Is there a geometric insight from this exercise? I am solving some exercises in a book I'm reading and so far all the exercises contained some insight. But then I got to the following exercise:
Let $z,a\in \mathbb C$. Show that
$$ (1-|z|^2)(1-|a|^2) =|1-z\overline{a}|^2 - |z-a|^2$$
Then deduce from this that if $|a|<1$ then
$$ |z|<1 \iff \left| {z-a \over \overline{a}z -1} \right|<1$$
and
$$ |z|=1 \iff \left| {z-a \over \overline{a}z -1} \right|=1$$

What is the insight to be had from this exercise?

All the other exercises were insightful but this one here seems to be just computational. 
Note that this is part b) of the exercise. Part a) was:
Let $\mathbb H := \{z \in \mathbb C \mid \operatorname{Im}{z}>0 \}$. Show that $z \in \mathbb H$ if and only if $-{1\over z}\in \mathbb H$.
I solved part a) and I feel that it should be somehow related to part b) but I don't know how.
 A: $\newcommand{\Cpx}{\mathbf{C}}$You're secretly describing certain automorphisms (holomorphic bijections with holomorphic inverse) of the open unit disk.
The open unit disk plays a starring role in complex analysis: It turns out that every connected, simply-connected, proper open subset of the complex plane is "biholomorphic" to the open disk (the so-called Riemann mapping theorem).

Edit: There was a mistake in my original answer, in that part (b) doesn't describe all automorphisms of the disk, but only a certain family of involutions (automorphisms that are their own inverse).
To fill in some of the background, here's a brief sketch of the Riemann sphere as the complex projective line, Möbius transformations as $2 \times 2$ complex matrices, and a more detailed interpretation of part (b).
The complex projective line is the set of (complex) lines through the origin in $\Cpx^{2}$. Algebraically, consider ordered pairs $(z_{0}, z_{1})$ of complex numbers (not both zero) modulo the equivalence relation
$$
(z_{0}, z_{1}) \sim (\lambda z_{0}, \lambda z_{1})\quad\text{for all $\lambda \neq 0$.}
$$
Associating $z$ in $\Cpx$ with the line through $(z, 1)$ (labeled $V_{0}$ in the figure below), we represent every line except the line through $(1, 0)$; since $(z, 1) \sim (1, \frac{1}{z})$ for $z \neq 0$, it's reasonable to view $(1, 0)$ as "$\infty$"; doing this identifies the complex projective line with the Riemann sphere. (There's a bit more detail in another post, from which the following figure is recycled.)

An "automorphism" of the projective line/Riemann sphere is a bijection that is meromorphic in the local coordinate $z$. To make a moderately lengthy story short, every automorphism of the projective line turns out to be induced by a linear automorphism of $\Cpx^{2}$, via the correspondence
$$
\left[\begin{array}{@{}rr@{}}
    a & b \\
    c & d \\
  \end{array}\right]
  \left[\begin{array}{@{}c@{}}z \\ 1\end{array}\right]
  = \left[\begin{array}{@{}c@{}} az + b \\ cz + d\end{array}\right]
  \sim \left[\begin{array}{@{}c@{}}\dfrac{az + b}{cz + d} \\ 1\end{array}\right],\quad ad - bc \neq 0.
$$
(This is the origin of the term "fractional linear" transformation.) In particular, composition of Möbius transformations corresponds to matrix multiplication.
The coefficient matrix can be multiplied by a non-zero constant without changing the associated Möbius transformation, and it's customary to normalize by taking $ad - bc = 1$, i.e., to identify projective automorphisms with $SL(2, \Cpx)/{\pm}$.

As to part (b) of your question: The Möbius transformation $f(z) = \dfrac{z - a}{\overline{a}z - 1}$ corresponds to the matrix
$$
A = \left[\begin{array}{@{}rr@{}}
    1 & -a \\
    \overline{a} & -1
  \end{array}\right].
$$
Part (b) asserts that if $|a| < 1$, then $|z| < 1$ if and only if $|f(z)| < 1$, and $|z| = 1$ if and only if $|f(z)| = 1$. In other words, $f$ maps the open unit disk to itself, and maps the unit circle to the unit circle. Moreover, $A^{2} = (1 - |a|^{2}) I \sim I$, so $f$ is an involution of the disk.
Incidentally, the mapping $g(z) = -1/z$ in part (a) is an involution of the upper half plane. It's not difficult to show that the Möbius transformations
$$
\phi(z) = \frac{z - i}{z + i}
  \leftrightarrow \left[\begin{array}{@{}rr@{}}
      1 & -i \\
      1 &  i \\
    \end{array}\right],\quad
\phi^{-1}(z) = i\frac{1 + z}{1 - z}
  \leftrightarrow \left[\begin{array}{@{}rr@{}}
      i & i \\
     -1 & 1 \\
    \end{array}\right]
$$
map the open upper half plane to the open disk (and vice versa), and it's an easy exercise to check that $(\phi \circ g \circ \phi^{-1})(z) = -z$, the "$a = 0$" mapping in part (b).
