Expand $X(z) = \frac{1}{1+az}$ into a causal sequence

Question

For a HW problem, I'm told to expand $\frac{1}{1+az}$ into a causal and noncausal sequence. I found the noncausal sequence by long division (the result is $1-az+(az)^2-\dots$) and found the region of convergence, but I'm not sure how to find the causal sequence. The full question is :

Answer 1

If $\frac{1}{1+az}=\sum b_n z^{-n}$, then plugging in $w=1/z$, we get $\frac{1}{1+a/w}=\sum b_n w^n$. So let us rewrite $\frac{1}{1+a/w}=\frac{1}{\frac{w+a}{w}}=\frac{w}{a+w}=(w/a)\frac{1}{1-(-w/a)}=(w/a)\sum (-w/a)^n$. Now just expand and plug back in for $z$.

Answer 2

HINT

You have $$ \frac{1}{1+az} = \frac{1}{az} \left[ \frac{1}{1 + 1/(az)} \right] $$