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

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?

• Yes you need more restrictions for this to work. – Simply Beautiful Art Sep 4 at 11:45

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 $$z$$.