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Plots in an Argand Diagram

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1. The problem statement, all variables and given/known data

Find the locus of points which satisfy

2π*|z −1| = Arg(z − 1) where |z −1| ≤ 2.

2. Relevant equations

n/a

3. The attempt at a solution

I know that |z-1| ≤ 2 is the 'inside' bits of a circle center (1,0) with a radius 2

After that I get confused surely with |z-1|≤2 then Arg(z-1) has to be between 0 and 4π... but then that's all space?

Thanks in advance
 
3,810
92
1. The problem statement, all variables and given/known data

Find the locus of points which satisfy

2π*|z −1| = Arg(z − 1) where |z −1| ≤ 2.

2. Relevant equations

n/a

3. The attempt at a solution

I know that |z-1| ≤ 2 is the 'inside' bits of a circle center (1,0) with a radius 2

After that I get confused surely with |z-1|≤2 then Arg(z-1) has to be between 0 and 4π... but then that's all space?

Thanks in advance
Since everything is with reference to ##1+0i##, it would be good idea to shift the origin here. Shifting the origin, your equations transform to:
$$2\pi |z|=\arg(z)$$
$$|z|\leq 2$$
Use the substitution ##z=re^{i\theta}## in the first equation, do you see where that leads to?
 
6
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Since everything is with reference to ##1+0i##, it would be good idea to shift the origin here. Shifting the origin, your equations transform to:
$$2\pi |z|=\arg(z)$$
$$|z|\leq 2$$
Use the substitution ##z=re^{i\theta}## in the first equation, do you see where that leads to?
So that means that ##2\pi r=\theta## which is a spiral beginning at (0,0) so to get the answer is it just
$$2\pi (r-1)=\theta$$
Thanks!
 
3,810
92
So that means that ##2\pi r=\theta## which is a spiral beginning at (0,0)
Yes.
so to get the answer is it just
$$2\pi (r-1)=\theta$$
Well, no. I don't think that transformation is correct. Once you plot the graph, you need to move everything by 1 unit towards right.

Look at the plots of ##2\pi r=\theta## and ##2\pi (r-1)=\theta##.
 
6
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Ah ok got you now :) thank you!
 

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