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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.

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    $\begingroup$ 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. $\endgroup$
    – dxiv
    Mar 31, 2018 at 21:16
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    $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Mar 31, 2018 at 21:19
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    $\begingroup$ 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. $\endgroup$
    – Arthur
    Mar 31, 2018 at 21:19
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    $\begingroup$ @user : just calculate 7th roots of unity that will be your answer $\endgroup$
    – user454960
    Mar 31, 2018 at 21:25
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    $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Mar 31, 2018 at 21:27

3 Answers 3

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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

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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.

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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

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