Prove that if a complex number raised to nth power equals one, then product of the solutions to this euqations is one

Question

I've thought of this "proof", but it seems lame to me (I'm not sure it is even a proof), I probably should have used other properties of complex numbers to write one.

So,

$z^n=1 \implies \sqrt[n]{z}=\sqrt[n]{1} \implies \sqrt[n]{1}_1 \times \sqrt[n]{1}_2 \times ... \times \sqrt[n]{1}_n = (\sqrt[n]{1})^n = 1$

And we know there are $n$ terms in the product because it is given by the fundamental theorem of algebra (which says that a polynomial of power $n$ has $n$ solutions). And the next step is given by the fact that the product of powers of $x$ is $x$ raised to the power of the sum of the powers of the factors.

This is obviously a homework, so, please don't assume much when replying! :)

Answer 1

No proof is possible. The conjecture is false.

Suppose $a_i$, $i = 1, ..., n$ are the solutions to $z^n - 1 = 0$. Then

$$z^n - 1 = (z - a_1)(z - a_2)...(z-a_n)$$

When you multiply out the brackets, you get a polynomial in $z$ with constant term $$(-1)^n a_1 ... a_n$$

So, by equating coefficients, we have

$$a_1 ... a_n = (-1)^{n+1}$$

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Possibly, the question is asking for the product of the non-real solutions to this equation.

• If $n$ is odd, the only real solution is $1$, so the product of the non-real solutions is $(-1)^{n+1} = 1$
• If $n$ is even, $+1$ and $-1$ are both solutions, so the product of the non-real solutions is $(-1)^{n+1} / (-1) = 1$ again.