What is meant by 'root' in complex number trigonometry?

In complex number trigonometry, there exists a theorem called De Moivre's theorem, which states that for any complex number, $Z$, in trigonometric form, $r(cos(\theta)+i\cdot sin(\theta))$, the following rule applies:

$Z^n=r^n(cos(n\theta)+i\cdot sin(n\theta))$

and from this (and here's the question) it follows that:

$r^{\frac{1}{n}}(cos(\frac{\theta}{n})+i\cdot sin(\frac{\theta}{n}))$

is an $n^{th}$ root of Z, and $Z = r(cos(\theta)+i\cdot sin(\theta))$

the question is, what is meant by a 'root' in this context and why is this information useful or what can it be used for? Obviously when graphing functions, at a very elementary level at least, the root implies where the graph crosses the x-axis, but in this context I can't quite figure it out. Any help is appreciated.

• In algebra, it is important to solve polynomial equations of the form $p(x)=0$, whose solutions are also called "roots" of the polynomial $p(x)$. Examples include $ax^2+bx+c=0$, whose roots are given by the quadratic formula. Solutions to the equation $x^n=0$ are given by the formula that you gave, and they are called "$n^{\text{th}}$ roots", generalizing "square roots" and "cube roots". – Lee Mosher Aug 9 '17 at 22:23
• It's a bit surprising that you are familiar with the terminology for roots of functions, but not with the terminology for square roots, cube roots etc. of numbers. In any case, the $n$-th root of a number $Z$ is defined to be a solution to $x^n = Z$, i.e., a root of the function $x^n - Z$. So a square root or $2$-nd root of $Z$ is a solution to $x^2 - Z = 0$. – Rob Arthan Aug 9 '17 at 22:43

A $n^{\text{nh}}$ root of a complex number $z$ is a complex number $w$ such that $w^n=z$. So, if $z=r\bigl(\cos(\theta)+i\sin(\theta)\bigr)$ and you want a $n^{\text{nh}}$ of $z$, it follows from de Moivre's formula that all you need is to take $w=\sqrt[n]r\bigl(\cos\bigl(\frac\theta n\bigr)+i\sin\bigl(\frac\theta n\bigr)\bigr)$. More generally, you can take$$w=\sqrt[n]r\left(\cos\left(\frac{\theta+2k\pi}n\right)+i\sin\left(\frac{\theta+2k\pi}n\right)\right)$$ ($k\in\mathbb Z$).
The $n^{th}$ root of a complex number is much like the $n^{th}$ root of a real number. If a complex number $w$ multiplied by itself $n$ times gives $Z$, then $w$ is a $n^{th}$ root of $Z$.
• You could do more to reconcile the two uses of "root" by explaining what polynomial has "roots" that correspond to (say) the $n$th "root" (in the sense of a radical) of a real (or a complex) number. – hardmath Aug 9 '17 at 22:33