If $1, z_{1}, z_{2}, z_{3}, . . ., z_{n - 1}$ are $n$ th roots of unity, then show that $(1 - z_{1}) (1 - z_{2}) . . . (1 - z_{n - 1}) = n$ .

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Solution

$z^{n} - 1 = (z - 1) (z - z_{1}) (z - z_{2}) . . . (z - z_{n - 1})$
or $\frac{z^{n} - 1}{z - 1} = (z - z_{1}) (z - z_{2}) . . . (z - z_{n - 1})$
or $1 + z + z^{2} + . . . + z^{n - 1} = (z - z_{1}) (z - z_{2}) . . . (z - z_{n - 1})$
Putting $z = 1$ , we get
$(1 - z_{1}) (1 - z_{2}) . . . (1 - z_{n - 1}) = 1 + 1 + . . . + 1 = n$
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