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In solving the equation $z^7-1=0$, the obvious route is to get the root $z=1$. The next step is to solve : $(z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=0$. Now, it is difficult for me to solve, and lack of experience is the key. However, have found out that an approach given is to have $a=z+1/z$ to get $a^3+a^2-2a-1=0$. I want to ask why on what basis this approach is made possible. If there were any geometrical reason also, then would be much better. All I could figure out was that the equation obtained in $a,z$ is quadratic, as : $z^2-az+1=0$, with $a$ being the coefficient of the linear term. Second question is for even power of initial expression, i.e. $z^6-1=0$ it would lead to $(z-1)(z^5+z^4+z^3+z^2+z+1)=0$. What can be suitable substitution now? And again what is the algebraic (& if possible also the geometric) reasoning for such substitution.

Also, would request a suitable source to see such substitution text, with either algebraic or geometric reasoning. I hope that Chrystal should serve for algebraic part, but could not find there.

jiten
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    For the first one, see reciprocal polynomials and in particular property #10. For the second one, factor as $,(z^2-1)(z^4+z^2+1),$ instead. – dxiv Mar 31 '18 at 21:16
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    In the first case $z=e^{2\pi i/7}$ is a root of that sextic. Therefore $a=z+1/z=2\cos(2\pi/7)$. The other roots of that cubic are $2\cos(4\pi/7)$ and $2\cos(8\pi/7)$. One reason to look for the minimal polynomial of $a$ comes from Galois theory. $a$ is fixed by complex conjugation (generating a subgroup of index three), so it is a zero of a cubic with rational coefficients. – Jyrki Lahtonen Mar 31 '18 at 21:19
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    The polynomial has real coefficients, so all non-real roots come in complex conjugate pairs. In this case, we know that all solutions have absolute value $1$, so complex conjugate is the same as inverse. So the substitution pairs them up so that you only get a third degree equation, exploiting that there are only three distinct real parts among the six solutions. – Arthur Mar 31 '18 at 21:19
  • Is it not enough that the roots of the equation are well-known? – user Mar 31 '18 at 21:24
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    @user : just calculate 7th roots of unity that will be your answer –  Mar 31 '18 at 21:25
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    Another way of looking at it is to observe that any palindromic polynomial $$p(z)=a_0z^{2n}+a_1z^{2n-1}+\cdots a_nz^n+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+\cdots+a_1z+a_0$$ can be written in the form $$p(z)=z^nq(z+\frac1z)$$ for some polynomial $q(z)$ of degree $n$. So if you can find the zeros of $q$, then you only need to solve a quadratic to find zeros of $p$. Here $p$ is simpler to handle, look up cyclotomic polynomials. – Jyrki Lahtonen Mar 31 '18 at 21:27
  • @user Please elaborate your approach. – jiten Mar 31 '18 at 21:32
  • @JyrkiLahtonen Please take both $z^7-1=0$ and $z^6-1=0$ to elaborate. If possible, please put an answer. – jiten Mar 31 '18 at 21:38
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    @JyrkiLahtonen Please help. Case 1: $z^7-1=0, => (z-1).p(z), p(z) = z^6+z^5+z^4+z^3+z^2+z+1= z^nq(z+\frac{1}{z})$. It is difficult to move further. – jiten Mar 31 '18 at 21:48
  • @user Please see chapter 1 of https://www.duo.uio.no/bitstream/handle/10852/45298/PiaLindstromMasteroppgave.pdf?sequence=1, to see why it is significant for me to delve into palindromic polynomials. – jiten Apr 01 '18 at 10:07
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    You accepted the answer I have hinted to. So I assume my guess was correct. – user Apr 01 '18 at 12:04
  • user -- I would have accepted more an answer that would have been on how the solution approach stated in the article by 'dxiv' for palindromic polynomials links to the complex roots' approach. Second, preference would have been one that simply details upon the solution of palindromic polynomials per-se, as by @JyrkiLahtonen . In face of no such answer, the third preference was selected. If some 'such' answer comes, the selection would change too. – jiten Apr 01 '18 at 12:27
  • @user Please see my last comment. – jiten Apr 01 '18 at 12:43
  • @JyrkiLahtonen Sorry for not being careful. Request name for the alternate approach that uses the fact that palindromic polynomial have even degree. So, for our example, $p(z)=(z^{2.3}+z^{2.3-1}+z^{2.3-2})+(z^3+z^{3-1}+z^{3-2})+1=(z^3+z^{3-1}+z^{3-2})(z^2+1)+1= z^3(1+\frac{1}{z}+\frac{1}{z^2})(z^2+1)+1= z^3(z^2+z+2+\frac{1}{z}+\frac{1}{z^2})+1=z^3q(z+\frac{1}{z})$. Need polynomial $q(z')$ of degree $n=3, z'=z+\frac{1}{z}$, but $(z^2+z+2+\frac{1}{z}+\frac{1}{z^2})$ not yields for any degree of $(z+\frac{1}{z})$. Also, additionally there is term of $a_0=1$. I hope you would help this time out. – jiten Apr 01 '18 at 14:31
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    In this case the trick with palindromic stuff leads exactly to the cubic you had. But, if you had instead, say, $p(z)=z^6+4z^5+2z^4+z^3+2z^2+4z+1$, then you could write $$p(z)=z^3[z^3+4z^2+2z+1+2z^{-1}+4z^{-2}+z^{-3}]=z^3[(z+\frac1z)^3+4(z+\frac1z)^2-(z+\frac1z)-7],$$ then find the roots of $u^3+4u^2-u-7$, and then solve for $z$ from $u=z+\dfrac1z$. When your equation is $z^n-1$. You don't want to do it like this. You just use the formula for complex roots of unity. That suggestion was for other similar occasions. – Jyrki Lahtonen Apr 01 '18 at 15:42

3 Answers3

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Using Euler's equation $$e^{ix}=\cos x+i\sin x$$ Setting $x=\pi$ givs us Euler's identity $$e^{i\pi}=-1$$ But $$e^{2i\pi k}=1,\forall k\in\mathbb Z$$

So $$\begin{align}z^7-1&=0\\z^7&=1\\z^7&=e^{2i\pi k}\\z&=e^{\frac 27i\pi k}\end{align}$$

Setting $k=0,1,2,3,4,5,6$ ends the solution

Awnon Bhowmik
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you can solve it using complex numbers

$z=(1)^{\frac{1}{7}}=e^{\frac{2k~i~\pi}{7}} $ where $k \in \{0,1,2.....6\}$

all the 7 roots can be be found by plugging values of k

and also these all roots will lie on the unit circle

ViktorStein
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Lets use the principles of the roots of unity. First of all,

$$ z^7 - 1 = 0 \iff z^7 = 1 = e^{i \cdot 2 k \pi} ~~\forall k \in \mathbb{Z} $$ We can than use that to obtain $$ z = (e^{i \cdot 2 k \pi})^{\frac{1}{7}} = e^{\frac{i \cdot 2k \pi}{7}} $$ We know that $k \in \{0,1,2,3,4,5,6\}$, because we can have a maximum of seven solutions. Substituting the different values of $k$ , we obtain all solutions.

ViktorStein
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