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I'm trying to understand why there are $n$ amount of roots in an equation of the form $z^n=$ complex equations.

I understand why there are several answers to a $ \sin(x)=$ equation but I can't wrap my head around the complex one.

Please don't answer with a big load of algebraic clutter like my book does. Try and answer with some reasoning if possible. Thank you.

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1  
Let $\omega_n = \exp \frac{2\pi i}{n}$. Then $\omega_n^n = 1$, so for all $z$ you have $z^n = (\omega_n z)^n$. –  Daniel Fischer Jan 25 at 18:46
    
Note that the equation $x^2=1$ has 2 solutions, 1 and -1. if you look at the solutions of $z^3=1$, you will find 1 real solution : 1. there are 2 otherswhich are "hidden" away from the real axis, and it is $e^{2i\pi/3}$ and $e^{-2i\pi/3}$ –  Glougloubarbaki Jan 25 at 18:51
    
...That's just a random fact for me, not an answer. –  Paze Jan 25 at 18:52

2 Answers 2

up vote 3 down vote accepted

In the complex plane, the function

$$ z \to z^n$$

has a simple geometric interpretation. A complex number is specified by two pieces of data. A distance $r$ from the origin, and angle $\theta$ relative to the horizontal.

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The function $z \to z^n$ sends a pair $(r,\theta)$ to the pair $(r^n, n\theta)$. In plain English, it raises the distance to the power $n$ and multiplies the angle by $n$. What is important to note here is that the angle takes values between $0$ and $2\pi$ and then loops back to $0$. To see why this gives rise to multiple roots, consider the equation

$$z^2 = 1$$

In our geometric understanding, this is asking: what complex numbers $(r,\theta)$ exist such that $r^2 = 1$ and $2\theta = 0$? Since $r$ is assumed positive, we must have $r=1$. But what about the angle? Clearly we can let $\theta = 0$, which corresponds to the solution $z=1$. But we can also let $\theta = \pi$, because if we double the angle $\pi$ we return to the angle $2\pi \equiv 0$. This corresponds to the solution $z=-1$. So the multiplicity of roots to the equation $z^2 = 1$ emerges from the fact that the angle resets every $2\pi$, and that $z \to z^2$ is composed of a stretching and a rotation. The same reasoning explains why $z^n = 1$ has multiple roots.

More generally, a polynomial like $z^8 - 3z^2 + 4$ can be thought of as describing some complicated geometric operation composed of stretching, translation and rotating. It is not then unreasonable to propose that there are multiple ways to arrive at the same destination via this operation.

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I'm assuming that you are talking about polynomial equations of order $n$. The idea here is that any polynomial (in the reals or complex plane) have at most $n$ solutions. This is because if $P(x)$ is a polynomial, and $c$ is a root of the polynomial then we can factor out this root like:

$$ P(x) = (x - c)Q(x) $$

This is true for both real and complex polynomials. Now in the real case you may have less than $n$ solutions. An example of this is the polynomial $x^2 + 1 = 0$ has no real solutions. But it does have a complex solutions, namely $x = \pm i$.

The big theorem here is that in the complex field, every polynomial has at least one solution. That is, there are no examples like $x^2 + 1 = 0$ for the complex field like there are for the real field. And what you can do is just keep factoring out solutions until you no longer have a polynomial, like:

$$ P(x) = (x - r_1)Q_1(x) = (x - r_1)(x - r_2)Q_2(x) = \cdots = (x - r_1)(x - r_2) \cdots (x - r_n) $$

I hope this is not too algebraic of an answer for you and I hope it kind of answers your question.

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