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I have a polynomial with coefficients that are integer polynomials in another (complex) variable. For example:

```
1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &
```

I want to solve for $v$ such that the second root of this polynomial is a root of unity. Notice that $v=0$ is such a solution:

```
In[67]:= Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2] /. {v -> 0}
Out[67]= 1
```

My two attempts to pull this information out of Mathematica have not worked. Namely,

```
In[68]:= Reduce[RootOfUnityQ[Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2]], {v}]
Out[68]= False
```

and

```
In[66]:= Solve[RootOfUnityQ[Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2]], {v}]
Out[66]= {}
```

both say that there are no solutions, which is not the answer that I am looking for.

# Question:

How do I solve for $v$ such that some expression is a root of unity?

Have you noticed that this function is

linearin $v^2$? This means that a very simple formula for $v$ can be found as the square root of a rational function of`#1`

. Replacing`#1`

by $\exp(2 k \pi i/n)$ for integral $k$ and nonzero integral $n$ finishes the job. – whuber – 2012-05-23T15:48:18.567