Complex Argument: Prove arg(z1*z2)=arg z1+arg z2

[dust]

Hello, it is my first post here, so dont blame me if this thread already exist somewhere.
I need some help with complex's.

[/b] Prove arg(z1*z2)=arg z1+arg z2 [/b]

Is there any analogy to arg (z1/z2)=arg z1-arg z2 ?

 

[daon2]

Hello, it is my first post here, so dont blame me if this thread already exist somewhere.
I need some help with complex's.

[/b] Prove arg(z1*z2)=arg z1+arg z2 [/b]

Is there any analogy to arg (z1/z2)=arg z1-arg z2 ?

It is just knowing what the argument is, and multiplying two things together.
\(\displaystyle z_1\cdot z_2 = |z_1|e^{i\text{Arg}(z_1)}\cdot |z_2|e^{i\text{Arg}(z_2)} = |z_1\cdot z_2| e^{i(\text{Arg}(z_1)+\text{Arg}(z_2))}\)

Yes, there is an obvious analogy.

\(\displaystyle \text{Arg}(z^{-1})=-\text{Arg}(z)\)

I think you should be able to show that now.

 

[dust]

Thanks, Good guy..
I already passed my paper, and i am so happy
Thank you one more tme

 

[SAMUELK]

I think using z = r exp(i theta) is kind of like cheating. (OK, OK, It's not, but I just don't like it.)

I like this better:

Prove arg(z1*z2)=arg z1+arg z2
IF z1 = r1 (cos t1 + i sin t1)
and z2 = r2 (cos t2 + i sin t2)
then z1 z2 = r1 r2(cos t1 cos t2 - sin t1 sin t2 + i(cos t1 sin t2 + cos t2 sin t1))

= r1 r2(cos (t1 + t2) + i sin(t1 + t2))
using your well-known sum identities.
So arg(z1 z2) = t1 + t2
etc.