When does the argument of a complex number follow the sum of angles formula?

[Snowfall]

When is it true that $\arg\left(z_1 z_2\right) = \arg(z_1) + \arg(z_2)$?

 

[chiro]

When is it true that $\arg\left(z_1 z_2\right) = \arg(z_1) + \arg(z_2)$?

Why don't you just write out the equations for argument and solve them?

 

[Snowfall]

Why don't you just write out the equations for argument and solve them?

You mean like defining, for real numbers a, b, c, d,

$$z_{1} := ai+b, ~ z_{2} := ci+d$$

then solving the following daunting looking equation:

$$\tan^{-1}\left(\frac{ad+bc}{bd-ac}\right) = \tan^{-1}\frac{b}{a}+\tan^{-1}\frac{d}{c}.$$

Please by all means show me how to solve that one!

 

[uart]

You mean like defining, for real numbers a, b, c, d,

$$z_{1} := ai+b, ~ z_{2} := ci+d$$

then solving the following daunting looking equation:

$$\tan^{-1}\left(\frac{ad+bc}{bd-ac}\right) = \tan^{-1}\frac{b}{a}+\tan^{-1}\frac{d}{c}.$$

Please by all means show me how to solve that one!

Divide num and denom by $bd$ and you get :

$$\tan(\arg(z_1 z_2)) = \frac{a/b+c/d}{1 - (a/b) \,\, (c/d)} = \frac{\tan(arg(z_1))+\tan(\arg(z_2))}{1 - \tan(\arg(z_1)) \tan(\arg(z_2))} = \tan(\arg(z_1) + \arg(z_2))$$

However it's easier to use the exponential or polar form.

$$r_1 e^{i \phi_1} \, r_2 e^{i \phi_2} = r_1 \, r_2 \, e^{i(\phi_1 + \phi_2)}$$

 

[dalcde]

Since you have got inverse tangents, it's wise to take tangents to get rid of them. For the right hand side, you can consider the sum of angle formula for the tangent:
$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$