Lerch-$\small \zeta(\varphi,0,-n)$ of integer *n* purely real and imaginary ($\small \zeta_\varphi (-n)^2 $ is real) for $\small n \ge 2$? Are the  Lerch-$\zeta(\varphi,0,-n) $ of integer n (for shortness I use the notation of my earlier question $\small \zeta_\varphi(-n)$) periodically purely real and imaginary: $\zeta_\varphi (-n)^2 $ is real, ($ n \ge 2$) ? And how to prove this?
I've a strange observation of periodicity which I would like to explain/derive/prove  analytically.  

Consider the triangle of eulerian numbers E (ideally of infinite size, row and col-indices beginning at zero, with elements $ \small e_{r,c}$)     

$ \qquad E = \small 
\begin{array} {rrrrr}
 1 & . & . & . & . & . \\
 1 & 0 & . & . & . & . \\
 1 & 1 & 0 & . & . & . \\
 1 & 4 & 1 & 0 & . & . \\
 1 & 11 & 11 & 1 & 0 & . \\
 1 & 26 & 66 & 26 & 1 & 0
 \end{array}  $
Assume some angular parameter $\varphi$, the associated complex number from the unit-circle $z=z_\varphi= \exp(i \varphi ) \qquad z \ne 1 $, $ \small {\varphi \over \pi} $ not necessarily rational.
Now consider the Eulerian polynomials, whose coefficients are taken from a row of E,
 $ \small E(z,r) = \sum_{c=0}^\infty ( e_{r,c} \cdot z^c ) $     
Then compute $ \small \zeta_\varphi(-n) = {z \over (1-z)} \cdot E(z,n) \cdot (1-z)^{-n} $ .   
Observation: I observe, that for $n>2$ the $ \small \zeta_\varphi(-n) $ lie either on the real or on the imaginary axis, in other words $ \small \zeta_\varphi(-n) ^2 $ are real..      
Q: I seem to be missing just the key idea, so my question here is, how I could derive that behave analytically, given the description via the Eulerian triangle.     
Addendum/Generalization: While it was already surprising, that this works with rational roots z of the complex unit, it seems to me even more interesting, that the observation holds for arbitrary $ \small \varphi \ne 0 $ and $ \small z_\varphi = \exp(I \cdot \varphi) \ne 1 $ .
 A: By the symmetry of the triangle of Eulerian numbers the rational function
$$R_{n}(z) = {z \over (1-z)} \cdot E(z,n) \cdot (1-z)^{-n} $$
satisfies$$R_{n}(z)=(-1)^{n+1}R_{n}(\frac{1}{z}).$$
Since the numerator and denominator polynomials of $R_{n}(z)$ have real (in fact,   integer) coefficients we have$$\overline{R_{n}(z)}=R_{n}(\bar z).$$
Your observation should now be explained (there is no need to restrict to $n>2$).
To clarify further:
Let $ z = \exp(i \cdot \varphi) \ne 1 $ lie on the unit circle so $\bar z = \frac {1}{ z}$ and let $n \geq 1$.
Suppose $R_{n}(z) =  a+ib$.
Then from the above$$\overline{R_{n}(z)}=R_{n}(\bar z) = R_{n}(\frac{1}{z})=(-1)^{n+1}R_{n}(z). $$  Hence $$a-ib = (-1)^{n+1}(a+ib).$$ If $n$ is even we get $a = 0,$ and so $R_{n}(z)$ is pure imaginary; if $n$ is odd we get $b= 0,$ and $R_{n}(z)$ is real.
A particularly interesting case is when $z = i$. The sequence $(2 \cdot R_{n}(i))n \geq 1$ equals $[-1, -1 \cdot i, 2, 5 \cdot i, -16, -61 \cdot i,...]$, which is the sequence of up-down numbers multiplied by powers of $i$. For more on this see my comment of Jan 2011 in A000111 in the OEIS. 
A: Unfortunately, I couldn't make Peter's answer fully transparent to me. So I tried two things; both led to a solution.
1) rewrite the $ \small \zeta _\varphi(-n) $ - expression based on the description via the Eulerian polynomials as taylor-series in terms of $ \small \varphi$, which is real. We can factor out $ \small i^n$ and stay with a series in $ \small \varphi ^2$, thus the result is $ \small i^n \cdot x$ where x is real.    
2) Rewrite the expression using the Eulerian polynomials using $ \small z=exp(i \varphi)$
in $ \small \zeta _\varphi(-n) = {z \over 1-z} E(z,n) \cdot (1-z)^{-n} $     
Factor out $ \small  {z \over (1-z)^2} $ to get as first cofactor
$\qquad \small  {z \over (1-z)^2} = {1 \over (1/\sqrt{z} - \sqrt z )^2} ={1 \over (\exp(- i \varphi/2) - \exp(i \varphi /2))^2 } 
 = {1 \over (-2 i\sin(\varphi /2))^2 } = -1/4 \cdot \sin(\varphi /2)^{-2} $.
This value is always real, and is also $ \zeta_\varphi(-1) $.     
Then , for instance, for n=3 we get for the remaining part of the formula:   
$ \qquad\small \begin{array} {ll} (1 + 4z + 1z^2) \cdot (1-z)^{-2} &=& (1/z + 4 + 1z) \cdot (1/\sqrt{z} - \sqrt z )^{-2} \\\ 
 & = &(1 \cdot (1/z + z) + 4 ) \cdot -1/4 \cdot \sin(\varphi /2)^{-2} \\\
 & = & - ( {1 \over 2} \cos(\varphi)) + 1)\cdot \sin(\varphi /2)^{-2} \\\
 \end{array}$     
Together this gives
$ \qquad \small \zeta_\varphi(-3)= 1/4 \cdot \sin(\varphi /2)^{-4} \cdot ( {1 \over 2} \cos(\varphi)) + 1)
$
Well, this is only for one n, but I begin to understand, how this works and why we get these purely imaginary/real results (and I even begin to see Peter's mentioned symmetries here.... )
