Complex solutions of equation

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I'm trying to find the roots of P(x) :$$x^3-3xenspace-4=0$$



First I tried $P(pm1),enspace P(pm2), enspace P(pm4)$ , and I found no rational solutions.



So I used the formula for the solutions of a third degree equation:



given $$x^3+px+q=0$$



you have $$x=sqrt[3]-fracq2+sqrtfracq^24+fracp^327enspace+enspace sqrt[3]-fracq2-sqrtfracq^24+fracp^327$$



with $p = -3,enspace q = -4enspace$
I found the real solution $$x = sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3$$



Now how do I find the two complex solutions?



Should I divide
$enspace x^3-3xenspace-4enspace$ by $Bigl(x - (sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3)Bigr) $?




complex-numbers roots cubic-equations




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  • I checked on wolfram
    – IDK
    Aug 11 at 22:30










  • See math.stackexchange.com/questions/2157643/…
    – lab bhattacharjee
    Aug 12 at 0:07














up vote
0
down vote

favorite












I'm trying to find the roots of P(x) :$$x^3-3xenspace-4=0$$



First I tried $P(pm1),enspace P(pm2), enspace P(pm4)$ , and I found no rational solutions.



So I used the formula for the solutions of a third degree equation:



given $$x^3+px+q=0$$



you have $$x=sqrt[3]-fracq2+sqrtfracq^24+fracp^327enspace+enspace sqrt[3]-fracq2-sqrtfracq^24+fracp^327$$



with $p = -3,enspace q = -4enspace$
I found the real solution $$x = sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3$$



Now how do I find the two complex solutions?



Should I divide
$enspace x^3-3xenspace-4enspace$ by $Bigl(x - (sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3)Bigr) $?




complex-numbers roots cubic-equations




share|cite|improve this question






















  • I checked on wolfram
    – IDK
    Aug 11 at 22:30










  • See math.stackexchange.com/questions/2157643/…
    – lab bhattacharjee
    Aug 12 at 0:07












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to find the roots of P(x) :$$x^3-3xenspace-4=0$$



First I tried $P(pm1),enspace P(pm2), enspace P(pm4)$ , and I found no rational solutions.



So I used the formula for the solutions of a third degree equation:



given $$x^3+px+q=0$$



you have $$x=sqrt[3]-fracq2+sqrtfracq^24+fracp^327enspace+enspace sqrt[3]-fracq2-sqrtfracq^24+fracp^327$$



with $p = -3,enspace q = -4enspace$
I found the real solution $$x = sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3$$



Now how do I find the two complex solutions?



Should I divide
$enspace x^3-3xenspace-4enspace$ by $Bigl(x - (sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3)Bigr) $?




complex-numbers roots cubic-equations




share|cite|improve this question














I'm trying to find the roots of P(x) :$$x^3-3xenspace-4=0$$



First I tried $P(pm1),enspace P(pm2), enspace P(pm4)$ , and I found no rational solutions.



So I used the formula for the solutions of a third degree equation:



given $$x^3+px+q=0$$



you have $$x=sqrt[3]-fracq2+sqrtfracq^24+fracp^327enspace+enspace sqrt[3]-fracq2-sqrtfracq^24+fracp^327$$



with $p = -3,enspace q = -4enspace$
I found the real solution $$x = sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3$$



Now how do I find the two complex solutions?



Should I divide
$enspace x^3-3xenspace-4enspace$ by $Bigl(x - (sqrt[3]2+sqrt3enspace +enspacesqrt[3]2 - sqrt3)Bigr) $?





complex-numbers roots cubic-equations





share|cite|improve this question













share|cite|improve this question




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edited Aug 11 at 22:59









José Carlos Santos

116k1699178




116k1699178










asked Aug 11 at 22:18









IDK

205




205











  • I checked on wolfram
    – IDK
    Aug 11 at 22:30










  • See math.stackexchange.com/questions/2157643/…
    – lab bhattacharjee
    Aug 12 at 0:07
















  • I checked on wolfram
    – IDK
    Aug 11 at 22:30










  • See math.stackexchange.com/questions/2157643/…
    – lab bhattacharjee
    Aug 12 at 0:07















I checked on wolfram
– IDK
Aug 11 at 22:30




I checked on wolfram
– IDK
Aug 11 at 22:30












See math.stackexchange.com/questions/2157643/…
– lab bhattacharjee
Aug 12 at 0:07




See math.stackexchange.com/questions/2157643/…
– lab bhattacharjee
Aug 12 at 0:07










1 Answer
1






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Let $omega=cosleft(frac2pi3right)+sinleft(frac2pi3right)i=-frac12+fracsqrt32i$. Then the roots of the equation are$$sqrt[3]2+sqrt3+sqrt[3]2-sqrt3, sqrt[3]2+sqrt3,omega+sqrt[3]2-sqrt3,omega^2text and sqrt[3]2+sqrt3,omega^2+sqrt[3]2-sqrt3,omega.$$You will find an explanation here.






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  • Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
    – IDK
    Aug 11 at 22:47










  • I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
    – José Carlos Santos
    Aug 11 at 22:55











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Let $omega=cosleft(frac2pi3right)+sinleft(frac2pi3right)i=-frac12+fracsqrt32i$. Then the roots of the equation are$$sqrt[3]2+sqrt3+sqrt[3]2-sqrt3, sqrt[3]2+sqrt3,omega+sqrt[3]2-sqrt3,omega^2text and sqrt[3]2+sqrt3,omega^2+sqrt[3]2-sqrt3,omega.$$You will find an explanation here.






share|cite|improve this answer






















  • Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
    – IDK
    Aug 11 at 22:47










  • I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
    – José Carlos Santos
    Aug 11 at 22:55















up vote
1
down vote



accepted










Let $omega=cosleft(frac2pi3right)+sinleft(frac2pi3right)i=-frac12+fracsqrt32i$. Then the roots of the equation are$$sqrt[3]2+sqrt3+sqrt[3]2-sqrt3, sqrt[3]2+sqrt3,omega+sqrt[3]2-sqrt3,omega^2text and sqrt[3]2+sqrt3,omega^2+sqrt[3]2-sqrt3,omega.$$You will find an explanation here.






share|cite|improve this answer






















  • Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
    – IDK
    Aug 11 at 22:47










  • I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
    – José Carlos Santos
    Aug 11 at 22:55













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Let $omega=cosleft(frac2pi3right)+sinleft(frac2pi3right)i=-frac12+fracsqrt32i$. Then the roots of the equation are$$sqrt[3]2+sqrt3+sqrt[3]2-sqrt3, sqrt[3]2+sqrt3,omega+sqrt[3]2-sqrt3,omega^2text and sqrt[3]2+sqrt3,omega^2+sqrt[3]2-sqrt3,omega.$$You will find an explanation here.






share|cite|improve this answer














Let $omega=cosleft(frac2pi3right)+sinleft(frac2pi3right)i=-frac12+fracsqrt32i$. Then the roots of the equation are$$sqrt[3]2+sqrt3+sqrt[3]2-sqrt3, sqrt[3]2+sqrt3,omega+sqrt[3]2-sqrt3,omega^2text and sqrt[3]2+sqrt3,omega^2+sqrt[3]2-sqrt3,omega.$$You will find an explanation here.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 11 at 22:38

























answered Aug 11 at 22:27









José Carlos Santos

116k1699178




116k1699178











  • Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
    – IDK
    Aug 11 at 22:47










  • I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
    – José Carlos Santos
    Aug 11 at 22:55

















  • Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
    – IDK
    Aug 11 at 22:47










  • I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
    – José Carlos Santos
    Aug 11 at 22:55
















Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
– IDK
Aug 11 at 22:47




Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it?
– IDK
Aug 11 at 22:47












I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
– José Carlos Santos
Aug 11 at 22:55





I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=sqrt[3]2+sqrt3$ and $v=sqrt[3]2-sqrt3$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $omega^3=1$ and therefore $3(uomega)(vomega^2)=3uv=-p$ and $(uomega)^3+(vomega^2)^3=u^3+v^3=-q$; so, $u,omega+v,omega^2$ is also a root. And the same argument applies to $u,omega^2+v,omega$.
– José Carlos Santos
Aug 11 at 22:55













 

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