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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: kenny_ on November 10, 2011, 10:29:55 pm

Title: Confirmation on Complex Numbers
Post by: kenny_ on November 10, 2011, 10:29:55 pm: Hey just wanted to know if this is correct.

If Arg(w) = x
then is Arg(-3w) = Arg(-3) + Arg(w)
= pi + x?
Title: Re: Confirmation on Complex Numbers
Post by: A-i-d-a-n-K on November 10, 2011, 10:52:13 pm: Yes, multiply any complex number by a negative real and its argument changes by pi. Think of it like rotating it 180 degrees around the Argand diagram
Title: Re: Confirmation on Complex Numbers
Post by: A-i-d-a-n-K on November 10, 2011, 10:56:18 pm: Just be aware that you are 'removing' the negative by doing this; Arg and the r value work in conjuction. For example 2cis(pi/3)=-2cis(-2pi/3) . I always try and keep my r positive...
Title: Re: Confirmation on Complex Numbers
Post by: Mao on November 11, 2011, 01:25:50 am: Quote from: A-i-d-a-n-K on November 10, 2011, 10:56:18 pm
Just be aware that you are 'removing' the negative by doing this; Arg and the r value work in conjuction. For example 2cis(pi/3)=-2cis(-2pi/3) . I always try and keep my r positive...

~~No. This is wrong.~~
actually, reading your post twice, I am now totally confused about what you are trying to say. I think you have the maths correct in your head, but gave an example of 'what not to do'.
Anyhow, I'll keep this here in case anyone else gets confused.

Two things:
1. Those two things are not equal:
$r cis(\theta) = r \cos(\theta) + r \sin(\theta) \imath$
$-r cis(-\theta) = -r \cos(-\theta) - r \sin(-\theta) \imath = -r \cos(\theta) + r \sin(\theta) \imath$
As you can see, not the same.

2. By convention (and definition), r is always positive.
When multiplying a complex number $z=r\, cis(\theta)$ by a negative real number, i.e. $-k\times z = -kr\, cis(\theta)$ , always try to convert the coefficient to a positive number. i.e.
$-kr\, cis(\theta) = -k \times (r\, cis(\theta)) = k\, cis(\pi) \times r\, cis(\theta) = kr\, cis(\theta+\pi)$
Title: Re: Confirmation on Complex Numbers
Post by: A-i-d-a-n-K on November 11, 2011, 10:00:10 am: yeah should have put the inbetween step:
-2cis(-2pi/3)=2cis(-2pi/3+pi)=2cis(pi/3)