Roots of $z^4 - 3z^2 + 1 = 0$.












0












$begingroup$



Show that the solutions of $z^4 - 3z^2 + 1 = 0$ are given by $$z = 2cos 36^{defo{mathrm{o}}o}, 2cos 72^o, 2cos 216^o, 2cos 252^o.$$ Hence show that




  • $cos 36^o = frac 14(sqrt{5}+1)$

  • $cos 72^o = frac 14(sqrt{5}-1)$




I can solve for the roots, however I could not express them as the question wanted. Is there a specific way to express the root in kinda polar form without the imaginary part.






complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2cos{36^{circ}} qquad Rightarrow qquad Im(z) = 2sin{36^{circ}} $$
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:20












  • $begingroup$
    The question clearly states that all roots are real. Why would you need a polar form ?
    $endgroup$
    – Yves Daoust
    Dec 20 '18 at 8:23










  • $begingroup$
    Is your problem about showing $z=2cos 36^circ$ etc are roots of the quartic, or about expressing roots of the quartic as $pm(1pmsqrt5)/4$ (independent $pm$s)?
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:33










  • $begingroup$
    Do you need to solve the equation? Maybe you can set $s = z^2$ .
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:38
















0












$begingroup$



Show that the solutions of $z^4 - 3z^2 + 1 = 0$ are given by $$z = 2cos 36^{defo{mathrm{o}}o}, 2cos 72^o, 2cos 216^o, 2cos 252^o.$$ Hence show that




  • $cos 36^o = frac 14(sqrt{5}+1)$

  • $cos 72^o = frac 14(sqrt{5}-1)$




I can solve for the roots, however I could not express them as the question wanted. Is there a specific way to express the root in kinda polar form without the imaginary part.






complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2cos{36^{circ}} qquad Rightarrow qquad Im(z) = 2sin{36^{circ}} $$
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:20












  • $begingroup$
    The question clearly states that all roots are real. Why would you need a polar form ?
    $endgroup$
    – Yves Daoust
    Dec 20 '18 at 8:23










  • $begingroup$
    Is your problem about showing $z=2cos 36^circ$ etc are roots of the quartic, or about expressing roots of the quartic as $pm(1pmsqrt5)/4$ (independent $pm$s)?
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:33










  • $begingroup$
    Do you need to solve the equation? Maybe you can set $s = z^2$ .
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:38














0












0








0





$begingroup$



Show that the solutions of $z^4 - 3z^2 + 1 = 0$ are given by $$z = 2cos 36^{defo{mathrm{o}}o}, 2cos 72^o, 2cos 216^o, 2cos 252^o.$$ Hence show that




  • $cos 36^o = frac 14(sqrt{5}+1)$

  • $cos 72^o = frac 14(sqrt{5}-1)$




I can solve for the roots, however I could not express them as the question wanted. Is there a specific way to express the root in kinda polar form without the imaginary part.






complex-numbers roots







share|cite|improve this question











$endgroup$





Show that the solutions of $z^4 - 3z^2 + 1 = 0$ are given by $$z = 2cos 36^{defo{mathrm{o}}o}, 2cos 72^o, 2cos 216^o, 2cos 252^o.$$ Hence show that




  • $cos 36^o = frac 14(sqrt{5}+1)$

  • $cos 72^o = frac 14(sqrt{5}-1)$




I can solve for the roots, however I could not express them as the question wanted. Is there a specific way to express the root in kinda polar form without the imaginary part.






complex-numbers roots




complex-numbers roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 13:42









Brahadeesh

6,42442363




6,42442363










asked Dec 20 '18 at 8:15









AbecAbec

145




145












  • $begingroup$
    I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2cos{36^{circ}} qquad Rightarrow qquad Im(z) = 2sin{36^{circ}} $$
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:20












  • $begingroup$
    The question clearly states that all roots are real. Why would you need a polar form ?
    $endgroup$
    – Yves Daoust
    Dec 20 '18 at 8:23










  • $begingroup$
    Is your problem about showing $z=2cos 36^circ$ etc are roots of the quartic, or about expressing roots of the quartic as $pm(1pmsqrt5)/4$ (independent $pm$s)?
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:33










  • $begingroup$
    Do you need to solve the equation? Maybe you can set $s = z^2$ .
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:38


















  • $begingroup$
    I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2cos{36^{circ}} qquad Rightarrow qquad Im(z) = 2sin{36^{circ}} $$
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:20












  • $begingroup$
    The question clearly states that all roots are real. Why would you need a polar form ?
    $endgroup$
    – Yves Daoust
    Dec 20 '18 at 8:23










  • $begingroup$
    Is your problem about showing $z=2cos 36^circ$ etc are roots of the quartic, or about expressing roots of the quartic as $pm(1pmsqrt5)/4$ (independent $pm$s)?
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:33










  • $begingroup$
    Do you need to solve the equation? Maybe you can set $s = z^2$ .
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:38
















$begingroup$
I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2cos{36^{circ}} qquad Rightarrow qquad Im(z) = 2sin{36^{circ}} $$
$endgroup$
– Matti P.
Dec 20 '18 at 8:20






$begingroup$
I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2cos{36^{circ}} qquad Rightarrow qquad Im(z) = 2sin{36^{circ}} $$
$endgroup$
– Matti P.
Dec 20 '18 at 8:20














$begingroup$
The question clearly states that all roots are real. Why would you need a polar form ?
$endgroup$
– Yves Daoust
Dec 20 '18 at 8:23




$begingroup$
The question clearly states that all roots are real. Why would you need a polar form ?
$endgroup$
– Yves Daoust
Dec 20 '18 at 8:23












$begingroup$
Is your problem about showing $z=2cos 36^circ$ etc are roots of the quartic, or about expressing roots of the quartic as $pm(1pmsqrt5)/4$ (independent $pm$s)?
$endgroup$
– user10354138
Dec 20 '18 at 8:33




$begingroup$
Is your problem about showing $z=2cos 36^circ$ etc are roots of the quartic, or about expressing roots of the quartic as $pm(1pmsqrt5)/4$ (independent $pm$s)?
$endgroup$
– user10354138
Dec 20 '18 at 8:33












$begingroup$
Do you need to solve the equation? Maybe you can set $s = z^2$ .
$endgroup$
– Matti P.
Dec 20 '18 at 8:38




$begingroup$
Do you need to solve the equation? Maybe you can set $s = z^2$ .
$endgroup$
– Matti P.
Dec 20 '18 at 8:38










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $w=e^{itheta}$. Try to search for $z=w+bar w=w+frac{1}{w}$. Setting it in the equation gives
$$
w^8+w^6+w^4+w^2+1=frac{w^{10}-1}{w^2-1}=0.
$$
Now you need to pick those $w$ that make $w^{10}=1$ and $w^2ne 1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
    $endgroup$
    – Abec
    Dec 21 '18 at 12:14










  • $begingroup$
    I agree, it is brilliant (+1)
    $endgroup$
    – user376343
    Dec 26 '18 at 12:02





















1












$begingroup$

I would solve the equation $$z^4-3z^2+1=0$$ by substituting $$z^2=t$$ we get
$$t^2-3t+1=0$$ and by the quadratic formula we obtain the solutions as
$$t_{1,2}=frac{3}{2}pmsqrt{frac{9}{4}-1}$$
Can you finish?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how do you get the trigonometric expressions for the roots?
    $endgroup$
    – Ovi
    Dec 21 '18 at 15:45










  • $begingroup$
    see here math-only-math.com/exact-value-of-cos-36-degree.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 21 '18 at 16:06











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let $w=e^{itheta}$. Try to search for $z=w+bar w=w+frac{1}{w}$. Setting it in the equation gives
$$
w^8+w^6+w^4+w^2+1=frac{w^{10}-1}{w^2-1}=0.
$$
Now you need to pick those $w$ that make $w^{10}=1$ and $w^2ne 1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
    $endgroup$
    – Abec
    Dec 21 '18 at 12:14










  • $begingroup$
    I agree, it is brilliant (+1)
    $endgroup$
    – user376343
    Dec 26 '18 at 12:02


















2












$begingroup$

Let $w=e^{itheta}$. Try to search for $z=w+bar w=w+frac{1}{w}$. Setting it in the equation gives
$$
w^8+w^6+w^4+w^2+1=frac{w^{10}-1}{w^2-1}=0.
$$
Now you need to pick those $w$ that make $w^{10}=1$ and $w^2ne 1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
    $endgroup$
    – Abec
    Dec 21 '18 at 12:14










  • $begingroup$
    I agree, it is brilliant (+1)
    $endgroup$
    – user376343
    Dec 26 '18 at 12:02
















2












2








2





$begingroup$

Let $w=e^{itheta}$. Try to search for $z=w+bar w=w+frac{1}{w}$. Setting it in the equation gives
$$
w^8+w^6+w^4+w^2+1=frac{w^{10}-1}{w^2-1}=0.
$$
Now you need to pick those $w$ that make $w^{10}=1$ and $w^2ne 1$.






share|cite|improve this answer









$endgroup$



Let $w=e^{itheta}$. Try to search for $z=w+bar w=w+frac{1}{w}$. Setting it in the equation gives
$$
w^8+w^6+w^4+w^2+1=frac{w^{10}-1}{w^2-1}=0.
$$
Now you need to pick those $w$ that make $w^{10}=1$ and $w^2ne 1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 10:20









A.Γ.A.Γ.

22.8k32656




22.8k32656












  • $begingroup$
    Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
    $endgroup$
    – Abec
    Dec 21 '18 at 12:14










  • $begingroup$
    I agree, it is brilliant (+1)
    $endgroup$
    – user376343
    Dec 26 '18 at 12:02




















  • $begingroup$
    Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
    $endgroup$
    – Abec
    Dec 21 '18 at 12:14










  • $begingroup$
    I agree, it is brilliant (+1)
    $endgroup$
    – user376343
    Dec 26 '18 at 12:02


















$begingroup$
Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
$endgroup$
– Abec
Dec 21 '18 at 12:14




$begingroup$
Thanks for the guide, i nvr thought of using conjugate to get 2Re(w) and use it for z. It is brilliant.
$endgroup$
– Abec
Dec 21 '18 at 12:14












$begingroup$
I agree, it is brilliant (+1)
$endgroup$
– user376343
Dec 26 '18 at 12:02






$begingroup$
I agree, it is brilliant (+1)
$endgroup$
– user376343
Dec 26 '18 at 12:02













1












$begingroup$

I would solve the equation $$z^4-3z^2+1=0$$ by substituting $$z^2=t$$ we get
$$t^2-3t+1=0$$ and by the quadratic formula we obtain the solutions as
$$t_{1,2}=frac{3}{2}pmsqrt{frac{9}{4}-1}$$
Can you finish?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how do you get the trigonometric expressions for the roots?
    $endgroup$
    – Ovi
    Dec 21 '18 at 15:45










  • $begingroup$
    see here math-only-math.com/exact-value-of-cos-36-degree.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 21 '18 at 16:06
















1












$begingroup$

I would solve the equation $$z^4-3z^2+1=0$$ by substituting $$z^2=t$$ we get
$$t^2-3t+1=0$$ and by the quadratic formula we obtain the solutions as
$$t_{1,2}=frac{3}{2}pmsqrt{frac{9}{4}-1}$$
Can you finish?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how do you get the trigonometric expressions for the roots?
    $endgroup$
    – Ovi
    Dec 21 '18 at 15:45










  • $begingroup$
    see here math-only-math.com/exact-value-of-cos-36-degree.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 21 '18 at 16:06














1












1








1





$begingroup$

I would solve the equation $$z^4-3z^2+1=0$$ by substituting $$z^2=t$$ we get
$$t^2-3t+1=0$$ and by the quadratic formula we obtain the solutions as
$$t_{1,2}=frac{3}{2}pmsqrt{frac{9}{4}-1}$$
Can you finish?






share|cite|improve this answer









$endgroup$



I would solve the equation $$z^4-3z^2+1=0$$ by substituting $$z^2=t$$ we get
$$t^2-3t+1=0$$ and by the quadratic formula we obtain the solutions as
$$t_{1,2}=frac{3}{2}pmsqrt{frac{9}{4}-1}$$
Can you finish?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 9:55









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.4k42866




76.4k42866












  • $begingroup$
    But how do you get the trigonometric expressions for the roots?
    $endgroup$
    – Ovi
    Dec 21 '18 at 15:45










  • $begingroup$
    see here math-only-math.com/exact-value-of-cos-36-degree.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 21 '18 at 16:06


















  • $begingroup$
    But how do you get the trigonometric expressions for the roots?
    $endgroup$
    – Ovi
    Dec 21 '18 at 15:45










  • $begingroup$
    see here math-only-math.com/exact-value-of-cos-36-degree.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 21 '18 at 16:06
















$begingroup$
But how do you get the trigonometric expressions for the roots?
$endgroup$
– Ovi
Dec 21 '18 at 15:45




$begingroup$
But how do you get the trigonometric expressions for the roots?
$endgroup$
– Ovi
Dec 21 '18 at 15:45












$begingroup$
see here math-only-math.com/exact-value-of-cos-36-degree.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 21 '18 at 16:06




$begingroup$
see here math-only-math.com/exact-value-of-cos-36-degree.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 21 '18 at 16:06


















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