What is the Locus of Points for 2π|z - 1| = Arg(z - 1) in an Argand Diagram?

[Gone]

Homework Statement

Find the locus of points which satisfy

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

Homework Equations

n/a

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

 

[Saitama]

Homework Statement

Find the locus of points which satisfy

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

Homework Equations

n/a

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?

 

[Gone]

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!

 

[Saitama]

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 $2\pi (r-1)=\theta$.

 

[Gone]

Ah ok got you now :) thank you!