ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on March 23, 2009, 11:59:29 pm

Title: Argument
Post by: /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
Title: Re: Argument
Post by: TonyHem on March 24, 2009, 12:21:10 am: Is it just me or latex isn't working?
Title: Re: Argument
Post by: Mao on March 24, 2009, 12:45:11 pm: what do you mean by disparities?
Title: Re: Argument
Post by: /0 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?
Title: Re: Argument
Post by: Mao 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.
Title: Re: Argument
Post by: hard on March 24, 2009, 05:29:24 pm: MAO
Title: Re: Argument
Post by: shinny 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...