If $\mathrm{z}=\mathrm{x}+$ iy satisfies the condition $|\mathrm{z}+1|=1$, then $\mathrm{z}$ lies on the
If $\mathrm{z}=\mathrm{x}+$ iy satisfies the condition $|\mathrm{z}+1|=1$, then $\mathrm{z}$ lies on the
parabola with vertex $(0,0)$
circle with centre $(-1,0)$ and radius 1
circle with centre $(1,0)$ and radius 1
Y-axis
Solution
Let $\mathrm{z}=\mathrm{x}+$ iy and $|\mathrm{z}+1|=1$
$\begin{aligned}
& \therefore|(\mathrm{x}+1)+\mathrm{iy}|=1 \\
& \therefore \sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=1 \Rightarrow(\mathrm{x}+1)^2+\mathrm{y}^2=1
\end{aligned}$
This is an equation of a circle with centre $(-1,0)$ and radius 1 .