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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

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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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    So I've got the roots of $z^6 -1$ as $\omega_k = e^{i\cdot k\cdot\frac{2\pi}{6}}$. How to conclude now $t$ for example from $\frac{t+1}{t-1}=\omega_1$?2012-02-02
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    Try multiplying by $t-1$...2012-02-02
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    I tried and edited the original post accordingly..2012-02-02
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    Looks right....2012-02-02
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    Well a collegue of mine tried the way mentioned by André Nicolas and got only five roots: $$t_{1}=0$$ $$t_{2,3} = \pm \sqrt{3} i, $$ $$t_{4,5} = \pm \sqrt{\frac{1}{3}} i$$But I got 6 roots (see original post).. Where's the problem?2012-02-02
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    By the way. There's a problem in my solution: For $k=0$ the numerator is $0$..2012-02-03
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    @meinzlein: You mean the denominator is $0$. That gives the "infinite" root, which kind of makes sense in a complex setting. You do indeed get only $5$ ordinary roots.2012-02-03
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    Yes, denominator. So when writing a exam I should restrict $k = 1,...5$?2012-02-03
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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.