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?

$$|z|=\frac1{|z|}\Rightarrow |z|=1,$$
$$|z|=|1-z|\Rightarrow \sqrt{x^2+y^2}=\sqrt{(1-x)^2+y^2}\Rightarrow x=\frac12,$$
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
$$z_1=\frac12+i\frac{\sqrt{3}}2,$$
$$z_2=\frac12-i\frac{\sqrt{3}}2.$$
