# Finite length truncated exponential sequence $\mathcal Z$-transform, zeros over circle explanation

I'm looking at an example on how to obtain the $\mathcal Z$-transform from a finite length truncated exponential sequence, namely: $$x[n] = \begin{cases} a^N &\text{for} & 0 \leq n \leq N-1\\[2ex] 0 & \text{otherwise} \end{cases}$$

The $\mathcal Z$-transform from that sequence is: $$\frac {1}{z^{N-1}} \cdot \frac {z^{N} - a^N}{z - a}$$

It can be seen that there are $N-1$ poles at the origin and $N-1$ zeros at $z = a$ (because the pole in $z=a$ is cancelled by a zero). My doubt is on the location of the zeros over the $z$-plane.

I know that the $N-1$ zeros at $z = a$ can be expressed as:

$z = a\cdot1\ \text{and}\ 1 = e^{j2\pi k}$ so: $$z_k = ae^{j2\pi k}\quad\text{with}\quad k = 1,2,\ldots,N-1$$ Yet in the example I'm looking at, they say that the zeros are located at $$z_k = ae^{j\frac{2\pi k}{N}}\quad\text{with}\quad k = 1,2, \ldots, N-1$$ So they are spaced at $\frac{2\pi}{N}$ instead of being all at the same place and I don't understand where that $N$ came from.

The $z$-transform is
$$\frac{1}{z^{N-1}}\cdot\frac{z^N-a^{N}}{z-a}$$ The zeros will therefore be given by \begin{align} z^N&=a^{N}\cdot1\\ &=\left(a^{N}\right)e^{j2k\pi}\\ &=\left(ae^{\frac{j2k\pi}{N}}\right)^{N},\,k=1,2,\cdots,N-1 \end{align} Hence, $$\boxed{z_k=ae^\frac{j2k\pi}{N},\,k=1,2,\cdots,N-1}$$
$$z^N = a^N \cdot 1$$ where $1$ is recognized equivalently as $$1=e^{j2\pi k}$$ allowing complex roots to be distinguished and hence, the N complex roots become : $$z^N = a^N \cdot e^{j2\pi k}$$ $$z_k = a \cdot e^{j \frac{2\pi}{N}k}$$ for $k = 0,1,...,N-1$
All the roots have the same magnitude; i.e., $|z_k| = a$ for all $k$, however they are complex numbers and as such they have phase values, which distinguishes them, by effectively distributing the roots along the circumference of a circle with radius $a$ equidistantly in angle.