You are probably dividing by $\alpha$ at some point to eliminate a global phase, leading to your divide by zero in some cases. It would be better to get the phase angles of $\alpha$ and $\beta$ with $\arg$, and set the relative phase $\phi=\arg(\beta)-\arg(\alpha)$. Angle $\theta$ is now simply extracted as $\theta = 2\cos^{-1}(|\alpha|)$ (note that the absolute value of $\alpha$ is used). This is all assuming that you want to get to

$$|\phi\rangle = \cos(\theta/2)|0\rangle + \mathrm{e}^{i\phi}\sin(\theta/2)|1\rangle\,,$$

which neglects global phase.

You are probably dividing by $\alpha$ at some point to eliminate a global phase, leading to your divide by zero in some cases. It would be better to get the phase angles of $\alpha$ and $\beta$ with $\arg$, and set the relative phase $\phi=\arg(\beta)-\arg(\alpha)$. Angle $\theta$ is now simply extracted as $\theta = 2\cos^{-1}(|\alpha|)$ (note that the absolute value of $\alpha$ is used). This is all assuming that you want to get to

$$|\psi\rangle = \cos(\theta/2)|0\rangle + \mathrm{e}^{i\phi}\sin(\theta/2)|1\rangle\,,$$

which neglects global phase.