Z-transform and ROC of two-sided exponential

x[n] = (−⅓)ⁿ u[n] − (⅓)ⁿ u[−n−1]. Take the two terms separately.

First term (right-sided): x₁[n] = (−⅓)ⁿ u[n]

X₁(z) = Σ(n=0 to ∞) (−⅓ z⁻¹)ⁿ = 1/(1 + ⅓z⁻¹), ROC: |z| > ⅓

Second term (left-sided): x₂[n] = −(⅓)ⁿ u[−n−1]

X₂(z) = −Σ(n = −∞ to −1) (⅓ z⁻¹)ⁿ = 1/(1 − ⅓z⁻¹), ROC: |z| < ⅓

Combined: x[n] = x₁[n] + x₂[n], so

X(z) = 1/(1 + ⅓z⁻¹) + 1/(1 − ⅓z⁻¹) = 2/(1 − (1/9)z⁻²) = 2z²/(z² − 1/9)

The poles are at z = ±⅓.

ROC: the ROC of the sum must lie in both |z| > ⅓ and |z| < ⅓. These two regions do not overlap.

 Im
  |      ROC1: |z| > 1/3  (outside)
  |      ROC2: |z| < 1/3  (inside)
 -x---o---x-- Re       no common ring

Answer: The algebraic form is X(z) = 2z²/(z² − 1/9), but the intersection of the ROCs is empty, so the z-transform of this x[n] does not converge for any z: the ROC does not exist (the signal has no z-transform, and no DTFT). (X(z) = 2z²/(z²−1/9) would correspond to the signal with ROC |z| > ⅓ only, i.e. to a different sequence.)