What are the rules for argument division in complex numbers?

[bonbon22]

Homework Statement

This argument is equal to , arg(z1/z2) = arg(z1) - arg(z2) as said by exam solutions need help understanding why.

Relevant Equations

arg(z1/z2) = arg(z1)-arg(z2)

above video states that arg(z1/z2) = arg(z1) - arg(z2) this is seems very similar to Log rules but these are inverse function for angles right? And log rules only apply to logarithms, not sure where he got this from? What am i missing?

 

[PeroK]

Homework Statement: This argument is equal to , arg(z1/z2) = arg(z1) - arg(z2) as said by exam solutions need help understanding why.
Homework Equations: arg(z1/z2) = arg(z1)-arg(z2)



above video states that arg(z1/z2) = arg(z1) - arg(z2) this is seems very similar to Log rules but these are inverse function for angles right? And log rules only apply to logarithms, not sure where he got this from? What am i missing?

Complex numbers can be expressed in polar form: $z_1 = r_1e^{i\theta_1}, z_2 = r_2 e^{i\theta_2}$, where $r$ is the modulus and $\theta$ is the argument.

$z_1/z_2 = (r_1/r_2)e^{i(\theta_1 - \theta_2)}$

This uses the properties of the exponential function.

 

[bonbon22]

Complex numbers can be expressed in polar form: $z_1 = r_1e^{i\theta_1}, z_2 = r_2 e^{i\theta_2}$, where $r$ is the modulus and $\theta$ is the argument.

$z_1/z_2 = (r_1/r_2)e^{i(\theta_1 - \theta_2)}$

This uses the properties of the exponential function.

i see that makes sense does the argument function cancel out the (r_1/r_2)e^{i] so you are just left with theta 1 - theta 2?

 

[PeroK]

i see that makes sense does the argument function cancel out the (r_1/r_2)e^{i] so you are just left with theta 1 - theta 2?

The argument is the angle $\theta$ in that expression. I would't say there is any cancelling out, as such. Would you say the "hair colour" function cancels out the rest of your body and just leaves the hair colour?

 

[WWGD]

Also $Arg(z \cdot w)= Argz+Argw -2k\pi$
As PeroK wrote, if $w = re^{it} \rightarrow 1/w= (1/r )e^{-it}$. Edit: We must assume $w \neq 0$.