What is the radius of the circle that inscribes the complex numbers treated as vectors in the series $\sum_{n=1}^Ne^{i2\pi n/N}$ i am studing Fourier transforms and in the book i am reading it recomends to prove that the following series is zero
$$\sum_{n=1}^{N}e^{i2\pi n/N}=0$$
But it says to do so imagining that the complex numbers are vectors and that they form a regular polygon, and that is ok and all, i thought that since $e^{i2\pi N/N}=1$ all the other vectors would have radii equal to 1 since they must have the same lenghts to be a regular polygon, but when i saw the solution it said that the radius of the circle that inscribed the vectors had radius $R=\frac{1}{\sin(\theta/2)} $ , with $\theta = 2\pi/N$ so that it seems that the radius changes with N, what is happening here?
 A: Seems that there is a double understanding about which polygon your book is referring to.


*

*If the oriented segments (represented by vectors) are placed in a star-shaped way, i.e. all starting from the origin, then their heads lie symmetrically on a circle of radius one, and the resultant
is null.

*If the segments are added, i.e placed head to tail (Delta shaped in electric engineering), then the heads and tails  will lie on a circle, starting at the origin and ending there, the resultant is again $0$. The center of circle will be on the $y$ axis, at the crossing of the bisecting axis of each  segment, hence with radius $1/\sin(\theta/2)$.


According to the goal of the analysis you want to do, you can use either representation. The latter is useful ,e.g., for calculating partial sums.
A: The radius of the polygon does change with $N$.  Since the sides all have length one, the polygons must get bigger and bigger with $N$.
The individual summands have magnitude $1$.  However, this is not what the text is discussing.  The regular polygon made by summing the vectors (in the usual "tail-to-head" way) has both an inscribed and a circumscribed circle.  The circumscribed circle has diameter (standard fact about the circumscribed circle of a regular polygon)
$$  \frac{s}{\sin \frac{\pi}{N}}  \text{,}  $$
where here $s=1$ is the length of the sides of the polygon and $N$ is the number of sides.  In terms of $\theta$, $\theta = 2 (\frac{\pi}{N})$, so we almost obtain the given radius, 
$$  \frac{1}{\sin \frac{\theta}{2}}  \text{.}  $$
So what's going on?  The diameter of a circle is not its radius (except for the trivial circle).  If you draw a few of these polygons, you will see that they lie entirely on one side of the origin, with one vertex on the origin.  This means that the point on the polygon farthest from the origin is no further away than the diameter of that circumscribed circle.  That is, we may draw a circle, centered at the origin, with radius, $R$, equal to the diameter of the circumscribed circle of the polygon, and this circle contains the entire polygon (and a lot of space outside the polygon).
A: This is not an answer but an illustration of the second interpretation (the good one), as given by @G Cab. You can see on the left figure the increasing size of the polygons, from $N=3$ (triangle, in blue) to $N=10$ (decagon, in red). On the right, one finds the set of "arrows" representing $e^{\tfrac{2i\pi k}{N}} \ \ (k=1\cdots N)$ in the case $N=10$ that are progressively summed up to generate the decagon. 

