http://www.math.grin.edu/~chamberl/papers/trig_products.pdf ]]>

The part:

(z^(n-1) + z^(n-2) + â€¦ 1) = (z â€“ Ï‰)(z â€“ Ï‰^2) â€¦ (z â€“ Ï‰^(n-1)).

Is only valid if z is not equal to 1, making the subsitution of z=1 into the equation impossible.

]]>Because **(z-1)(z-w)(z-w^2)…(z-w^(n-1))** is an alternate factorization of **z^(n-1)**, since all the different values of **w^k** are roots of **z^(n-1) – 1 = 0**. Dividing by **(z – 1)** on both sides results in the last equation in the blog post.

Essentially, this proof is setting two different (but equally valid) factorizations of **z^(n – 1)** against each other, and plugging in **z = 1**.

z^n-1 /(zâ€“1)

to

(z-w)(z-w^2 )(z-w^3 )â€¦(z-w^(n-1)) ]]>

Thanks for the answer in advance! ðŸ™‚ ]]>

(z^n-1 + z^n-2 + â€¦ 1) = (z â€“ Ï‰)(z â€“ Ï‰^2) â€¦ (z â€“ Ï‰n-1) and you get 1 + 1 + … + 1 = n on the left side and the product of the diagonal lenghts on the right. ]]>

“Since

(z^n â€“ 1) = (z â€“ 1)(z^n-1 + z^n-2 + â€¦ 1)

it follows that

(z^n-1 + z^n-2 + â€¦ 1) = (z â€“ Ï‰)(z â€“ Ï‰2) â€¦ (z â€“ Ï‰n-1).

The result follows from evaluating the expression above at z = 1.”

if you put z=1 into the expression you just prove that 0 = 0

since (z^n – 1) = 1 – 1 = 0

and (z â€“ 1)(z^n-1 + z^n-2 + â€¦ 1) = (0)(z^n-1 + z^n-2 + â€¦ 1) = 0

Maybe my maths is a bit rusty after 50 years!

]]>Thanks ðŸ™‚

]]>Just to clarify, |a-b| means the modulus of the complex number formed by subtracting complex number b from another complex number a?

]]>