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If $z^6=1$ ,find all the possible values of $z$.

My Attempt: $$z^6=1$$ $$z^6-1=0$$ $$(z^3+1)(z^3-1)=0$$

Now, what should I do?

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    $z^3+1=(z+1)(z^2-z+1)$ and $z^3-1=(z-1)(z^2+z+1)$. You can solve quadratics by known formula – 2017-02-07
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    Is $z$ a complex number? – 2017-02-07
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    Alternatively, if you know that we can write any complex number $z$ as $z = r e^{i\theta}$ for some real $r, \theta$ (this is the *polar form*), we can substitute to give $r^6 e^{6 i \theta} = 1$. Take the norm of both sides to determine a restriction on $r$, then substitute to produce an equation in $\theta$. – 2017-02-07

5 Answers 5

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Use De Moivre's theorem.Let z=$ e^{i \Theta}$. Then $z^6$=1 implies $e^{6* i \Theta}=1$.Then $(\cos \Theta + i \sin \Theta)=1=\cos 0+ i \sin 0$. So comparing real and imaginary parts,we get that $6* \Theta=2*n* \pi$,where n=0,1,2,...5. We take n such because the equation is of degree 6 and thus,has 6 roots.

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The point of this question is probably to look at the so called roots of unity (http://artofproblemsolving.com/wiki/index.php/Roots_of_unity) This comes up in algebra quite a bit and they are quite interesting.

The solutions of $x^n=1$ are given by the formula

$$ cos(\frac{2\pi k}{n})+i\sin(\frac {2\pi k}{n}). $$ For $k=0,1,...,n-1$

Also it is importaint to note that when $x^n$ has root x=1 for n=0.

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Assuming $z \in \mathbb{C}$, $$(z^3+1)(z^3-1)=0$$ $$\implies (z+1)(z^2-z+1)(z-1)(z^2+z+1)=0$$

So we have $2$ real roots $z=\pm1$ and $2$ pairs of complex conjugate roots, i.e. $z=\frac{1\pm i\sqrt{3}}{2}$ and $z=\frac{-1\pm i\sqrt{3}}{2}$, obtained on solving the $2$ quadratic equations.

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$(z-1)(z^2 + z +1)(z+1)(z^2 + z +1) = 0 $. So you get 1 ,-1 as roots .Now solve both the quadratic equations and using the formula of roots , you get all the four other roots. Basically , these are known as 6 th roots of unity.

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Well find the other complex roots. Use Moivre theorem: $z^n = r^n(cos(n\alpha) + isin(n\alpha))$, where $r$ is the "magnitude"... $r = |z| = {\sqrt {x^2+y^2}}$ if you know how to work with the complex numbers.