Phase factors for an 32 point fft 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
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.