Mod-arg identities/proofs (1 Viewer)

[.ben]

hi how do you prove these:

arg (z¹z²)=arg(z¹)-arg(z²)

LHS=arg (z¹z²)
=a+b
=arg(z¹)+arg(z²)
=RHS

i know what to proof looks like i just don't understand it.

 

[noah]

sub in r(cos@ + isin@) and then expand and simplify

you will have to use the cosine and sine addition of angle laws.

 

[KFunk]

If you let Z₁ = cisθ and Z₂ = cisφ then

Z₁.Z₂ = cosθcosφ - sinθsinφ + i(sinθcosφ + cosθsinφ )

= cos(θ + φ ) + isin(θ + φ )
= cis(θ + φ )

∴ arg(Z₁Z₂) = θ + φ = arg(Z₁) + arg(Z₂)

 

[.ben]

ooOO00h i get it now! thanks ppl