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

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 $z = 1 - i \sqrt{3}.$
