Author Topic: Argument (Read 1541 times) Tweet Share

/0 · « **on:** March 23, 2009, 11:59:29 pm »

If $\frac{-\pi}{2} < Arg(z_1), Arg(z_2) < \frac{\pi}{2}$ , show that

$Arg(z_1z_2) = Arg(z_1)+Arg(z_2)$ and $Arg\left(\frac{z_1}{z_2}\right) = Arg(z_1)-Arg(z_2)$

Is this sufficient proof?

$Arg(z_1z_2) = Arg(z_1)+Arg(z_2)+2k\pi$ and $Arg\left(\frac{z_1}{z_2}\right) = Arg(z_1)-Arg(z_2) + 2k\pi$ from definitions.

$Arg(z_1z_2) \in (-\pi, \pi]$ and $Arg(z_1)+Arg(z_2) \in (-\pi, \pi)$

$Arg\left(\frac{z_1}{z_2}\right) \in (-\pi, \pi]$ and $Arg(z_1)-Arg(z_2) \in (-\pi, \pi)$

Therefore the values will be in the same domain.

Also, what about the disparities in domain? What effect does this have on the proof? Thanks

TonyHem · « **Reply #1 on:** March 24, 2009, 12:21:10 am »

Is it just me or latex isn't working?

Mao · « **Reply #2 on:** March 24, 2009, 12:45:11 pm »

what do you mean by disparities?

/0 · « **Reply #3 on:** March 24, 2009, 04:18:08 pm »

Like, $Arg(z_1z_2)$ is defined between $(-\pi, pi]$ , but when you add up the arguments with domain $(-\frac{\pi}{2}, \frac{\pi}{2})$ you get $(-\pi, \pi)$ , does that mean the proof is incomplete?

Mao · « **Reply #4 on:** March 24, 2009, 05:11:14 pm »

nope, $Arg(z_1)+Arg(z_2)\in \left(-\pi,\pi \right)$ , which is a subset of $(-\pi,\pi ]$
so it's all good.

hard · « **Reply #5 on:** March 24, 2009, 05:29:24 pm »

shinny · « **Reply #6 on:** March 24, 2009, 06:55:29 pm »

Uhh...I don't think you should be using the 'by definition' step...they're asking you to prove the definition, not the domains I'm pretty certain. I remember doing an algebraic way to prove that definition, but not doing maths for this long has pretty much killed all ability to do Specialist...

Search the forums now!

We have moved!

Author Topic: Argument (Read 1541 times) Tweet Share

/0

Argument

TonyHem

Re: Argument

Mao

Re: Argument

/0

Re: Argument

Mao

Re: Argument

hard

Re: Argument

shinny

Re: Argument

Recent Posts

User:
Password: