Find all 4 complex roots to $3z^4-z^3+2z^2-z+3=0$ using a trig substitution
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Find all 4 complex roots to $3z^4-z^3+2z^2-z+3=0$ using a trig substitution
A previous part of the question made us prove that $2cos ntheta=z^n+z^{-n}$ and it wants us to prove that the polynomial can be expressed as $6cos 2theta -2costheta +2=0$. I'm assuming that using $6cos 2theta -2costheta +2=0$ we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation. Any help would be appreciated.
trigonometry polynomials complex-numbers
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
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add a comment |
$begingroup$
Find all 4 complex roots to $3z^4-z^3+2z^2-z+3=0$ using a trig substitution
A previous part of the question made us prove that $2cos ntheta=z^n+z^{-n}$ and it wants us to prove that the polynomial can be expressed as $6cos 2theta -2costheta +2=0$. I'm assuming that using $6cos 2theta -2costheta +2=0$ we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation. Any help would be appreciated.
trigonometry polynomials complex-numbers
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
$endgroup$
add a comment |
$begingroup$
Find all 4 complex roots to $3z^4-z^3+2z^2-z+3=0$ using a trig substitution
A previous part of the question made us prove that $2cos ntheta=z^n+z^{-n}$ and it wants us to prove that the polynomial can be expressed as $6cos 2theta -2costheta +2=0$. I'm assuming that using $6cos 2theta -2costheta +2=0$ we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation. Any help would be appreciated.
trigonometry polynomials complex-numbers
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
$endgroup$
Find all 4 complex roots to $3z^4-z^3+2z^2-z+3=0$ using a trig substitution
A previous part of the question made us prove that $2cos ntheta=z^n+z^{-n}$ and it wants us to prove that the polynomial can be expressed as $6cos 2theta -2costheta +2=0$. I'm assuming that using $6cos 2theta -2costheta +2=0$ we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation. Any help would be appreciated.
trigonometry polynomials complex-numbers
trigonometry polynomials complex-numbers
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
edited Jan 13 at 16:16
H.Linkhorn
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
asked Jan 13 at 13:47
H.LinkhornH.Linkhorn
366113
366113
add a comment |
add a comment |
2 Answers
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Hint: The given equation is equivalent to
$$
3(z^2+frac{1}{z^2})-(z+frac{1}{z})+2=0.
$$ By letting $z+frac{1}{z}=w$, we get
$$
3(w^2-2)-w+2=0.
$$ Solve this quadratic equation for $w$ and subsequently solve $z+frac{1}{z}=w$. Since $w$ is real and $|w|le 2$, we can substitute $z=e^{itheta}$, $w=2cos theta$ and solve for $theta$.
answered Jan 13 at 16:20
SongSong
11.1k628
$endgroup$
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
add a comment |
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Use that $$cos(2x)=2cos^2(x)-1$$
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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active
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votes
$begingroup$
Hint: The given equation is equivalent to
$$
3(z^2+frac{1}{z^2})-(z+frac{1}{z})+2=0.
$$ By letting $z+frac{1}{z}=w$, we get
$$
3(w^2-2)-w+2=0.
$$ Solve this quadratic equation for $w$ and subsequently solve $z+frac{1}{z}=w$. Since $w$ is real and $|w|le 2$, we can substitute $z=e^{itheta}$, $w=2cos theta$ and solve for $theta$.
answered Jan 13 at 16:20
SongSong
11.1k628
$endgroup$
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
add a comment |
$begingroup$
Hint: The given equation is equivalent to
$$
3(z^2+frac{1}{z^2})-(z+frac{1}{z})+2=0.
$$ By letting $z+frac{1}{z}=w$, we get
$$
3(w^2-2)-w+2=0.
$$ Solve this quadratic equation for $w$ and subsequently solve $z+frac{1}{z}=w$. Since $w$ is real and $|w|le 2$, we can substitute $z=e^{itheta}$, $w=2cos theta$ and solve for $theta$.
answered Jan 13 at 16:20
SongSong
11.1k628
$endgroup$
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
add a comment |
$begingroup$
Hint: The given equation is equivalent to
$$
3(z^2+frac{1}{z^2})-(z+frac{1}{z})+2=0.
$$ By letting $z+frac{1}{z}=w$, we get
$$
3(w^2-2)-w+2=0.
$$ Solve this quadratic equation for $w$ and subsequently solve $z+frac{1}{z}=w$. Since $w$ is real and $|w|le 2$, we can substitute $z=e^{itheta}$, $w=2cos theta$ and solve for $theta$.
answered Jan 13 at 16:20
SongSong
11.1k628
$endgroup$
Hint: The given equation is equivalent to
$$
3(z^2+frac{1}{z^2})-(z+frac{1}{z})+2=0.
$$ By letting $z+frac{1}{z}=w$, we get
$$
3(w^2-2)-w+2=0.
$$ Solve this quadratic equation for $w$ and subsequently solve $z+frac{1}{z}=w$. Since $w$ is real and $|w|le 2$, we can substitute $z=e^{itheta}$, $w=2cos theta$ and solve for $theta$.
answered Jan 13 at 16:20
SongSong
11.1k628
edited Jan 13 at 16:26
answered Jan 13 at 16:20
SongSong
11.1k628
answered Jan 13 at 16:20
SongSong
11.1k628
answered Jan 13 at 16:20
SongSong
11.1k628
11.1k628
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
add a comment |
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
$begingroup$
Such polynomials are called "reciprocal" or "palindromic" ; see en.wikipedia.org/wiki/Reciprocal_polynomial
$endgroup$
– Jean Marie
Jan 15 at 1:02
add a comment |
$begingroup$
Use that $$cos(2x)=2cos^2(x)-1$$
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
add a comment |
$begingroup$
Use that $$cos(2x)=2cos^2(x)-1$$
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
add a comment |
$begingroup$
Use that $$cos(2x)=2cos^2(x)-1$$
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
Use that $$cos(2x)=2cos^2(x)-1$$
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Jan 13 at 13:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
74.6k42865
add a comment |
add a comment |
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