Z-transform of k a^(k−1)

Use $\mathcal{Z}[a^k] = \frac{z}{z-a}$ and the property $\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z)$.

$$\begin{aligned} \mathcal{Z}[k a^k] &= -z\frac{d}{dz}\left(\frac{z}{z-a}\right) = -z\cdot\frac{(z-a) - z}{(z-a)^2} = \frac{az}{(z-a)^2} \end{aligned}$$

Dividing by $a$:

$$\mathcal{Z}[k a^{k-1}] = \frac{z}{(z-a)^2}, \quad |z| > |a|$$

Alternative (differentiate with respect to $a$): $\sum_{k=0}^{\infty} a^k z^{-k} = \frac{z}{z-a}$; differentiating both sides with respect to $a$ gives $\sum k a^{k-1}z^{-k} = \frac{z}{(z-a)^2}$.

Check: $\frac{z}{(z-a)^2} = z^{-1} + 2az^{-2} + 3a^2z^{-3} + \cdots$, i.e. coefficients $k a^{k-1}$ ✔.

Answer: $\mathcal{Z}[k a^{k-1}] = \dfrac{z}{(z-a)^2} = \dfrac{z^{-1}}{(1-az^{-1})^2}$.