# roots of complex polynomial - tricks

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 $-i\cot(k\pi/6)$. – André Nicolas Feb 2 '12 at 23:57

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 $\frac{t+1}{t-1}=\omega_1$? – meinzlein Feb 2 '12 at 23:24
Try multiplying by $t-1$... – Aryabhata Feb 2 '12 at 23:32
I tried and edited the original post accordingly.. – meinzlein Feb 2 '12 at 23:42
Looks right.... – Aryabhata Feb 2 '12 at 23:54
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? – meinzlein Feb 2 '12 at 23:58

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.

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