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z-4-4z-3-6z-2-4z-1-z-4-4z-3-6z-2-4z-1-z-1-z-1-Find-z-R-

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Question Number 203474 by ajfour last updated on 20/Jan/24
((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((z+1)/(z−1)) Find z∈R.
$$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}=\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}} \\ $$$${Find}\:{z}\in\mathbb{R}. \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/24
Using Componendo-Dividendo (((z^4 +4z^3 −6z^2 −4z+1)+(z^4 −4z^3 −6z^2 +4z+1))/((z^4 +4z^3 −6z^2 −4z+1)−(z^4 −4z^3 −6z^2 +4z+1))) =(((z+1)+(z−1))/((z+1)−(z−1))) ((2z^4 −12z^2 +2)/(8z^3 −8z))=((2z)/2) ((z^4 −6z^2 +1)/(4z^3 −4z))=z z^4 −6z^2 +1=4z^4 −4z^2 3z^4 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 z^2 =−1,(1/3) z=±i, ±(1/( (√3))) ✓
$${Using}\:{Componendo}-{Dividendo} \\ $$$$\frac{\left({z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}\right)+\left({z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}\right)}{\left({z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}\right)−\left({z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({z}+\mathrm{1}\right)+\left({z}−\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)−\left({z}−\mathrm{1}\right)}\: \\ $$$$\frac{\mathrm{2}{z}^{\mathrm{4}} −\mathrm{12}{z}^{\mathrm{2}} +\mathrm{2}}{\mathrm{8}{z}^{\mathrm{3}} −\mathrm{8}{z}}=\frac{\mathrm{2}{z}}{\mathrm{2}} \\ $$$$\frac{{z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}}={z} \\ $$$${z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{4}{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{2}} \\ $$$$\mathrm{3}{z}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${z}^{\mathrm{2}} =−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}=\pm{i},\:\:\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/24
Using Cross Multiplication ▶z^5 +4z^4 −6z^3 −4z^2 +z−z^4 −4z^3 +6z^2 +4z−1 =z^5 −4z^4 −6z^3 +4z^2 +z+z^4 −4z^3 −6z^2 +4z+1 ▶8z^4 −2z^4 +4z^2 −2=0 ▶3z^4 +2z^2 −1=0 ▶(z^2 +1)(3z^2 −1)=0 ▶ z^2 =−1,(1/3) ▶z=±i, ±(1/( (√3))) ✓
$${Using}\:{Cross}\:{Multiplication} \\ $$$$\blacktriangleright{z}^{\mathrm{5}} +\mathrm{4}{z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{3}} −\mathrm{4}{z}^{\mathrm{2}} +{z}−{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} +\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}−\mathrm{1} \\ $$$$\:\:\:\:\:={z}^{\mathrm{5}} −\mathrm{4}{z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{3}} +\mathrm{4}{z}^{\mathrm{2}} +{z}+{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1} \\ $$$$\blacktriangleright\mathrm{8}{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\blacktriangleright\mathrm{3}{z}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\blacktriangleright\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\blacktriangleright\:{z}^{\mathrm{2}} =−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\blacktriangleright{z}=\pm{i},\:\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/24
((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((k(z+1))/(k(z−1))) (let) where k is such that: { ((z^4 +4z^3 −6z^2 −4z+1=k(z+1))),(( &)),((z^4 −4z^3 −6z^2 +4z+1=k(z−1))) :} { ((z^4 +4z^3 −6z^2 −4z+1−kz−k=0)),((z^4 −4z^3 −6z^2 +4z+1−kz+k=0)) :} Adding & Subtracting: { ((z^4 −6z^2 −kz+1=0)),((4z^3 −4z−k=0⇒k=4z^3 −4z)) :} ⇒z^4 −6z^2 −(4z^3 −4z)z+1=0 z^4 −6z^2 −4z^4 +4z^2 +1=0 3z^4 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 z^2 =−1,(1/3) z=±i, ±((√3)/3) ✓
$$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}=\frac{{k}\left({z}+\mathrm{1}\right)}{{k}\left({z}−\mathrm{1}\right)}\:\left({let}\right) \\ $$$${where}\:{k}\:{is}\:{such}\:{that}: \\ $$$$\begin{cases}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}={k}\left({z}+\mathrm{1}\right)}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\&}\\{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}={k}\left({z}−\mathrm{1}\right)}\end{cases}\: \\ $$$$\begin{cases}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}−{kz}−{k}=\mathrm{0}}\\{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}−{kz}+{k}=\mathrm{0}}\end{cases}\: \\ $$$${Adding}\:\&\:{Subtracting}: \\ $$$$\begin{cases}{{z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} −{kz}+\mathrm{1}=\mathrm{0}}\\{\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}−{k}=\mathrm{0}\Rightarrow{k}=\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}}\end{cases} \\ $$$$\Rightarrow{z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} −\left(\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}\right){z}+\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}{z}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${z}^{\mathrm{2}} =−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}=\pm{i},\:\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/24
Using k-method ((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((z+1)/(z−1))=k (let) z+1=kz−k kz−z=k+1 z=((k+1)/(k−1)) ((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=k z^4 +4z^3 −6z^2 −4z+1−k(z^4 −4z^3 −6z^2 +4z+1)=0 z^4 +4z^3 −6z^2 −4z+1−kz^4 +4kz^3 +6kz^2 −4kz−k=0 (1−k)z^4 +4(1+k)z^3 −6(1−k)z^2 −4(1+k)z+1−k=0 z^4 +4(((1+k)/(1−k)))z^3 −6z^2 −4(((1+k)/(1−k)))z+1=0 z^4 −4(((k+1)/(k−1)))z^3 −6z^2 +4(((k+1)/(k−1)))z+1=0 z^4 −4(z)z^3 −6z^2 +4(z)z+1=0 −3z^3 −2z^2 +1=0 3z^3 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 z^2 =−1,(1/3) z=±i, ±((√3)/3)
$${Using}\:{k}-{method} \\ $$$$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}=\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}}={k}\:\left({let}\right) \\ $$$${z}+\mathrm{1}={kz}−{k} \\ $$$${kz}−{z}={k}+\mathrm{1} \\ $$$${z}=\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\ $$$$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}={k} \\ $$$${z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}−{k}\left({z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}\right)=\mathrm{0} \\ $$$${z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}−{kz}^{\mathrm{4}} +\mathrm{4}{kz}^{\mathrm{3}} +\mathrm{6}{kz}^{\mathrm{2}} −\mathrm{4}{kz}−{k}=\mathrm{0} \\ $$$$\left(\mathrm{1}−{k}\right){z}^{\mathrm{4}} +\mathrm{4}\left(\mathrm{1}+{k}\right){z}^{\mathrm{3}} −\mathrm{6}\left(\mathrm{1}−{k}\right){z}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+{k}\right){z}+\mathrm{1}−{k}=\mathrm{0} \\ $$$${z}^{\mathrm{4}} +\mathrm{4}\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right){z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right){z}+\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{4}} −\mathrm{4}\left(\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}\right){z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}\left(\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}\right){z}+\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{4}} −\mathrm{4}\left({z}\right){z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}\left({z}\right){z}+\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{3}{z}^{\mathrm{3}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}{z}^{\mathrm{3}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\:\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${z}^{\mathrm{2}} =−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}=\pm{i},\:\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jan/24
Componedo-Dividendo version 2 ((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((z+1)/(z−1)) (((z^4 −6z^2 +1)+(4z^3 −4z))/((z^4 −6z^2 +1)−(4z^3 −4z)))=(((z)+(1))/((z)−(1))) determinant (((((a+b)/(a−b))=((c+d)/(c−d))⇔(a/b)=(c/d)))) ((z^4 −6z^2 +1)/(4z^3 −4z))=(z/1) z^4 −6z^2 +1=4z^4 −4z^2 3z^4 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 { ((z^2 +1=0 ⇒z∉R)),((3z^2 −1=0⇒z=±(1/( (√3))))) :}
$${Componedo}-{Dividendo}\:{version}\:\mathrm{2} \\ $$$$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}=\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}} \\ $$$$\frac{\left({z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}\right)}{\left({z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{1}\right)−\left(\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}\right)}=\frac{\left({z}\right)+\left(\mathrm{1}\right)}{\left({z}\right)−\left(\mathrm{1}\right)} \\ $$$$\begin{array}{|c|}{\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{d}}{{c}−{d}}\Leftrightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}}}\\\hline\end{array} \\ $$$$\frac{{z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{z}^{\mathrm{3}} −\mathrm{4}{z}}=\frac{{z}}{\mathrm{1}} \\ $$$${z}^{\mathrm{4}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{4}{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{2}} \\ $$$$\mathrm{3}{z}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{cases}{{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:\Rightarrow{z}\notin\mathbb{R}}\\{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow{z}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\end{cases} \\ $$
Commented by ajfour last updated on 20/Jan/24
Thanks sir. Good ways.
$${Thanks}\:{sir}.\:{Good}\:{ways}. \\ $$

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