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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TrueTears on October 29, 2009, 06:13:57 pm

Title: A quick question
Post by: TrueTears on October 29, 2009, 06:13:57 pm: How do you sketch $Arg(iz+2) = \frac{2\pi}{3}$

So normally for these questions I just apply transformations.

First let $Z = iz+2$

$\frac{(Z-2)}{i} = -i(Z-2)$

So it's the ray that makes an angle of $\frac{2\pi}{3}$ radians with horizontal then shifted 2 to the left in the real axis and then rotated 90 degrees clockwise. However that's wrong... how does it actually look like?
Title: Re: A quick question
Post by: mano91 on October 29, 2009, 06:22:41 pm: does the solution look like the line y= x/sqrt(3) +2 ?
Title: Re: A quick question
Post by: TrueTears on October 29, 2009, 06:24:18 pm: Quote from: mano91 on October 29, 2009, 06:22:41 pm
does the solution look like the line y= x/sqrt(3) +2 ?
I never use $\tan^{-1}(x)$ and change into cartesian simply because $\tan^{-1}(x)$ doesn't work all the time, I've always applied transformations like this and never failed me, I don't know why this one it doesn't work.
Title: Re: A quick question
Post by: TrueTears on October 29, 2009, 06:25:20 pm: Btw I know you can split up the $Arg(iz+2)$ into $Arg(i) + Arg(z+\frac{2}{i})$

But I want a way where you don't have to split it up.
Title: Re: A quick question
Post by: NE2000 on October 29, 2009, 06:46:39 pm: Maybe translation ought to be $\frac {2}{i}$ which comes out to be $-2i$ so you shift it up 2 and then you do the rotation.
Title: Re: A quick question
Post by: TrueTears on October 29, 2009, 07:08:02 pm: Thanks, I realised the question actually asked for something different lol.

Another Q is sketching $Arg(z) + i = \frac{\pi}{3}$ even possible?
Title: Re: A quick question
Post by: /0 on October 29, 2009, 07:48:52 pm: it's impossible
Title: Re: A quick question
Post by: TrueTears on October 29, 2009, 07:49:42 pm: Quote from: /0 on October 29, 2009, 07:48:52 pm
it's impossible
Yeah I asked kamil, says you need taylor series. GG
Title: Re: A quick question
Post by: /0 on October 29, 2009, 07:51:21 pm: Arg(z) is real though, so it won't be able to cancel out the $i$ , leaving the left side complex with an imaginary part and the right side real.
Title: Re: A quick question
Post by: addikaye03 on November 17, 2009, 08:17:30 pm: Quote from: TrueTears on October 29, 2009, 06:13:57 pm
How do you sketch $Arg(iz+2) = \frac{2\pi}{3}$

So normally for these questions I just apply transformations.

First let $Z = iz+2$

$\frac{(Z-2)}{i} = -i(Z-2)$

So it's the ray that makes an angle of $\frac{2\pi}{3}$ radians with horizontal then shifted 2 to the left in the real axis and then rotated 90 degrees clockwise. However that's wrong... how does it actually look like?

arg(iz+2)=2pi/3

let z=x+iy

arg(i(x+iy)+2)=2pi/3

pi/2+arg[x+2+i(y)]=2pi/3

arg[(x+2)+iy]=pi/6

so now sketch from (-2,0) ray at angle pi/6 (@)

The equation of the line would just be y=arctan(@)x-2

(rt3)y=x-2(rt3)

I know this Q has been answered kinda but would that be right?