# Necessary and sufficient x^3=1

So we have ##x^3=1##. I have a really simple question. Why isn't it true that ##x^3 = 1## if and only if ##x = 1##, when we consider that if we take cube root of both sides and we can then take the cube again? Of course there are two other complex roots, but what am I missing in my naive logical argument?

andrewkirk
Homework Helper
Gold Member
The cube root is not a function, as it is a one-to-many mapping, in fact one-to-three. So taking the cube root of both sides gives three separate equations, not one: ##x=1,\ x=\omega## and ##x=\omega^2##, where ##\omega=e^{2\pi/3}##.

Expressed in formal logic, we have
$$x^3=1 \Leftrightarrow (x=1\vee x=\omega\vee x=\omega^2)$$

Given the properties of ##\vee## (OR), this allows us to deduce that
$$x=1\Rightarrow x^3=1$$
but not that
$$x^3=1 \Rightarrow x=1$$

The cube root is not a function, as it is a one-to-many mapping, in fact one-to-three. So taking the cube root of both sides gives three separate equations, not one: ##x=1,\ x=\omega## and ##x=\omega^2##, where ##\omega=e^{2\pi/3}##.

Expressed in formal logic, we have
$$x^3=1 \Leftrightarrow (x=1\vee x=\omega\vee x=\omega^2)$$

Given the properties of ##\vee## (OR), this allows us to deduce that
$$x=1\Rightarrow x^3=1$$
but not that
$$x^3=1 \Rightarrow x=1$$
I think I see. So with square roots, ##+ \sqrt[2]{x}## is the principal square root which defines a function, while ##\pm \sqrt[2]{x}## is the value of both roots. So analogously ##\sqrt[3]{x}## is the principal cube root, while... What's the analogous way of retrieving the two complex roots easily?

lurflurf
Homework Helper
^the n solutions of $$x^n=y$$ are
##exp(2k \pi i /n)\sqrt[n]{y}##
for
k=0,1,...,n-2,n-1

thus the complex cube roots of a positive real y are
##exp(2 \pi i /n)\sqrt[n]{y}=\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\sqrt[n]{y}##
and
##exp(4 \pi i /n)\sqrt[n]{y}=\left(-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)\sqrt[n]{y}##

fresh_42
Mentor
What's the analogous way of retrieving the two complex roots easily?
Polynomials split over the complex numbers. So we have to look at complex multiplication. If we represent complex numbers as points in the plane, then multiplication means to multiply the real length of two numbers (= distance to the origin), and add there angles (= towards the real axis). This means in return, that numbers of length one are all on the unit circle, and multiplication of those is adding angles, i.e. in order to solve ##x^n=x \cdot \ldots \cdot x = 1##, we have to divide the unit circle in ##n## equal angles and all possible ##x## are those points on the circle which @lurflurf listed.