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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:

$\frac{ \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} } + 1 }{ 1 - \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} }}$

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    Yes, it is right. And divide top and bottom by $e^{\pi i k/6}$. On top you get $2\cos(k\pi/6)$. On the bottom you get $2i\sin(k\pi/6)$. The answers simplify to $-i\cot(k\pi/6)$.2012-02-02

2 Answers 2

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Notice that $t=1$ is not a root. Divide by $(t-1)^6$.

If $\omega$ is a root of $z^6 - 1$, then a root of the original equation is given by $\frac{t+1}{t-1} = \omega$.

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    @meinzlein: Yes, on exam/homework, the safe thing is to restrict, the grader may not be happy with the infinite root. Division by $0$ is frowned on. If you *imagine* expanding $(t+1)^6$ and $(t-1)^6$, note that the $t^6$ terms cancel, so you are really finding the roots of a polynomial of degree $5$, and therefore the number of roots (counting multiplicity, though here there are no multiple roots) is $5$.2012-02-03
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Note that $(t+1)^6 - (t-1)^6=((t+1)^3-(t-1)^3)((t+1)^3+(t-1)^3)$ (difference of squares).

When you simplify the first term in the product on the right, there is no $t^3$ term and no $t$ term! The second term in the product simplifies to $2t^3+6t$.

Remark: The solution by Arhabhata is the right one, it works if we replace $6$ by $n$. And when we set $\frac{t-1}{t+1}=e^{2\pi i k/n}$, where $k=1,2,\dots,n-1$, and solve for $t$, we get $-i$ times cotangents.