For example, it is known that if |c| = |d|, then the linear fractional transformation:
$w(z)=\frac{az+b}{cz+d}, ad-bc \neq 0$ should map the unit circle ($z=e^{i \theta})$ to a line. How would I know that, and see what line that the function maps to. $w(z),z \in \mathbb{C}$ of course. Any help is appreciated.
The following is an elementary proof which assumes no prior knowledge of Möbius transformations. From the given relation:
$$w(cz+d)=az+b \quad\iff\quad z(wc-a)=b-wd \tag{1}$$
Taking the complex conjugate of both sides:
$$\bar z(\bar w \bar c- \bar a)=\bar b-\bar w \bar d \tag{2}$$
Multiplying $(1)\cdot (2)$ and dropping $z \bar z = |z|^2=1\,$:
$$\require{cancel} (wc-a)(\bar w \bar c- \bar a)=(b-wd)(\bar b-\bar w \bar d) \\ \iff\quad |w|^2|c|^2 - wc\bar a - \bar w \bar c a+|a|^2=|b|^2-w \bar b d - \bar w b \bar d + |w|^2|d|^2 \\ \iff\quad \cancel{|w|^2\left(|c|^2-|d|^2\right)} +w(\bar b d - \bar a c) +\bar w(b \bar d - a \bar c) + |a|^2-|b^2| = 0 $$
The first term cancels out in the latter equation because $|c|=|d|\,$, then what remains is an equation of the form $\alpha w + \bar \alpha \bar w + \lambda = 0$ with $\alpha=\bar b d - \bar a c$ and $\lambda = |a|^2-|b^2| \in \mathbb{R}$ which represents a line in the complex plane.
[ EDIT ] Plugging in values of $z$ on the unit circle $|z|=1$ will give corresponding points on the $w$ line. For example $z=\pm 1$ shows that the line passes through $\,\frac{a+b}{c+d}\,$ and $\,\frac{a-b}{c-d}\,$ (when $\,c \ne \pm d\,$).
The intercepts with the axes can be determined by solving for $w = \pm \bar w\,$. For example the real axis intercept must satisfy $w=\bar w$ and calculates to $w=\frac{-\lambda}{\alpha + \bar \alpha}=\frac{-|a|^2+|b|^2}{\bar b d - \bar a c+b \bar d - a \bar c}\,$ (when $\Re(b \bar d-a \bar c)\ne0$).