Consider the infinite-length, discrete-time sequence x[n] = 0.9^|n|, where n is an integer. The region of convergence of its Z-transform X(z) is given by: (Note: z is a complex variable)

The correct option is:
0.9 < |z| < 1/0.9

Step-by-Step Explanation:

We are given the discrete-time sequence:
$$x\left[n\right]={0.9}^{\left|n\right|}$$
where $n$ is an integer ranging from $-\infty$ to $\infty$.

To find the Z-transform $X\left(z\right)$ and its Region of Convergence (ROC), we split the bilateral sum into two parts: one for $n\ge 0$ (causal part) and one for $n<0$ (anti-causal part).

Let $x\left[n\right]=x_1\left[n\right]+x_2\left[n\right]$, where:
1. $x_1\left[n\right]={0.9}^{n}u\left[n\right]$
2. $x_2\left[n\right]={0.9}^{-n}u[-n-1]$

1. Analysis of the Causal Part $x_1\left[n\right]$:
The Z-transform of $x_1\left[n\right]={0.9}^{n}u\left[n\right]$ is given by:
$$X_1\left(z\right)=\sum _{n=0}^{\infty }{0.9}^n{z}^{-n}=\sum _{n=0}^{\infty }{\left(0.9z^{-1}\right)}^n$$
This infinite geometric series converges if and only if:
$$\left|0.9z^{-1}\right|<1\Rightarrow \left|z\right|>0.9$$
So, the ROC of the causal part is ${\text{ROC}}_1:\left|z\right|>0.9$.

2. Analysis of the Anti-causal Part $x_2\left[n\right]$:
The Z-transform of $x_2\left[n\right]={0.9}^{-n}u[-n-1]$ is given by:
$$X_2\left(z\right)=\sum _{n=-\infty }^{-1}{0.9}^{-n}{z}^{-n}$$
Letting $m=-n$, the index $m$ ranges from $1$ to $\infty$:
$$X_2\left(z\right)=\sum _{m=1}^{\infty }{0.9}^m{z}^m=\sum _{m=1}^{\infty }{\left(0.9z\right)}^m$$
This geometric series converges if and only if:
$$\left|0.9z\right|<1\Rightarrow \left|z\right|<\frac{1}{0.9}$$
So, the ROC of the anti-causal part is ${\text{ROC}}_2:\left|z\right|<\frac{1}{0.9}$.

3. Combined Region of Convergence:
For the overall Z-transform $X\left(z\right)$ to exist, both series must converge simultaneously. Therefore, the overall ROC is the intersection of ${\text{ROC}}_1$ and ${\text{ROC}}_2$:
$$\text{ROC}={\text{ROC}}_1\cap {\text{ROC}}_2$$
$$0.9<\left|z\right|<\frac{1}{0.9}$$