Different answers with the property $\arg (z_1 z_2)= \arg z_1 + \arg z_2$

Question

I know the property $\arg (z_1z_2)= \arg z_1 + \arg z_2$.

So if I have $z_1=-1$ och $z_2= a+ib$ I get \begin{align} \arg (z_1 z_2) &= \arg(-1) + \arg (a+ib)\tag 1\\ &=\arctan \Big(\frac{0}{-1}\Big )+\pi + \arctan \Big (\frac{b}{a}\Big )\tag 2 \end{align} I guess $\pi$ is correct because $z_1=-1$ lies on the negative real axis.

However, if I first muliply $z_1$ och $z_2$ I get $$ z_1 z_2 = (-1)(a+ib) = -a -ib $$ But now the argument is wrong, e.g. \begin{align} \arg(z_1 z_2)= \arg (-a-ib) = \arctan\Big ( \frac{-b}{-a} \Big ) = \arctan \Big (\frac{b}{a} \Big )\tag 3 \end{align}

So equation $2$ and $3$ aren't the same. Shouldn't they be?

Answer 1

We have $\arg z = \operatorname{Arg} z + k2\pi ,k \in \mathbb{Z}$. From $\arg \left( {{z_1}{z_2}} \right) = \arg {z_1} + \arg {z_2}$, we have $$\arg \left( {{z_1}{z_2}} \right) = \arg {z_1} + \arg {z_2} \Leftrightarrow \operatorname{Arg} \left( {{z_1}{z_2}} \right) = \operatorname{Arg} \left( {{z_1}} \right) + \operatorname{Arg} \left( {{z_2}} \right) + k2\pi ,\,k \in \mathbb{Z}.$$ $\operatorname{Arg} z$ is principle argument of $z$.