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 $y \leq 0.$

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?
