Find the locus of points |z-1|= -Im(z).

Question

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:

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

Answer 1

HINT

You are on the right track, but you forgot to square $\text{Im}(z)$. If we let $z = x + yi$, it results that \begin{align*} |z - 1| = -\text{Im}(z) = -y & \Longleftrightarrow \begin{cases} |z - 1|^{2} = y^{2}\\\\ y\leq 0 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} |z|^{2} - 2\text{Re}(z) + 1 = y^{2}\\\\ y\leq 0 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x^{2} + y^{2} - 2x + 1 = y^{2}\\\\ y\leq 0 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} (x - 1)^{2} = 0\\\\ y\leq 0 \end{cases} \end{align*}

Can you take it from here?

Answer 2

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.