$0$ in Polar form

Question

I want to prove that for $z_1, z_2 \in \mathbb{C}$, then $$ \text{arg}_I (z_1 z_2) = \text{arg}_I (z_1 )+ \text{arg}_I (z_2) $$

The proof is staighforward, but I was wondering what to do about $z = 0$, because I solved the general case using polar form of $z_1$ and $z_2$, namely $z= re^{i \theta} $, where $r > 0 $.

The statement is supposed to hold for all $z \in \mathbb{C}$, but $0 \in \mathbb{C}$ and is not included when talking about complex numbers in polar form.

Answer 1

Without using the De Moivre's theorem, you may approach it as \begin{align*} z_{1}z_{2} & = \rho_{1}\rho_{2}[\cos(\theta_{1}) + i\sin(\theta_{1})][\cos(\theta_{2}) + i\sin(\theta_{2})]\\ & = \rho_{1}\rho_{2}[(\cos(\theta_{1})\cos(\theta_{2}) - \sin(\theta_{1})\sin(\theta_{2}) + i(\cos(\theta_{1})\sin(\theta_{2}) + \sin(\theta_{1})\cos(\theta_{2})]\\ & = \rho_{1}\rho_{2}[\cos(\theta_{1}+\theta_{2}) + i \sin(\theta_{1} +\theta_{2}) \Longrightarrow \arg(z_{1}z_{2}) = \arg(z_{1}) + \arg(z_{2}) \end{align*}

If you are allowed to use complex analysis' results, you have \begin{align*} z_{1}z_{2} = \rho_{1}\rho_{2}\exp(i\theta_{1})\exp(i\theta_{2}) = \rho_{1}\rho_{2}\exp(i(\theta_{1}+\theta_{2})) \Longrightarrow \arg(z_{1}z_{2}) = \arg(z_{1})+\arg(z_{2}) \end{align*}