Finding PMF of geometric random variable from PGF I need help understanding the solution to this question.

The PGF of a geometric random variable with mean $\frac{1}{p}$ is
$$P_N(z) = \frac{pz}{1-qz}, |z| < \frac{1}{q}$$
where $q = 1-p$.
Consider 2 independent geometric random variables $N_1$ and $N_2$,
  with probabilities $p_1$ and $p_2$ respectively. If $p_1 \neq p_2$,
  and given the PGF of $M = N_1 + N_2$ as
$$P_M(z) = \frac{p_1 p_2 z^2}{p_2 -p_1}(\frac{q_1}{1-q_1z}-\frac{q_2}{1-q_2z})$$
Find the PMF of M.

The solution is as follows.

Recall that
$$\sum_{k=0}^{\infty}a^kz^k = \frac{1}{1-az}, |z| < \frac{1}{a}$$ and
  that $Z^{-1}[z^NX(z)] = x_{k-N}$ if $x_k \leftrightarrow X(z)$. Then
  we see that
$$Z^{-1}[\frac{q_1}{1-q_1z}] = q_1 \cdot q_1^k = q_1^{k+1}, k =
0,1,2,...$$ And the linearity of the z transform then yields $$p_M(k)
= \frac{p_1p_2(q_1^{k-1}-q_2^{k-1})}{p_2-p_1}, k=2,3,4,...$$

My lecturer said the that z-transform of $a^k$ is $\frac{1}{1-az}$, and the inverse transform of $\frac{1}{1-az}$ is $\sum_{k=0}^{\infty}a^kz^k$ (not sure if I copied this correctly). How do I relate this to the "$Z^{-1}[z^NX(z)] = x_{k-N}$ if $x_k \leftrightarrow X(z)$" in the solution?
Also, how do you get the $p_M(k)$? Where did the $z^2$ disappear to?
 A: I'm not familiar with the notation you use, but here's what is going on.
First of all, let's try and understand what this means:
$$ Z^{-1}[z^NX(z)] = x_{k-N}, \quad \mbox{if } x_k \leftrightarrow X(z).$$
Going back to definitions, I assume that $X(z)$ is equal to
$$ X(z) = \sum_{k = 0}^\infty x_k z^k.$$
If you multiply by $Z^N$, you get
$$ z^N X(z) = z^N\sum_{k = 0}^\infty x_k z^k = \sum_{k = 0}^\infty x_k z^{k + N} = \sum_{k = 0}^\infty x_{k - N} z^k,$$
with the convention that $x_{j} = 0$ whenever $j < 0$. In other words, the $k$-th element of the generating function $z^N X(z)$ is the $(k - N)$-th element of the generating function $X(z)$. 
What about the rest? Well, let's expand your probability generating function (PGF) above:
\begin{align*}
P_M(z) &= \frac{p_1 p_2 z^2}{p_2 -p_1}\left(\frac{q_1}{1-q_1z}-\frac{q_2}{1-q_2z}\right)\\
&= \frac{p_1 p_2 z^2}{p_2 -p_1}\left(q_1\sum_{k=0}^{\infty}q_1^kz^k-q_1\sum_{k=0}^{\infty}q_1^kz^k\right)\\
&= \frac{p_1 p_2}{p_2 -p_1}\left(\sum_{k=0}^{\infty}q_1^{k + 1}z^{k + 2} - \sum_{k=0}^{\infty}q_2^{k + 1}z^{k + 2}\right)\\
&= \frac{p_1 p_2}{p_2 -p_1}\sum_{k=0}^{\infty}(q_1^{k + 1} - q_2^{k + 1}) z^{k + 2}\\
&= \frac{p_1 p_2}{p_2 -p_1}\sum_{k=0}^{\infty}(q_1^{k - 1} - q_2^{k - 1}) z^{k}\\
&= \sum_{k=0}^{\infty}\left(\frac{p_1 p_2(q_1^{k - 1} - q_2^{k - 1})}{p_2 -p_1}\right) z^{k}.\\
\end{align*}
Again, we need to follow the convention (which, I agree, is akward in this situation), that $q_j^{-1} = 0$. In any case, you now get a valid probability generating function which can easily be understood in terms of  a probability mass function:
$$ p_M(k)
= \frac{p_1p_2(q_1^{k-1}-q_2^{k-1})}{p_2-p_1}, k= 1,2,3,4,\ldots $$
(Note that the formula in your prof's solutions is also valid for $k=1$. 
Finally, to get the PGF $P_M(z)$ in the first place, you need to use the fact that the PGF of a sum of two independent random variable is the product of their respective PGFs:
$$P_M(z) = P_{N_1}(z)P_{N_2}(z).$$
