To find Z-transform of given sequence How to find the $z$-transform of $\left[a^{n}\sin\left(bn\right)\right]/n!$ where "!" denotes factorial of a number and b is constant??
 A: Hints:
i) 

$$\sin( bn )=\frac{1}{2i}(e^{ibn}-e^{-ibn}).$$

ii) Z-Transform of $a^ne^{ibn}$ is given by 

$$F(z) = \sum_{n=0}^{\infty} a^n (e^{ib})^nz^{-n}.$$

iii) The following is known as the geometric series which you need to find a closed form for $F(z)$ 

$$ \sum_{n=0}^{\infty} t^n = \frac{1}{1-t}. $$

I think you can finish it now.
A: I'm going to find Z Transform with the use of Euler's formula
$$
sin(x)= \frac{e^{ix} - e^{-ix}}{2i}
$$
$$
x(n)= \frac{a^n sin(bn)}{n!}=\frac{a^n (e^{ibn} - e^{-ibn})}{(2i)n!}
$$
$$
X(z) = \sum_{0}^{\infty}\frac{(ae^{ib})^n - (ae^{-ib})^n}{(2i)n!}z^{-n}=\frac{1}{2i}(\sum_{0}^{\infty} \frac{1}{n!}(\frac{ae^{ib}}{z})^n -\sum_{0}^{\infty} \frac{1}{n!}(\frac{ae^{-ib}}{z})^n )
$$
$$
X(z) = \frac{1}{2i}(exp{(\frac{ae^{ib}}{z})} -exp{(\frac{ae^{-ib}}{z})} )
$$
$$
X(z) = \frac{e^{\frac{a}{z}}}{2i}(exp{(cos(b) + i sin(b))} -exp{(cos(b) - i sin(b))} )
$$
$$
X(z) = \frac{e^{\frac{a}{z}}exp{(cos(b))}}{2i}(exp{(i sin(b))} -exp{(- i sin(b))} )
$$
$$
X(z) = {e^{\frac{a}{z}}exp{(cos(b))}}sin{( sin(b))}
$$
