The derivative of $\rho e^{it}$ Why is 
$${df \over dz} \rho e^{it} = i \rho e^{it} \text{?}$$
The product rule states that
$$
{df\over dz}(f_1 \cdot f_2) = f_1 f'_2 + f'_1 \cdot f_2
$$
so why doesn't this imply that
$$
{df\over dz}(\rho \cdot e^{it}) = \rho e^{it} + 0 \cdot e^{it} \text{?} = \rho e^{it} \ne i \rho e^{it} \text{?}
$$

You're confusing three different things with each other:
$$
\frac{df}{dz}, \qquad \frac{d}{dz}, \qquad\frac{d}{dt}
$$
If you had written
$$
\frac{d}{dt} \rho e^{it} = \rho ie^{it}
$$
then it would be correct, but what you have written is at best a misunderstanding of notation.
Applying the product rule, one gets
$$
\begin{align}
\frac{d}{dt} \rho e^{it} & = \rho\frac{d}{dt} e^{it} + e^{it}\frac{d\rho}{dt} \\[12pt]
& = \rho\frac{d}{dt}e^{it} + e^{it}\cdot 0 \\[12pt]
& = \rho\frac{d}{dt}e^{it}.
\end{align}
$$
Next, use the chain rule:
$$
\rho\frac{d}{dt} e^{it} = \rho e^{it} \frac{d}{dt}(it).
$$
