Why is $\sum_{n=0}^{N-1}e^{-2i\pi(k+k_0)n/N}=N\delta(k-N+k_0)$?

Question

EDIT: $\delta$ is the Dirac delta function and in the context it is defined as $\delta(0)=1$ and $0$ for all $n\neq 0$.

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I am having trouble concluding that

$\sum_{n=0}^{N-1}e^{-2i\pi(k+k_0)n/N}=N\delta(k-N+k_0)$.

Any help clarifying would be highly appreciated.

Answer 1

Hint: Consider two cases:

Case I is when $k = -k_0 + \alpha N$ for an integer $\alpha$. In this case $$e^{-2i\pi(k+k_0)/N} = ?$$ Case II is when $k \neq -k_0 + \alpha N$ for an integer $\alpha$. Use $$\sum_{n=0}^{N-1} q^n = \frac{1-q^{N}}{1-q}$$ to show that the sum is zero.

Answer 2

Whenever $k - N + k_0 = 0$ then our sum is $\sum_{n=0}^{N-1}e^{-2\pi i n} = N$.

Otherwise, it's the sum of the roots of unity of order $N$ which is zero.