<s>[INST] Unilateral Z-Transform of sen($\frac{\pi}{4}n$) I was calculating the Unilateral Z-Transform of $sen(\frac{\pi}{4}n)$
I calculated it this way, but then the calculation become very difficult and i think i have made some mistakes.
It is : $\sum_{n=0}^{+\infty} sen(\frac{\pi}{4}n)z^{-n}$
$sen(\frac{\pi}{4}n) $ is periodic of a period 8.
And if $a_n=sen(\frac{\pi}{4}n)$
$a_0=0, \, a_1=\frac{\sqrt{2}}{2}, \, a_2= 1, \, a_3 = \frac{\sqrt{2}}{2},
\, a_4= 0, \, a_5 = -\frac{\sqrt{2}}{2}, \, a_6= -1, \, a_7 = -\frac{\sqrt{2}}{2}$
Then $a_{n+8k}=a_n, k \in Z$
Then:
\begin{align}
\sum_{n=0}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} &= (\sum_{n=4k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=1+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=2+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=3+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} \\
& \hspace{5mm} + \sum_{n=5+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=6+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n}+ \sum_{n=7+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} )
\\
&= 0 + \frac{\sqrt{2}}{2} \sum_{n=1+8k}^{+\infty} z^{-n}+ \sum_{n=2+8k}^{+\infty} z^{-n}+ \frac{\sqrt{2}}{2} \sum_{n=3+8k}^{+\infty} z^{-n} \\
& \hspace{5mm} -\frac{\sqrt{2}}{2} \sum_{n=5+8k}^{+\infty} z^{-n} - \sum_{n=6+8k}^{+\infty} z^{-n} - \frac{\sqrt{2}}{2} \sum_{n=7+8k}^{+\infty} z^{-n}
\end{align}
Those Z-Transforms are not difficult (as $\sum_{n=0}^{+\infty} z^{-n}= \frac{z}{z-1} $ but very long ones. 
Wolframs shows a simplier Z-Transform of this function.
Is there any more efficient way to calculate it?
 [/INST] For the series 
\begin{align}
f(t) = \sum_{n=0}^{\infty} \sin\left(\frac{n \pi}{4} \right) \, t^{n}
\end{align}
then it is seen that
\begin{align}
f(t) &= \sin\left(\frac{\pi}{4} \right) \, t + \sin\left(\frac{2 \pi}{4} \right) \, t^{2} + \sin\left(\frac{3 \pi}{4} \right) \, t^{3} + \cdots \\
&= \sum_{n=0}^{\infty} \left[ \sin\left(\frac{(4n+1) \pi}{4} \right) \, t^{4n+1} + \sin\left(\frac{(4n+2) \pi}{4} \right) \, t^{4n+2} + \sin\left(\frac{(4n+3) \pi}{4} \right) \, t^{4n+3}\right] \\
&= \left[\sin\left(\frac{\pi}{4} \right) \, t + \sin\left(\frac{2 \pi}{4} \right) \, t^{2} + \sin\left(\frac{3 \pi}{4} \right) \, t^{3} \right] \, \frac{1}{1+t^{4}} \\
&= \frac{t(1 + \sqrt{2} t + t^2)}{\sqrt{2} \, (1+t^4)}.
\end{align}
By making the change $t \to z^{-1}$ the desired result is obtained. 
</s>