}{m} \frac{2}{m + 1} = \frac{2}{m (m + 1)}$ $\underset{k = 2}{\prod^{m}} \frac{k^{3} - 1}{k^{3} + 1} = \frac{2 (m^{2} + m + 1)}{3 m (m + 1)}$ $\Rightarrow \underset{k = 2}{\prod^{\infty}} \frac{k^{3} - 1}{k^{3} + 1} = \underset{m \to \infty}{l i m} \frac{2 (m^{2} + m + 1)}{3 m (m + 1)} = \frac{2}{3}$ $\Rightarrow Ω = \underset{n = 1}{\sum^{\infty}} {(\frac{2}{3})}^{n} = \frac{2}{3} \underset{n = 0}{\sum^{\infty}} {(\frac{2}{3})}^{n} =$ $= \frac{2}{3} \frac{1}{1 - \frac{2}{3}} = 2$ $z^{4} + 3 z^{3} + Ω z^{2} + 3 z + 1 = 0$ $z^{4} + 3 z^{3} + 2 z^{2} + 3 z + 1 = 0$ $z = 0 \Rightarrow n o t s o l .$ $z^{2} + 3 z + 2 + 3 \frac{1}{z} + \frac{1}{z^{2}} = 0$ $w = z + \frac{1}{z} \Rightarrow w^{2} = 2 + z^{2} + \frac{1}{z^{2}} \Rightarrow z^{2} + \frac{1}{z^{2}} = w^{2} - 2$ $w^{2} - 2 + 3 w + 2 = 0$ $\Rightarrow w^{2} + 3 w = 0 \Rightarrow w = 0; - 3$ $\Rightarrow z + \frac{1}{z} = 0 \Rightarrow z^{2} + 1 = 0 \Rightarrow z_{1, 2} = \pm i$ $z + \frac{1}{z} = - 3 \Rightarrow z^{2} + 3 z + 1 = 0 \Rightarrow z_{3, 4} = \frac{- 3 \pm \sqrt{5}}{2}$ $\Rightarrow z \in {i, - i, \frac{\sqrt{5} - 3}{2}, - \frac{3 + \sqrt{5}}{2}}; Ω = 2$
	Commented by Shrinava last updated on 28/Feb/23

	$perfect dear professor thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum