Question

If $1,{\alpha }_1,{\alpha }_2,{\alpha }_3,...,{\alpha }_{n-1}$ are the nth roots of unity, then the value of $1.{\alpha }_1.{\alpha }_2.{\alpha }_3...{\alpha }_{n-1}$ is

Answer 1

For any complex number
$z=R.e^{i(2k\pi +\theta )}$

$z^{\frac{1}{n}}=R^{\frac{1}{n}}.e^{i\frac{2k\pi +\theta }{n}}$

$=r.e^{i\frac{2k\pi }{n}}.e^{i\frac{\theta }{n}}$

Where $k=1,2,\mathrm{3..}n$

Thus
$z^{\frac{1}{n}}=\left(r{e}^{i\frac{\theta }{n}}\right).e^{i\frac{2k\pi }{n}}$

$z^{\frac{1}{n}}=z_1.e^{i\frac{2k\pi }{n}}$ where $k=1,2,\mathrm{3..}n...\left(i\right)$
Hence
$z^{\frac{1}{n}}$ has exactly n roots.

Now for the roots of $z^n=1$

We substitute $z_1=1$

Hence
$z^{\frac{1}{n}}=e^{i\frac{2k\pi }{n}}$ where $k=1,2,\mathrm{3..}n$

Hence
$z_1.z_2....z_{n-1}.z_n$ where $z_n=e^{\frac{i2n\pi }{n}}=1$

$=e^{i\frac{2\pi }{n}}.e^{i\frac{4\pi }{n}}.e^{i\frac{6\pi }{n}}....e^{i\frac{2(n-1)\pi }{n}}.e^{\frac{i2n\pi }{n}}$

$=e^{i\frac{2\pi }{n}(1+2+...n)}$

$=e^{i\frac{2\pi }{n}\left(\frac{n(n+1)}{2}\right)}$

$=e^{i\pi (n+1)}$

$=1$ if n is odd.
Now if n is even, then $-1$ is also a root.
Hence $-e^{i\pi (n+1)}$
$=1$