Let $z=\rho e^{i\phi}$ be a complex number and $\alpha$ some parameter. I determined the following ODE $$ \dot{\rho}e^{i\phi}+i\rho\dot{\phi}e^{i\phi}=\rho e^{i\phi}(\alpha+i-\rho^2). $$
How to get the polar form $$ \dot{\rho}=\rho(\alpha-\rho^2),\qquad\dot{\phi}=1 $$ from this? I do not see it...
If you assume $\rho$ and $\phi$ are real (and you do), then you're given a complex equation for two real variables. First, you divide both sides by $e^{i \phi}$. Then, you write both sides in the form $a+b i$, yielding \begin{equation} \big(\dot{\rho}\big) + i \big(\rho \dot{\phi}\big) = \big(\alpha \rho- \rho^3\big) + i \big(\rho\big). \end{equation} The complex number on the left hand side must equal the complex number on the right hand side, so both the real and imaginary parts must match. Therefore, you get two equations: \begin{align} \dot{\rho} &= \alpha \rho - \rho^3, \\ \rho \dot{\phi} &= \rho. \end{align} I'm sure you can take it from here.