Find all complex numbers $z$ such that $|z|=\frac{1}{|z|}=|1-z|$ 
Problem: Find all complex numbers $z$ such that $|z|=\frac{1}{|z|}=|1-z|$.

Basically I have an idea how to solve this and I get $x=\frac12$ but how should I express it mathematically? Should I go and find $y$ also?
 A: $$|z|=\frac1{|z|}\Rightarrow |z|=1,$$
$$|z|=|1-z|\Rightarrow \sqrt{x^2+y^2}=\sqrt{(1-x)^2+y^2}\Rightarrow x=\frac12,$$
where $z=x+iy$. Next, we will determine the imaginary part
$$\sqrt{\left(\frac12\right)^2+y^2}=1\Rightarrow y=\pm \frac{\sqrt{3}}2.$$
Therefore solutions are
$$z_1=\frac12+i\frac{\sqrt{3}}2,$$
$$z_2=\frac12-i\frac{\sqrt{3}}2.$$
A: Hint: $\require{cancel}$ $|z|^2=|1-z|^2$ $\iff$ $z \bar z = (1-z)(1-\bar z)$ $\iff$ $\cancel{z \bar z}=1-z-\bar z+\cancel{z \bar z}\,$
Since $1=|z|^2=z \bar z$ it follows that $\bar z = \cfrac{1}{z}$ so the last equality becomes $z^2-z+1=0\,$, which is a simple quadratic having the two possible values of $z\,$ as roots.
A: Hint: $|z|=1$ represents a circle in the complex plane where the x-axis is the real part and y-axis is the imaginary part of the complex number(all points are equidistant from the origin.)
And $|z|=|1-z|$ represents the line $x=\frac{1}{2}$ (set of all points equidistant from 0 and 1 should be the perpendicular bisector of the segment joining 1 and 0). Now you are looking for intersection of the two curves.
