Substitution for solving $z^n-1=0$

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.