ROC of x(k) = −a^k u(−k−1)

$x(k) = -a^k u(-k-1)$ is a left-sided (anti-causal) sequence. It is nonzero only for $k \le -1$, where it equals $-a^k$. Its z-transform is $\dfrac{z}{z-a}$ with ROC $|z| < |a|$.

Step 1: Apply the definition

$$X(z) = \sum_{k=-\infty}^{\infty}x(k)z^{-k} = -\sum_{k=-\infty}^{-1}a^k z^{-k}$$

Put $m = -k$, so $m$ runs from 1 to $\infty$:

$$X(z) = -\sum_{m=1}^{\infty}a^{-m}z^{m} = -\sum_{m=1}^{\infty}\left(\frac{z}{a}\right)^m$$

Step 2: Sum the geometric series

The series converges only when $\left|\dfrac{z}{a}\right| < 1$, i.e. $|z| < |a|$. Then

$$X(z) = -\frac{z/a}{1 - z/a} = -\frac{z}{a - z} = \frac{z}{z-a} = \frac{1}{1 - az^{-1}}$$

Step 3: Region of convergence

$$\text{ROC: } |z| < |a|$$

This is the inside of a circle of radius $|a|$, centred at the origin. The pole at $z = a$ lies on the outer boundary.

           Im z
            |
        .---+---.
      /     |     \
     |  ROC: |z|<|a|
  ---+------0------x---- Re z
     |      |      a  (pole on
      \     |     /    boundary)
        '---+---'

Comparison with the right-sided sequence

Both sequences have the same algebraic $X(z)$. Only the ROC tells them apart, so an ROC must always be stated with a z-transform. For a left-sided sequence the ROC is always the inside of a circle bounded by the innermost pole.

Answer: $X(z) = \dfrac{z}{z-a}$, ROC $|z| < |a|$.