Inverting density in favour of fugacity In these notes on pages 80 and 81 the following step was used
The density in terms of fugacity is 
$$
\frac{N}{V} = \frac{z}{\lambda^3}\left ( 1+ \frac{z}{2 \sqrt{2}} + \ldots \right )
$$
and this is inverted to 
$$
z = \frac{\lambda^3 N}{V} \left ( 1 - \frac{1}{2 \sqrt{2}} \frac{\lambda^3 N}{V} +\ldots \right )
$$
Using the approximation that $z$ and $\lambda^3N/V$ are much less than 1. Frankly I'm bewildered by what they did. Would someone be able to explain? 
 A: Ok. After experimenting a bit I think I answered the question, although confirmation would be useful. 
I will write $N\lambda^3/V = \alpha$ and $1/2 \sqrt{2} = \gamma$ so that we have 
$$ 
\alpha = z + \gamma z^2
$$
or that
$$ 
z = \frac{-1 + \sqrt{1+ 4\gamma \alpha}}{2\gamma}
$$
Taylor expanding to second order, 
$$
-\frac{1}{2 \gamma} + \frac{1}{2 \gamma}+ \frac{\frac12 4 \gamma \alpha}{2 \gamma} -\frac{ \frac14 (4\gamma)^2 \alpha^2}{4 \gamma} 
$$
or
$$
\alpha -\gamma \alpha^2 = \frac{\lambda^3 N}{V} \left ( 1 - \frac{1}{2 \sqrt{2}} \frac{\lambda^3 N}{V} \right)
$$
A: Let $\phi(z)=\sum_{k\geq 1} a_k z^k$ be absolutely convergent and invertible in a neighborhood of $z=0$, with $a_1\neq 0$. Let us denote its (compositional) inverse by $\phi^{-1}(u)=\sum_{k\geq 1} b_k u^k$.
It is not difficult to check that the following relations hold (see below):
$$
\sum_{n=1}^m a_n\sum_{\substack{k_1,\dots,k_n\geq 1\\ k_1+\cdots +k_n=m}}
\prod_{i=1}^n b_{k_i}
=
\begin{cases}
1&\text{ if }m=1\,,\\
0&\text{ otherwise.}
\end{cases}
\tag{1}
$$
From this, it is easy to express the coefficients $(b_k)_{k\geq 1}$ in terms of the coefficients $(a_k)_{k\geq 1}$. One thus obtains, for example:
$$
b_1=\frac{1}{a_1},\quad
b_2=-\frac{a_2}{a_1^3},\quad
b_3=2\frac{a_2^2}{a_1^5}-\frac{a_3}{a_1^4},
\quad\text { etc. }
$$
In your case, you have $u=N/V$ and
$$
a_1=1/\lambda^3,\quad a_2=1/(2\sqrt{2}\lambda^3).
$$
Appying the above, one then gets
$$
b_1 = \lambda^3,\quad b_2=-\lambda^6/(2\sqrt{2}),
$$
as desired. Note that this approach can be used to extract the series for the (compositional) inverse to arbitrary order.

Let me briefly explain how relation (1) is derived. On the one 
hand, by definition,
$$
\phi(\phi^{-1}(u)) = u.
\tag{2}
$$
On the other hand,
\begin{eqnarray}
\phi(\phi^{-1}(u))
&=&
\sum_{n\geq 1} a_n \Bigl( \sum_{k\geq 1} b_k u^k \Bigr)^n \\
&=&
a_1 b_1 u + (a_1 b_2 + a_2 b_1^2) u^2 + \ldots \\
&=&
\sum_{m\geq 1} \Bigl\{ \sum_{n=1}^m a_n 
\sum_{\substack{k_1,\ldots,k_n\geq 1\\k_1+\cdots+k_n=m}} \prod_{i=1}^n b_{k_i} 
\Bigr\} u^m.
\tag{3}
\end{eqnarray}
It only remains to match the coefficients in the two expressions (2) and (3): the coefficient of $u$ must be $1$, and therefore
$$
a_1 b_1 = 1 \implies b_1 = 1/a_1.
$$
The coefficient of $u^2$ (and all higher powers of $u$) must be $0$, from which we deduce that
$$
a_1 b_2 + a_2 b_1^2 = 0 \implies b_2 = - a_2 b_1^2 / a_1 = - a_2 / a_1^3. 
$$
Proceeding in the same way for the coefficient of $u^m$ 
yields (1).
