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Sketch the curves in the complex plane given by
(a) $\operatorname{Im}(z)=-1$;
(b) $|z-1|=|z+i|$;
(c) $2|z|=|z-2|$.
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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 $4 / 3$.
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