What tricks are there for calculating the roots of complex polynomials like
$p(t) = (t+1)^6 - (t-1)^6$
$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$\left( \frac{t+1}{t-1} \right)^6 = 1$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which brings us to
$\omega_k = e^{i \cdot k \cdot \frac{2 \pi}{6}}$
So now we need to get the values from t for $k = 0,...5$.
How to get the values of t from the following identity then?
$ \begin{align} \frac{t+1}{t-1} &= e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\ (t+1) &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\ 1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - t \\ 1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t \cdot (e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1) \\ \end{align} $
And now?
$ t = \frac{1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}}}{e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1} $
So I've got six roots for $k = 0,...5$ as follows
$ t = \frac{1+e^{i \cdot k \cdot \frac{2 \pi}{6}}}{e^{i \cdot k \cdot \frac{2 \pi}{6}}-1} $
Is this right? But how can it be that the bottom equals $0$ for $k=0$?
I don't exactly know how to simplify this:
$\frac{ \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} } + 1 }{ 1 - \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} }}$