Find the locus of points |z-1|= -Im(z). If I wish to find the locus of complex points satisfying $ |z-1|= -\text{Im}(z)$, then would I be right in supposing it represents the half-circle
$(x-1)^2 + (y-1/2)^2 = 1/4, y \leq 0$?
My work follows:

*

*First, notice $|z-1| \geq 0 \Rightarrow \text{Im}(z) \leq 0.$

*Second,
$$|z-1|^2 = (-\text{Im}(z))^2 \Rightarrow |z|^2 - 2\text{Re}(z)+ 1 = \text{Im}(z) \\ \Rightarrow x^2 +y^2 - 2x - y + 1 =0  \\ \Rightarrow (x-1)^2 + (y-1/2)^2 = 1/4,$$
in completing the square.
As $y = \text{Im}(z)$, we further restrict $y \leq 0.$

From the given definition, we know that $y$ must be negative. Thus, $z$ must lie on the lower half plane.
A geometric approach would be like : Distance between $z$ and $(1,0)$ is equal to it's distance from real axis. Thus, $z$ must lie on the straight line $Re(z)= 1$ because a hypotenuse is always greater than other sides of right triangle.

