The principle argument of the product of two complex numbers in the second quadrant

Question

I would like some help to prove the following:

Show that, if Re $z_1<0$, Im $z_1>0$, Re $z_2<0$ and Im $z_2 >0$, then

Arg$(z_1z_2)=$Arg$(z_1)+$Arg$(z_2)-2\pi$.

Thanks for any help in advance.

Answer 1

Here we get that: $$z_1,z_2 \in Quad_2 \Rightarrow 0<\frac{\pi}{2} < Arg(z_1),Arg(z_2) < \pi$$

Express $$z_1 = r_1e^{\theta_1 i}, z_2 = r_2e^{ \theta_2 i} \Rightarrow z_1z_2 = r_1r_2e^{(\theta_1 + \theta_2)i}$$

From our relation it follows that $$Arg(z_1z_2)= Arz(z_1)+Arg(z_2)=\theta_1 + \theta_ 2 $$

We also have that $$0<\pi< Arg(z_1z_2) < 2 \pi$$ \

However, the above is not in the range of $Arg(z)$, which is typically defined as $(-\pi, \pi]$ so we need to subtract $2\pi$ allowing the $Arg(z_1z_2) = Arg(z_1)+Arg(z_2) - 2\pi$ to be properly defined.

$\textbf{Additional}$: Then, $-\pi<Arg(z_1z_2)<0$ which is in the range of Arg.