Question

Let $x\left[n\right]={(-\frac{1}{9})}^{n}u\left(n\right)-(-\frac{1}{3})u(-n-1)$
The Region of Convergence (ROC) of the $z-$ transform of $x\left[n\right]$

Answer 1

$x\left(n\right)=\frac{{(-\frac{1}{9})}^{n}u\left(n\right)}{x_1\left(n\right)}-\frac{{(-\frac{1}{3})}^{n}u(-n-1)}{x_2\left(n\right)}$

$x_1\left(n\right)={(-\frac{1}{9})}^{n}u\left(n\right)\to \text{Right sided signal}$

$X_1\left(z\right)=\frac{1}{1-(-\frac{1}{9})z^{-1}}$

$ROC|z|\ge \left|\frac{1}{9}\right|$ or $|z|>\frac{1}{9}$

$x_2\left(n\right)={(-\frac{1}{3})}^{n}u(-n-1)\to \text{Left sided signal}$

$X_2\left(z\right)=\frac{1}{1-(-\frac{1}{3})z^{-1}};|z|\le \frac{1}{3}$

Now, when both the transforms are added using linearity property,

ROC must be between $\frac{1}{3}>|z|>\frac{1}{9}$