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My problem is finding the locus of points which satisfy this relationship $$|z − 1| = Arg(z − 1)/2π.$$ I'm struggling to deal with both a modulus and an arg. Thank you for any help :)

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    try editing your question using MathJax – 2017-01-01

2 Answers 2

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Let Arg$(z-1)=\phi$

WLOG $z-1=\dfrac\phi{2\pi}e^{i\phi}$

If $z=x+iy, x-1=\dfrac\phi{2\pi}\cos\phi,y=\dfrac\phi{2\pi}\sin\phi$

What if $\phi=0?$

Else on division $\tan\phi=\dfrac y{x-1},$

$\dfrac{\cos\phi}{x-1}=\dfrac{\sin\phi}y=\pm\dfrac1{\sqrt{y^2+(x-1)^2}}$

$x-1=\dfrac\phi{2\pi}\cos\phi=\dfrac{\pm\phi (x-1)}{2\pi\sqrt{y^2+(x-1)^2}}\implies \sqrt{y^2+(x-1)^2}=\pm\dfrac\phi{2\pi}$

Now by atan2 $\phi\le\pi$

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    It seems that $ z -1 = \frac {1}{8\sqrt 2} + \frac {i}{8\sqrt 2} $ is a solution for which the argument is not $2 \pi$. – 2017-01-01
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    @Swapnil, Please pinpoint the mistake. – 2017-01-01
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    Mistake in Second last equation: $\dfrac\phi{2\pi}\cos\phi≠\dfrac{\pm\phi (x-1)}{2\pi} $ – 2017-01-01
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    @Swapnil, Thanks for observation, Just could not make out the flaw – 2017-01-01
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I assume you mean principal argument: -$-\pi \lt arg(z-1) \leq \pi$

If all you need is the locus and not an idea about the points themselves, then let $z=x+iy$

I assume you know how to find the argument of a complex number.

So that gives: $$ 2\pi (\sqrt{(x-1)^2 + y^2)}) = \arctan (\frac {y}{x-1}) + k \pi $$

Where k is either 0, 1 or -1

From this, since the RHS is less than $\pi$ so at least we know:

$(\sqrt{(x-1)^2 + y^2)} \leq \frac {1}{2} $