Is there is difference in proving Arg(zw)=Arg(z)+Arg(w) and arg(zw)=arg(z)+arg(w)?

Question

I am trying to prove/disprove $\operatorname{Arg}(zw)=\operatorname{Arg}(z)+\operatorname{Arg}(w)$. Apparently $\operatorname{Arg}(zw)=\operatorname{Arg}(z)+\operatorname{Arg}(w)+2k\pi$ where $k=0,1,\text{ or }-1$, but I have no idea why. I keep on finding that answer online. I am very lost on how to prove this statement, any help would be great. Thank you.

Answer 1

Let's say we have two complex numbers $\text{z}_1$ and $\text{z}_2$ we can write:

• $$\text{z}_1=\left|\text{z}_1\right|\cdot\exp\left(\left(\arg\left(\text{z}_1\right)+2\pi\cdot\text{k}_1\right)\cdot i\right)\tag1$$

Where $0\le\arg\left(\text{z}_1\right)<2\pi$ and $\text{k}_1\in\mathbb{Z}$

• $$\text{z}_2=\left|\text{z}_2\right|\cdot\exp\left(\left(\arg\left(\text{z}_2\right)+2\pi\cdot\text{k}_2\right)\cdot i\right)\tag1$$

Where $0\le\arg\left(\text{z}_2\right)<2\pi$ and $\text{k}_2\in\mathbb{Z}$

So, we get:

$$\text{z}_1\cdot\text{z}_2=\left|\text{z}_1\right|\cdot\exp\left(\left(\arg\left(\text{z}_1\right)+2\pi\cdot\text{k}_1\right)\cdot i\right)\cdot\left|\text{z}_2\right|\cdot\exp\left(\left(\arg\left(\text{z}_2\right)+2\pi\cdot\text{k}_2\right)\cdot i\right)=$$ $$\left|\text{z}_1\right|\cdot\left|\text{z}_2\right|\cdot\exp\left(\left(\arg\left(\text{z}_1\right)+2\pi\cdot\text{k}_1\right)\cdot i+\left(\arg\left(\text{z}_2\right)+2\pi\cdot\text{k}_2\right)\cdot i\right)=$$ $$\left|\text{z}_1\right|\cdot\left|\text{z}_2\right|\cdot\exp\left(2\pi\cdot\left(\arg\left(\text{z}_1\right)+\arg\left(\text{z}_2\right)+\text{k}_1+\text{k}_2\right)\cdot i\right)\tag3$$

Answer 2

It can be shown in many ways. The simplest is to consider the exponential form of complex numbers $$z=\rho e^{i\theta}$$

with $|z|=\rho$ and $Arg(z)=\theta$

Answer 3

Denote $z=r_1\cdot e^{i\theta_1}$, $z=r_2\cdot e^{i\theta_2}$.

Then $zw=r_1\cdot e^{i\theta_1}\cdot r_2\cdot e^{i\theta_2}=r_1r_2e^{i(\theta_2+\theta_2)}$, and since the argument of a complex number $re^{i\theta}$ is $\theta$, we receive that $Arg(z)+Arg(w)=\theta_1+\theta_2=Arg(zw)$ up to $\pm 2\pi$.

Answer 4

"but I have no idea why" ... Because exponential is periodic with period $2 \pi i$ over the complex plane. This is a picture from a Mathcad worksheet, but try this with your calculator: