Evaluate $\sum_{y=0}^{2^n-1}e^{2\pi iy(z-x)/2^n}$

Question

I am trying to evaluate $$\sum_{y=0}^{2^n-1}e^{2\pi iy(z-x)/2^n}$$ where $n\in\mathbb{Z}$ and $x,z\in\mathbb{Z}_{2^n}$.

Clearly, if $x=z$, then $\sum=2^n$. But I am unsure about when $x\neq z$. I believe it to be $=0$ as that would make my problem self-consistent.

Answer 1

We have two cases:

Case $1$: $z \neq x$ (mod. $2^n$) $$\sum_{y=0}^{2^n-1}e^{2\pi iy(z-x)/2^n} = \sum_{y=0}^{2^n-1} a^y = \frac{1 - a^{2^n}}{1-a}$$ where $a = \Big( e^{2\pi i(z-x)/2^n} \Big)$ Replacing as @lab bhattacharjee did, we get $0$.

and

Case $2$: $z = x$ (mod. $2^n$) $$\sum_{y=0}^{2^n-1}e^{2\pi iy(z-x)/2^n} = \sum_{y=0}^{2^n-1} e^0 = 2^n$$

Answer 2

$$\sum_{y=0}^{2^n-1}(e^{2\pi i(z-x)/2^n})^y$$

$$=\dfrac{(e^{2\pi i(z-x)/2^n})^{2^n}-1}{e^{2\pi i(z-x)/2^n}-1}$$

$$=\dfrac{e^{2\pi i(z-x)}-1}{e^{2\pi i(z-x)/2^n}-1}$$

If $z-x$ is integer $$e^{2\pi i(z-x)}=(e^{2\pi i})^{z-x}=?$$ provided $e^{2\pi i(z-x)/2^n}\ne1\iff2^n\nmid(z-x)$