Solve $\arg(-1/z)=-2\pi/3$ and $|1-\frac2z|=1$ $\arg(\frac{-1}z)=\frac{-2\pi}3$
what does this mean? how to i get the $\arg(z)$ from this.
I'm thinking of reciprocals.
$\left|1-\frac2z\right|=1$ how do i solve for this as well.. i'm confused when i negative sign appears
 A: Note that:
$$\frac{1}{z} = \frac{x -i y}{x^2+y^2}$$
Now you have:
$$\arg\left(-\frac{1}{z}\right) = -\frac{2\pi}{3}\implies \arg \left(\frac{-x +i y}{x^2+y^2}\right) = \tan^{-1}{\left(-\frac{y}{x}\right)} = -\frac{2\pi}{3}$$
$$\implies \frac{y}{x} = \tan(\frac{2\pi}{3})$$
The other condition you can write as:
$$ \vert z - 2 \vert = \vert z \vert \implies (x-2)^2 + y^2 = x^2 + y^2$$
$$\implies 4 x = 4 \implies x = 1 \implies y = \tan(\frac{2\pi}{3}) = -\sqrt{3}.$$
Hence $z = 1 - i \sqrt{3}.$
A: Set
$z = r(\cos \theta + i \sin \theta); \tag{1}$
then $\arg(z) = \theta$, and
$z^{-1} = r^{-1}(\cos \theta - i \sin \theta), \tag{2}$
as may easily be checked by a simple complex multiplication:
$(r(\cos \theta + i \sin \theta))(r^{-1}(\cos \theta - i \sin \theta)) = rr^{-1}(\cos^2 \theta + \sin^2 \theta) = rr^{-1}(1)  1. \tag{3}$
It follows that
$-z^{-1} = -r^{-1}\cos \theta + r^{-1} i\sin \theta, \tag{4}$
whence
$\tan (-\dfrac{2\pi}{3}) = \tan (\arg(-z^{-1})) = \dfrac{\sin \theta}{-\cos \theta} = \dfrac{\sin (-\theta)}{\cos (-\theta)} = \tan (-\theta) \tag{5}$
or
$\theta = \dfrac{2\pi}{3}. \tag{6}$
The modulus or magnitude $r = \vert z \vert$ of $z$ varies independently of $\arg(z) = \theta$; thus we may take any
$z = r(\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}) \tag{7}$
with $0 < r < \infty$ as a solution in this case.  And not altogether incidentally, we have
$\cos \dfrac{2\pi}{3} = -\dfrac{1}{2};  \;\; \sin \theta = \dfrac{\sqrt{3}}{2}, \tag{8}$
whence we may write
$z = r(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}); \tag{9}$
the set of all $z \in \Bbb C$ such that
$\arg(z^{-1}) = -\dfrac{2\pi}{3} \tag{10}$
is precisely the ray emanating from, but not including, the origin in the direction $e^{2\pi i/3}$.
As for the other problem,
$\vert 1 - \dfrac{2}{z} \vert = 1, \tag{11}$
both my approach and answer differ from those of incognito.  Looking at equation (11) geometrically, we ask what it tells us, and we see that it says that $2/z$ differs from $1$ by a complex number of unit modulus; that is, we must have
$\dfrac{2}{z} = 1 - e^{i\theta}; \tag{12}$
it is easily seen that $z$ satisfying (12) also obeys (11), provided we forbid the value $\theta = 0$;  (12) yields
$z = \dfrac{2}{1 - e^{i\theta}}, \tag{13}$
in which we allow $\theta \in (0, 2\pi)$.  We can in fact write such $z$ in "$x + iy$" form:
$z = \dfrac{2(1 - e^{-i\theta})}{(1 - e^{-i\theta})(1 -  e^{i\theta})} = \dfrac{1 - e^{-i\theta}}{1 - \cos \theta} = \dfrac{1 - \cos \theta + i \sin \theta}{1 - \cos \theta} = 1 + i\dfrac{\sin \theta}{1 - \cos \theta}; \tag {14}$
we note from (14) that the real part of $z$ is the constant $1$; as for the imaginary part, $\sin \theta / (1 - \cos \theta)$ takes on every real value precisely once as $0 \to \theta \to 2\pi$; to see this, obverve that the derivative of 
$y(\theta) = \dfrac{\sin \theta}{1 - \cos \theta} \tag{15}$
is
$y'(\theta) = \dfrac{\cos \theta (1 - \cos \theta) - \sin^2 \theta}{(1 - \cos \theta)^2} = \dfrac{\cos \theta - 1}{(1 - \cos \theta)^2} = \dfrac{1}{\cos \theta - 1 }< 0 \tag{16}$
since $\cos \theta < 1$ for $\theta \in (0, 2\pi)$; $y(\theta)$ is monotonically decreasing, hence takes on any value at most once; to see that $y(\theta)$ takes on every real value at least once, we note that
$\lim_{\theta \to 0^+} y(\theta) = \lim_{\theta \to 0^+} \dfrac{\cos \theta}{\sin \theta} = \infty \tag{17}$
by a simple application of L'Hopital's rule, which may similarly be used to show that
$\lim_{\theta \to 2\pi^{-}} y(\theta) = -\infty; \tag{18}$
thus the range of $y(\theta)$ is indeed all of $\Bbb R$.  As the differential topologists might put it, $y(\theta)$ is a diffeomorphism 'twixt $(0, 2\pi)$ and $\Bbb R$.  But the real point, for the present purposes, is that the solutions of (11) are in fact precisely the set of complex numbers $z$ with $\Re(z) = 1$.
We check that $z = 1 + iy$ solves (11); we have:
$z^{-1} = \dfrac{1 - iy}{1 + y^2}; \tag{19}$
$\dfrac{2}{z} = \dfrac{2 - 2iy}{1 + y^2}; \tag{20}$
$1 - \dfrac{2}{z} = 1 - \dfrac{2 - 2iy}{1 + y^2} = \dfrac{1 + y^2 - 2 + 2iy}{1 + y^2} = \dfrac{(y + i)^2}{(y + i)(y - i)} = \dfrac{y + i}{y - i}; \tag{21}$
finally,
$\vert 1 - \dfrac{2}{z} \vert = \dfrac{\vert y + i \vert}{\vert y - i \vert} = 1 \tag{22}$
since $y - i = \overline{y + i}$.  Things check out.
This answer differs from incognito's solution to (11).  Apparently, he (and here I use the masculine form of the pronoun since the relevance of gender distinction seems to disappear when one anonymously adopts a name like incognito!) assumed both (11) and (10) were meant to be solved concurrently, wheras I treated them as separate issues.  This makes each of us, in our own way, correct.   But it is true that there are more solutions to (11) than $x = 1$, $y = -\sqrt{3}$.  Indeed, $y$ is not determined by the relation $\vert z - 2 \vert = \vert z \vert$, as his own calculations reveal.  Indeed, his work on this question may be seen as an elegant indication that $1 + iy$ gives all $z$ satisfying (11).
