For $z\in \mathbb{C}\backslash \{i\}$.
- How can i go from this line : $|z|^2-Re[(1-i)z]=0$
To that one : $$\left|z-\dfrac{1+i}{2}\right|^2=\left|\dfrac{1+i}{2}\right|^2=\dfrac{1}{2}$$
indeed,
$$|z|^2-Re[(1-i)z]=|z|^2-\left[\dfrac{(1-i)z+(1+i)\bar{z}}{2}\right]=0$$ any help will be appreciated
You may want to start with $z=x+iy$, with $x \neq 0$ and $x,y \in \mathbb{R}$. Then the equation $|z|^2-Re[(1-i)z]=0$ results in $$(x^2+y^2)-(x+y)=0$$ This is a circle with center $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $=\frac{1}{\sqrt{2}}$. Now write the equation of this circle in $z \in \mathbb{C}$.