summation and product of sin and cos

Question

I wonder how to find summation for $\displaystyle \sum_{k=0}^{n-1}(\cos{\frac{2\pi k}{n}+i \sin\frac{2\pi k}{n}})$

and the same for product $\displaystyle \prod_{k=0}^{n-1}(cos{\frac{2\pi k}{n}+i \sin\frac{2\pi k}{n}})$

I know that sum is equal to $0$ and product to $(-1)^{n+1}$ but have no idea how to show it

Answer 1

Hint: Those are roots of unity. That is, they are roots of the equation: $$x^n-1=0$$ So use Vieta.

Answer 2

Hint:

Use Euler formula $e^{ix}=\cos x+i\sin x$

$$\sum_{k=0}^{n-1}e^{\dfrac{2k\pi i}n}=\sum_{k=0}^{n-1}\left(e^{\dfrac{2\pi i}n}\right)^k$$ which is a Geometric Series

and $$\prod_{k=0}^{n-1}e^{\dfrac{2k\pi i}n}=e^{\dfrac{2\pi i}n\left(\sum_{k=0}^{n-1}k\right)}$$

Answer 3

First sum is geometric series:

$$\sum_{k=0}^{n-1}(\cos{\frac{2\pi k}{n}+i \sin\frac{2\pi k}{n}})=\sum_{k=0}^{n-1}e^{2i\pi k/n}=\frac{1-e^{2 i\pi n /n }}{1-e^{2i\pi /n}}=\frac{1-1}{1-e^{2i\pi /n}}=0$$

Now product:

$$\prod_{k=0}^{n-1}(cos{\frac{2\pi k}{n}+i \sin\frac{2\pi k}{n}})=\prod_{k=0}^{n-1}e^{2i\pi k /n}=e^{2 i\pi \frac{(n-1)n}{2n}}=e^{i\pi(n-1)}=(-1)^{n-1}$$