Conformal Mapping defined by Cubic Polynomial Ahlfors studies the mapping defined by $\omega = a_0 z^3 + a_1 z^2 + a_2 z + a_3$ in Complex Analysis 3rd Edition at the bottom of page 95.
First he says we can get rid of the quadratic term by the substitution $z = z_1 - a_1/ 3 a_0$. Indeed, when I do the substitution, I get $\omega = a_0 z_1^3 + a_2 z_1 + K$ where $K$ is some messy constant. I wonder if there is some intuition why this would work?
I don't really understand what happens afterwards. He says "by obvious normalizations we can reduce the polynomial to the form $\omega = z^3 - 3z.$" What does normalization mean? Should I be trying to come up with another substitution that could produce this equation?
Then "we introduce an auxiliary variable $\zeta$ defined by $z = \zeta + 1/\zeta$. Our cubic polynomial then takes the simple form $\omega = \zeta^3 + 1/\zeta^3$" This substitution works out given the previous $\omega = z^3 - 3z$ but it was kind of surprising. Again, is there some intuition here?
From here Ahlfors analyzes the behavior of $\omega$ against $\zeta$
 A: Understanding $\omega = a_0 z^3 + a_21 + K$ is the same as understanding it shifted by $-K$, i.e. understanding $\omega = a_0z^3 + a_2z$. 
Now we can normalize $z$ and $\omega$ multiplicatively to handle $a_0$ and $a_2$ to be anything we choose - two constraints, two goals.
We will replace $\omega$ by $b \omega$ (where I abuse notation and refer to $\omega$ as both new and old variable) and $z$ by $cz$. This leads to $b\omega = a_0 c^3 z^3 + a_2 cz$, whose behaviour is completely contained within $\omega = \dfrac{a_0 c^3}{b}z^3 + \dfrac{a_2c}{b}z$.
So we want $\dfrac{a_0c^3}{b} = 1$ and $\dfrac{a_2c}{b} = -3$. The latter is more useful. It means that $b = -\dfrac{a_2c}{3}$, so that $-\dfrac{3a_0c^2}{a_2} = 1$, or rather $c = i\sqrt{\dfrac{a_2}{3a_0}}$. With these normalizations, we are reduced to $\omega = z^3 - 3z$.
It's hidden within Ahlfors's exposition, but the reasoning is a little backwards. We could have chosen anything instead of $1$ and $-3$, but we chose these precisely because they are what makes the substitution $z = \zeta + \zeta^{-1}$ work.
As an aside, if you have studied Cardano's solution to the cubic, you may notice thematic similarities (except that Ahlfors has a far more sophisticated approach and viewpoint).
