Maximum Number of Regions in the Plane Using Zig Lines? I'm having trouble understanding how the book came up with the formula for calculating $Z_n$, the maximum number of regions created using $n$ zig lines (a zig line is a line containing just one sharp bend). Earlier in the book (Page 5), we derived $L_n$, the maximum number of regions created using $n$ straight lines. What follows is the derivation of $Z_n$:




First, I'm not exactly sure what the underlined sentence means. From the picture of $Z_2$, my only guess at its meaning is that none of the zigs can contain the zig point of another zig.
My other question is about the assumptions that the book seems to be making in its argument (written in color):
1) Imagine $n$ zigs, and assume they split the plane into $Z_n$ regions. Extend all the lines as in the example. This creates a plane split by $2n$ lines $\color{red}{\text{into } L_{2n} \text{regions}}.$
2) Now, erase the extensions. For each extension we erase, $\color{blue}{\text{we lose 2 regions}}.$
3) Therefore $Z_n$ zigs split the plane into $L_{2n}-2n$ regions.
The blue assumption surely has something to do with my underlined sentence, but its not obvious to me that in very complicated figure with many intersections, we lose exactly two regions per deletion of one extensions.
Could someone help make the justification for these assumptions more clear? 
Thanks.
 A: Suppose that you already have $n$ zig lines in place, for a total of $2n$ rays. You draw a line $\ell_1$ that intersects all $2n$ rays. Then you draw a line $\ell_2$ that intersects that line and all $2n$ rays. Let $p$ be the point of intersection of $\ell_1$ and $\ell_2$, and for $i=1,2$ let $\{p_{i,k}:k=1,\ldots,2n\}$ be the set of points of intersection of $\ell_i$ with the $2n$ original rays. The underlined statement means that the points $p_{1,k}$ must all lie on the same side of $p$ on the line $\ell_1$, and the points $p_{2,k}$ must all lie on the same side of $p$ on the line $\ell_2$. When we erase half of $\ell_1$ and half of $\ell_2$ to convert them to a zig line with its bend at $p$, we’ll keep the half of $\ell_1$ that contains the points $p_{1,k}$ and the half of $\ell_2$ that contains the points $p_{2,k}$.
I don’t think that it’s immediately clear that this ensures that $Z_n=L_{2n}-2n$ for each $n\in\Bbb N$, but we can show this by induction on $n$.
The line $\ell_1$ cut each of the original $2n$ rays, so it split $2n+1$ regions in two and therefore added $2n+1$ regions. Then $\ell_2$ cut each of the $2n$ original rays and $\ell_1$, so it split $2n+2$ regions in two and so added another $2n+2$ regions. However, when we erase the halves of $\ell_1$ and $\ell_2$ that don’t contain the points of intersection with the original rays, we collapse three regions into one: the region between the two rays that we’re erasing merges with the two regions adjacent two it across those erased rays, and of course that means that they merge into each other as well. (Note that the original $2n$ rays never enter the region between the two rays that we’re erasing, and exactly what they do outside of it is irrelevant to this point.)
Assuming as an induction hypothesis that $Z_n=L_{2n}-2n$, this shows that 
$$\begin{align*}
Z_{n+1}&=Z_n+(2n+1)+(2n+2)-2\\
&=L_{2n}-2n+(2n+1)+(2n+2)-2\\
&=L_{2n}+2n+1\\
&=L_{2n+1}\\
&=L_{2n+2}-(2n+2)\;,
\end{align*}$$
and since $Z_0=1=L_0-2\cdot0\cdot1$, we have a proof by induction that $Z_n=L_{2n}-2n$ for each $n\in\Bbb N$.
A: Alternate Solution :
Suppose we already have the number of regions formed by first $n-1$ bent lines i.e. $Z_{n-1}$.
Now we add the $n^{th}$ bent line, the point of intersection(let it be $P$) will lie in certain arbitrary region and now the 2 half lines can be made to intersect all currently existing line which are L = $2*(n-1)$ in number. Since each intersection will result in a new region being formed just as the case with $n$ straight lines. Since there are 2 new lines this will create $2*L = 2*2*(n-1)$ new region and one extra region formed by intersection at the bent (point $P$). So recurrence can be written as :
$$Z_n = Z_{n-1} + 2*2*(n-1) + 1$$
$$Z_n = Z_{n-1} + 4n - 3$$
with base case $Z_0 = 1$
$$Z_n = 4\sum_{i=1}^{n}n - 3\sum_{i=1}^{n}1 + Z_0$$
$$Z_n = 4.\frac{n(n+1)}{2} -3n + 1$$
$$Z_n = 2n^2 - n+1$$
