The Real And Complex Number Systems

So, let z = 1, we obtain that
n−1
∏ (
n =
) n−1
∏
[(
1 − e
2πik/n
= 1 − cos 2πk
n
k=1
n−1
∏
=
k=1
n−1
∏
=
k=1
n−1
∏
= 2 n−1
(
2 sin 2 πk
n
(
2 sin πk
n
k=1
n−1
∏
= 2 n−1
=
[
k=1
∏
2 n−1 n−1
k=1
n−1
∏
= 2 n−1
k=1
)
− i
(
sin πk
n
(
sin πk
n
k=1
) (
sin πk
n
(
2 sin πk πk
cos
n n
)
) (
cos
)
e i( 3π 2 + πk
n )
(
sin πk ) ] e ∑ n−1 3π
k=1 2 + πk
n
n
(
sin πk
n
)
.
)
)
πk
− i cos
n
( 3π
2 + πk )
+ i sin
n
(
− i sin 2πk )]
n
( 3π
2 + πk ))
n
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