ROC of (1/2)^k u(k) + (−1/3)^k u(k)

Both terms are right-sided (causal) exponentials. Each converges outside a circle, so the ROC of the sum is the overlap: $|z| > \tfrac12$.

$u(k)$ is taken as the unit step. The printed "$u(k) = 2$" would only multiply $X(z)$ by 2 and does not change the ROC.

Step 1: First term

$$\mathcal{Z}\left[\left(\tfrac12\right)^k u(k)\right] = \sum_{k=0}^{\infty}\left(\tfrac12 z^{-1}\right)^k = \frac{1}{1 - \frac12 z^{-1}} = \frac{z}{z - \frac12}$$

This converges when $\left|\tfrac12 z^{-1}\right| < 1$, i.e. $|z| > \tfrac12$.

Step 2: Second term

$$\mathcal{Z}\left[\left(-\tfrac13\right)^k u(k)\right] = \sum_{k=0}^{\infty}\left(-\tfrac13 z^{-1}\right)^k = \frac{z}{z + \frac13}$$

This converges when $|z| > \tfrac13$.

Step 3: Combine

$$\begin{aligned} X(z) &= \frac{z}{z - \frac12} + \frac{z}{z + \frac13} = \frac{z\left(z + \frac13\right) + z\left(z - \frac12\right)}{\left(z - \frac12\right)\left(z + \frac13\right)} \\ &= \frac{2z\left(z - \frac{1}{12}\right)}{\left(z - \frac12\right)\left(z + \frac13\right)} \end{aligned}$$

• Poles: $z = \tfrac12$, $z = -\tfrac13$
• Zeros: $z = 0$, $z = \tfrac{1}{12}$

Step 4: ROC of the sum

The ROC is the intersection of the two ROCs:

$$\{|z| > \tfrac12\} \cap \{|z| > \tfrac13\} = \{|z| > \tfrac12\}$$

            Im z
             |   ROC: outside
        .----+----.   circle r=1/2
       /     |     \
  ----x------o--x---+---- Re z
     -1/3    0  1/2  1
       \     |     /
        '----+----'

Remarks

• For a causal sequence the ROC lies outside the circle through the outermost pole ($|z| = \tfrac12$ here).
• The ROC contains the unit circle, so the sequence is absolutely summable (stable).

Answer: $X(z) = \dfrac{z}{z-\frac12} + \dfrac{z}{z+\frac13}$, ROC $|z| > \tfrac12$.