Question

Let z = cos((2pi)/n) + i sin((2pi)/n) and n ≥ 2. Show that 1 + z + · · · + z^(n−1) = 0.

Answer 1

can you find z^n ?

Answer 2

then we use the formula
\(\large z^n-1= (z-1)(1+z+z^2+...+z^{n-1}) \\ z \ne1\)

Answer 3

see whether you get z^n=1 or not..

Answer 4

z^n can be found using deMoivre's theorem, in which case z^n would become cos((2pi)) + i sin((2pi)) = 1.

Answer 5

correct :)

Answer 6

and then from the formula that you said it shows that it is 0. Thanks.

Answer 7

welcome ^_^