# Tricky Identity

1. Feb 8, 2014

### ggb123

I'd like to figure out why

$$f(x) = x^n - r e^{i \theta} = \prod_{j = 0}^{n - 1} ( x - r^{1/n} e^{i \theta / n} e^{2 \pi i j / n} )$$,

as I've seen it used as an identity in a few courses but I cannot figure it out. Could somebody shed some light? I understand they are the roots of the function, but I'd like some kind of analytic description.

Thanks a lot!

2. Feb 9, 2014

### dodo

Hi,
I think you may have answered your own question: if you are willing to accept that the RHS above can be rewritten as
$$\prod_{j = 0}^{n - 1} ( x - a_j )$$
where $a_j$ (with $j=0,1,...,n-1$) are the n-th roots of the complex number in polar form $r e^{i \theta}$, then your question can be rephrased as: "can somebody point me to a proof of the Fundamental Theorem of Algebra"?

My 2 cents.

Last edited: Feb 9, 2014
3. Feb 9, 2014

### AlephZero

You have $x^n = re^{i\theta}$, so taking the n'th root of each side you get
$$x = \sqrt[n]{re^{i\theta}}\,\omega_k, \quad k = 0,\dots,n-1$$
where the $\omega_k$ are the n'th roots of unity.