Solution. Let $z=x+y i$.

(a) $\{\operatorname{Im}(z)=-1\}=\{y=-1\}$ is the horizontal line passing through the point $-i$.

(b) Since

$$

\begin{aligned}

|z-1|=|z+i| & \Leftrightarrow|(x-1)+y i|=|x+(y+1) i| \\

& \Leftrightarrow|(x-1)+y i|^{2}=|x+(y+1) i|^{2} \\

& \Leftrightarrow(x-1)^{2}+y^{2}=x^{2}+(y+1)^{2} \\

& \Leftrightarrow x+y=0,

\end{aligned}

$$

the curve is the line $x+y=0$.

(c) Since

$$

\begin{aligned}

2|z|=|z-2| & \Leftrightarrow 2|x+y i|=|(x-2)+y i| \\

& \Leftrightarrow 4|x+y i|^{2}=|(x-2)+y i|^{2} \\

& \Leftrightarrow 4\left(x^{2}+y^{2}\right)=(x-2)^{2}+y^{2} \\

& \Leftrightarrow 3 x^{2}+4 x+3 y^{2}=4 \\

& \Leftrightarrow\left(x+\frac{2}{3}\right)^{2}+y^{2}=\frac{16}{9} \\

& \Leftrightarrow\left|z+\frac{2}{3}\right|=\frac{4}{3}

\end{aligned}

$$

the curve is the circle with centre at $-2 / 3$ and radius $4 / 3$.