Consider a discrete time signal given by
x[n]=(-0.25)ⁿu[n]+(0.5)ⁿu[-n-1]
The region of convergence of its Z-transform would be

A discrete system is represented by the difference equation $$$\begin{bmatrix}X_1\left(k+1\right)\\X_2\left(k+2\right)\end{bmatrix}=\begin{bmatrix}a&a-1\\a+1&a\end{bmatrix}\begin{bmatrix}X_1\left(k\right)\\X_2\left(k\right)\end{bmatrix}$$$ It has initial condition $$X_1\left(0\right)=1;\;X_2\left(0\right)=0$$. The pole location of the system for a = 1, are

Let $$X\left(z\right)=\frac1{1-z^{-3}}$$ be the Z–transform of a causal signal x[n]. Then, the values of x[2] and x[3] are

Given X(z)=$$\frac z{\left(z-a\right)^2}$$ with $$\left|z\right|$$ > a, the residue of X(z)z^n-1 at z = a for n $$\geq$$ 0 will be