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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.
z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=z+1z−1Findz∈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))) ✓
UsingComponendo−Dividendo(z4+4z3−6z2−4z+1)+(z4−4z3−6z2+4z+1)(z4+4z3−6z2−4z+1)−(z4−4z3−6z2+4z+1)=(z+1)+(z−1)(z+1)−(z−1)2z4−12z2+28z3−8z=2z2z4−6z2+14z3−4z=zz4−6z2+1=4z4−4z23z4+2z2−1=0(z2+1)(3z2−1)=0z2=−1,13z=±i,±13✓
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))) ✓
UsingCrossMultiplication▸z5+4z4−6z3−4z2+z−z4−4z3+6z2+4z−1=z5−4z4−6z3+4z2+z+z4−4z3−6z2+4z+1▸8z4−2z4+4z2−2=0▸3z4+2z2−1=0▸(z2+1)(3z2−1)=0▸z2=−1,13▸z=±i,±13✓
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) ✓
z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=k(z+1)k(z−1)(let)wherekissuchthat:{z4+4z3−6z2−4z+1=k(z+1)&z4−4z3−6z2+4z+1=k(z−1){z4+4z3−6z2−4z+1−kz−k=0z4−4z3−6z2+4z+1−kz+k=0Adding&Subtracting:{z4−6z2−kz+1=04z3−4z−k=0⇒k=4z3−4z⇒z4−6z2−(4z3−4z)z+1=0z4−6z2−4z4+4z2+1=03z4+2z2−1=0(z2+1)(3z2−1)=0z2=−1,13z=±i,±33✓
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)
Usingk−methodz4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=z+1z−1=k(let)z+1=kz−kkz−z=k+1z=k+1k−1z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=kz4+4z3−6z2−4z+1−k(z4−4z3−6z2+4z+1)=0z4+4z3−6z2−4z+1−kz4+4kz3+6kz2−4kz−k=0(1−k)z4+4(1+k)z3−6(1−k)z2−4(1+k)z+1−k=0z4+4(1+k1−k)z3−6z2−4(1+k1−k)z+1=0z4−4(k+1k−1)z3−6z2+4(k+1k−1)z+1=0z4−4(z)z3−6z2+4(z)z+1=0−3z3−2z2+1=03z3+2z2−1=0(z2+1)(3z2−1)=0z2=−1,13z=±i,±33
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−Dividendoversion2z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=z+1z−1(z4−6z2+1)+(4z3−4z)(z4−6z2+1)−(4z3−4z)=(z)+(1)(z)−(1)a+ba−b=c+dc−d⇔ab=cdz4−6z2+14z3−4z=z1z4−6z2+1=4z4−4z23z4+2z2−1=0(z2+1)(3z2−1)=0{z2+1=0⇒z∉R3z2−1=0⇒z=±13
Commented by ajfour last updated on 20/Jan/24
Thanks sir. Good ways.
Thankssir.Goodways.

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