$Φ\_n(1)$ and $Φ\_n(−1)$ for the cyclotomic polynomials are well-known.


I am now looking for 
$$Φ\_n(i)$$
 and/or 
$$Φ\_n(−i)$$
with i the complex unit.


At this moments my endeavours result in intricate categorizing of values 
for $n$ e.g. as $4$, $2^k$, $4p$, $p^k$ with $p$ prime and considerations about
the prime factors of an odd $n$ being e.g. $\equiv1\mod4$ or/and $\equiv3\mod4$.



Looking for ways how to calculate $\Phi_n(i)$
The question has already been answered by Vladimir Dotsenko, but as Wolfgang Tintemann is still interested in a solution using Möbius inversion, let me expand the comments I made above.


Since $\Phi\_n(-i)=\overline{\Phi\_n(i)}$, it is enough to deal with $i$. We are going to prove
$$\Phi\_n(i)=\begin{cases}
i-1&n=1,\\
i+1&n=2,\\
0&n=4,\\
p&n=4p^k,\\
i&n=p^k,\ p\equiv3\pod4,\text{ $k$ odd, or}\\
&n=2p^k,\ p\equiv3\pod 4,\text{ $k$ even,}\\
-i&n=p^k,\ p\equiv3\pod4,\text{ $k$ even, or}\\
&n=2p^k,\ p\equiv3\pod 4,\text{ $k$ odd,}\\
-1&n=p^kq^l\text{ or }n=2p^kq^l,\ p,q\equiv3\pod4,\\
1&\text{if $n$ is not of any of the forms above,}
\end{cases}$$
where $p,q$ are prime, and $k,l\ge1$. The first three cases can be checked immediately, hence let me assume that $n\ne1,2,4$ below.


By Möbius inversion of the formula $x^n-1=\prod\_{d|n}\Phi\_d(x)$, we have
$$\Phi\_n(x)=\prod\_{d|n}(x^d-1)^{\mu(n/d)}.$$
We cannot directly plug in $x=i$ as we might get $0/0$, but we can compute it as $\lim\_{\epsilon\to0}\Phi\_n(i+\epsilon)$. 


Since $(i+\epsilon)^d=i^d+O(\epsilon)$ for $4\nmid d$, and $(i+\epsilon)^d=1-id\epsilon+O(\epsilon^2)$ for $4\mid d$, we see that $\Phi\_n(i+\epsilon)$ is
$$(1+O(\epsilon))
\prod\_{\substack{d|n\\d\equiv1\pod4}}(i-1)^{\mu(\frac nd)}
 \prod\_{\substack{d|n\\d\equiv3\pod4}}(-i-1)^{\mu(\frac nd)}
 \prod\_{\substack{d|n\\d\equiv2\pod4}}(-2)^{\mu(\frac nd)}
 \prod\_{\substack{d|n\\d\equiv0\pod4}}(-id\epsilon)^{\mu(\frac nd)}.$$
We can express $\pm i-1$ in terms of $\zeta\_8=(1+i)/\sqrt2$ and reshuffle the terms to get
$$\tag{$\*$}(1+O(\epsilon))
 (-\sqrt2)^{\sum\_{d|n}\mu(\frac nd)}
 (\sqrt2)^{\sum\_{2|d|n}\mu(\frac nd)}
 (2i\epsilon)^{\sum\_{4|d|n}\mu(\frac nd)}
 \zeta\_8^{s(n)}
 \prod\_{4|d|n}(d/4)^{\mu(\frac nd)},$$
where
$$s(n)=\sum\_{\substack{d|n\\d\equiv3\pod4}}\mu(n/d)
 -\sum\_{\substack{d|n\\d\equiv1\pod4}}\mu(n/d).$$
Since $n\ne1$, $\sum\_{d|n}\mu(n/d)=0$. Moreover, $\sum\_{2|d|n}\mu(n/d)=\sum\_{d\mid\frac n2}\mu(\frac n2/d)=0$ for $n$ even, and trivially $\sum\_{2|d|n}\mu(n/d)=0$ for $n$ odd; likewise for $4$. Thus the first four terms of $(\*)$ disappear as $\epsilon\to0$. If $4\nmid n$, the last term is vacuously $1$. On the other hand, if $4\mid n$, we have $s(n)=0$: all the $\mu(n/d)$ summands vanish on account of $n/d$ being divisible by $4$. Thus we arrive at
$$\tag{$\*{\*}$}
\Phi\_n(i)=\begin{cases}
\zeta\_8^{s(n)}&4\nmid n,\\
\prod\_{d|m}d^{\mu(m/d)}&n=4m.
\end{cases}$$
Assume first $4\mid n$. We can simplify $(\*{\*})$ further to
$$\Phi\_n(i)=m^{\sum\_{d|m}\mu(m/d)}\prod\_{d|m}(m/d)^{-\mu(m/d)}
=\prod\_{d|m}d^{-\mu(d)}.$$
If $m=p^k$ is a prime power, this gives immediately $\Phi\_n(i)=p$. On the other hand, if $m=p\_1^{k\_1}\cdots p\_r^{k\_r}$ with $r\ge2$, we have
$$\Phi\_n(i)=\prod\_{j=1}^rp\_j^{\sum\_{d|m\_j}\mu(d)}=1,$$
where $m\_j=m/p\_j^{k\_j}$.


Now assume $4\nmid n$, we need to compute $s(n)$ modulo $8$. Observe that if $n$ is odd, then $s(2n)=-s(n)$, hence $\Phi\_{2n}(i)=\Phi\_n(i)^{-1}$. It thus suffices to consider odd $n$.


If $n$ has a prime divisor $p\equiv1\pod4$, then each $d\mid n$ with $d\equiv3\pod4$ is paired with another one with opposite $\mu(n/d)$ by multiplying or dividing $d$ by $p$; likewise for $d\equiv1\pod4$. Thus in this case $s(n)=0$ and $\Phi\_n(i)=1$.


Finally, assume that $n=p\_1^{k\_1}\cdots p\_r^{k\_r}$, where each $p\_j\equiv3\pod4$. Let $t=\pm1$ be such that $n\equiv-t\pod4$. Then for every $I\subseteq\{1,\dots,r\}$, $d=n\prod\_{i\in I}p\_i^{-1}$ contributes $t(-1)^{|I|}\mu(n/d)=t$ to $s(n)$, hence $s(n)=t2^r$. Thus, $\Phi\_n(i)=1$ if $r\ge3$, $\Phi\_n(i)=-1$ if $r=2$, and $\Phi\_n(i)=i^t=(-1)^{k\_1+1}i$ if $r=1$.


