Proving $\arg(zw)=\arg(z)+\arg(w)$

Question

This is my attempt I know this is incomplete or may even be wrong.

Let $θ_1 \in \arg(z)$ and $θ_2 \in \arg(w)$. Then, $θ_1+θ_2 \in \arg(z)+\arg(w)$.

Also, $θ_1+θ_2 \in \arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.

EDIT: Just to avoid any confusions I will add how $\arg z$ is defined.

$\arg z=\{\text{Arg}\, z+2\pi k\mid k\text{ is an integer}\}$

Answer 1

Here's a proof: Let $z=r e^{i\theta}$ and $w=s e^{i\phi}$. Then $$\text{Arg}(zw)=\text{Arg}\left(rse^{i\theta}e^{i\phi}\right)=\text{Arg}\left(rse^{i(\theta+\phi)}\right)=\text{Arg}(z)+\text{Arg}(w)\quad(\text{mod }2\pi),$$ where $\text{Arg}(z)\in[0,2\pi)$ is the principal argument of $z$.

However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^{i(\theta + 2\pi n)}$$ and $$w=se^{i(\phi+2\pi m)},$$ where $n,m\in\mathbb{Z}$.

Answer 2

I'm not sure we can use the $w = e^{i\theta{}_w}$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.

So we have $\text{arg}(w) \in \theta{}_w$ and $\text{arg}(z) \in \theta{}_z$.

This means $w = |w|(\cos{\theta{}_w}+\sin{\theta{}_w})$ and $z = |z|(\cos{\theta{}_z}+\sin{\theta{}_z})$.

By multiplication : $wz = |{w}| |{z}| ((\cos \theta_w \cos \theta_z - \sin \theta_w \sin \theta_z) + i (\cos \theta_w \sin \theta_z + \sin \theta_w \cos \theta_z))$ which happens to be equal to $|{w}| |{z}| (\cos (\theta_w + \theta_z) + i \sin (\theta_w + \theta_z)).$

Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $\dfrac x {|wz|} = \cos (\theta_w + \theta_z)$ and $\dfrac y {|wz|} = \sin (\theta_w + \theta_z)$.

This is the definition of $(\theta_w + \theta_z)$ belonging to $\text{arg}(wz)$.

Answer 3

Let $z=r_1 e^{i \theta_1}$ and $w=r_2 e^{i \theta_2}$, then $ zw=r_1 r_2 e^{i (\theta_1+\theta_2)}. $ Hence $$ \arg(z)=\theta_1 + 2k\pi, k\in \mathbf{Z}\\ \arg(w)=\theta_2+2k\pi, k\in \mathbf{Z}\\ \arg(zw)=\theta_1+\theta_2+2k\pi, k\in \mathbf{Z}. $$ There it is: $\arg(zw)=\arg(z)+\arg(w)$.