Phase factors for an 32 point fft

Question

I have just started working on FFT and its related phase factors.

I have read for an N=2 point FFT, We have two phase factors, one on the positive axis and one on negative axis.

similarly for an 4 point FFT, we have 4 phase factors as theta=(pi/2), at

 0'  W 40=1
-90' W 41=-j
180' W 42=-1
 90' W 43= j

here we can see the values that are 180 degree apart are negative to each other.

Here I wanted to know how it works for an 32 point FFT.

As theta=(pi/16) do we have 16 phase factors on the positive axis and 16 phase factors on the negative axis with the interval of (pi/16) ?

so will it follow the pattern as I mentioned below?

 0'       W16 0 =1
-11.25'   W16 1 =0.98 -j0.195
-22.5'    W16 2 =0.92 -j-0.38
and so on
....
....
till
-168.75'  W16 15=
and the negative parts of the above values from **W16 16** to **W16 32**

Am I correct? please verify the above details and let me know how the phase factors works for an 32 point FFT

Thanks in advance

Answer 1

The base functions of the discrete time Fourier transform takes on values from the set $$\{z\in\mathbb{C}:z^N=1\}$$ where $N$ is the dimension of the vector space. The solutions can be parametrized to be in the form $$z_k = \exp(2 \pi i \frac{k}{N})$$ for $k=0..N-1$ because $$z_k^N = \exp(2 \pi i k \frac{N}{N}) = \exp(2\pi i k) = 1$$

In other words, the values the base functions components are chosen from are the $N$ non-identical $N$-th roots of 1, which are equidistantly distributed on the unit circle and include $1$ itself.