Why are the roots of the polynomial $z^N = a^N$ equal to $z_k = a \ e^{j\frac{2 \pi k}{N}}$? I am trying to understand equation 3.28 from this image in my book. 

I get everything that the author is saying, except for when he finds the roots, (zeros), of $z^N = a^N$. Of course, there are going to be $N$ roots. He says the roots are given by $z_k = a \ e^{j \frac{2 \pi k}{N}}$. 
What I do not understand is why they HAVE to be evenly spaced around a circle here? Where does it say that they have to be evenly spaced here? Why arent they all squished to one part of the circle, or why cant they be randomly spaced within the circle? I get that yes, those are solutions, but I can think of an infinitely more number of solutions that are all on the circle, but not evenly spaced...
Would appreciate someome elucidating this for me. Thank you.
 A: To find the solutions of $z^n=a^n$ you write the equation in polar coordinates. If $z=re^{j\theta}$, then we have
$$r^ne^{jn\theta}=a^ne^{j0}.$$
The magnitudes must be equal so we get $r^n=a^n$, which implies $r=a$. We also have $e^{jn\theta}=e^{i0}(=1).$ This is satisfied only if $n\theta=2\pi k$ for some integer $k$. So the solutions to the equation are $$z=ae^{j\frac{2\pi k}{n}}$$ where $k$ is an integer. Now you can check that you only get $n$ different complex numbers that correspond to $k=0\ldots n-1$, any other value of $k$ will just be a repeated solution.
Edit: We are not forcing the solutions to be evenly spread around the circle, we are just trying to find all the possible solutions. A priori we have a solution for each integer $k$. What I claim is that we have a lot of repeated solutions. For example, $k=0$ and $k=n$ give you the same solution
$$ae^{j\frac{2\pi 0}{n}}=a=ae^{j\frac{2\pi n}{n}}.$$
In fact, it is enough to consider $k=1\ldots n$ because any other $k$ would give you a solution which is already represented in ${1,\ldots,n}$. One way you can think about it is that you start with the solution given by $k=1$ and then increasing $k$ by one corresponds to rotating your solution by a $\frac{2\pi}{n}$ angle. Thus, when your reach $k=n+1$ you have completed a full turn in the circle and you start going through the same solutions.
A: Your equation can be rewritten as
$$\left(\dfrac{z}{a}\right)^N =1$$
Treating $\frac{z}{a}$ as an unknown, we know that the solutions to the equation are the $N^{th}$ roots of unity, i.e. $$\dfrac{z}{a} = e^{j\frac{2\pi k}{N}}$$ where $k = 1, \, 2, \, ..., \, N$.
Multiplying both sides by $a$ gives the desired result.
