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
$$
{df\over dz}(\rho \cdot e^{it}) = \rho e^{it} + 0 \cdot e^{it} \text{?} = \rho e^{it} \ne i \rho e^{it} \text{?}
$$

The product rule actually states:
$${\mathrm{d}\over \mathrm{d}z}(f_1 \cdot f_2) = f_1 \frac{\mathrm{d}f_2}{\mathrm{d}z} + \frac{\mathrm{d}f_1}{\mathrm{d}z} \cdot f_2$$
Put this together with $\frac{\mathrm{d}\rho}{\mathrm{d}z} \neq 0$ and $\frac{\mathrm{d}\mathrm{e}^{\mathrm{i}t}}{\mathrm{d}z} \neq \mathrm{e}^{\mathrm{i}t}$ and the answer should be evident.
