Towards a formula for the Euler $\phi$ function? 
*

*$\Phi_n(1)$ and $\Phi_n(-1)$ for the cyclotomic polynomials are well-known.
I am now looking for
$$\Phi_n(i)$$ and/or $$\Phi_n(-i)$$ with $i$ the complex unit.

*The reason is :  
I suppose it is true that $(1-i)^{\phi(n)}\Phi_n(i)$ is either zero or of the form
$$(-1)^{f(n)}2^{g(n)}$$ for some rational valued functions. 
Just now $f(n)$ and $g(n)$ are unknown to me. But if all calculations would succeed
by taking logarithms $\phi(n)$ may perhaps be calculated.
The ideas behind that rely on the polynomials defined by William E Heierman published   on his Web Site. 
 A: After a long thought, my conjecture is instead the following. Let $\nu=\nu_2(n)$, $\omega_1$ be the number of distinct prime numbers $\equiv 1\pmod{4}$ that divide $n$, $\omega_3$ be the number of distinct prime numbers $\equiv 3\pmod{4}$ that divide $n$. I claim:
$$\begin{array}{rl}\textbf{If...}&\textbf{then...}\\
n=2&\Phi_n(i)=1+i\\
n=4&\Phi_n(i)=0\\
n=2^{k},k\geq 2&\Phi_n(i)=2\\
n=4p&\Phi_n(i)=p\\
\omega_1\geq 1,n\neq 4p & \Phi_n(i)=1\\
\nu\geq 3&\Phi_n(i)=1\\
\omega_3\geq 3&\Phi_n(i)=1\\
\omega_1=0,\omega_3=1,\nu=0&\Phi_n(i)=i\\
\omega_1=0,\omega_3=1,\nu=1&\Phi_n(i)=-i\\
\omega_1=0,\omega_3=2,\nu<2&\Phi_n(i)=-1\\
\omega_1=0,\omega_3=2,\nu=2&\Phi_n(i)=1.\\
\end{array}$$
Please check this conjecture. I am too lazy now to prove that it holds in virtue of:
$$\Phi_n(i)=\prod_{d\mid n}\left(i^{n/d}-1\right)^{\mu(d)}$$
by an induction depending on eleven different cases, but I can find somewhere the strength to do my math if this is really useful.
A: I made a calculation for $n\ge 3$ and get
$$\prod_{E_n}\cos\left(\frac{k\pi}{n}\right)=\frac{\Phi_n(-1)}{2^{\phi(n)}}$$
This formula may be totally wrong at this moment.
I have not the CAS Mathematica available and also not the programming skills to check it by other CAS freely available.
A: I recalculated the formula and it is indeed  
$$\prod_{E_n}\cos\left(\frac{k\pi}{n}\right)=(-1)^{\frac{\phi(n)}{2}}\frac{\Phi_n(-1)}{2^{\phi(n)}}$$
for $n>2$. Taking absolute values on both sides allows one then to calculate $\phi(n)$.
