Question

If the complex number $z=x+iy$ satisfies the condition $\left|z+1\right|=1$, then z lies on

(a) x−axis
(b) circle with centre (−1, 0) and radius 1
(c) y−axis
(d) none of these

Answer 1

$$\begin{gathered} \left|z+1\right|=1 \\ \Rightarrow {\left|z+1\right|}^2=1^2 \\ \Rightarrow \left(z+1\right)\overline{\left(z+1\right)}=1 \\ \Rightarrow \left(z+1\right)\left(\overline{z}+1\right)=1 \\ \Rightarrow z\overline{z}+z+\overline{z}+1=1 \\ \Rightarrow z\overline{z}+z+\overline{z}=0 \\ \mathrm{Since},z=x+iy \\ \therefore z\overline{z}+z+\overline{z}=0 \\ \Rightarrow \left(x+iy\right)\left(x-iy\right)+x+iy+x-iy=0 \\ \Rightarrow x^2+y^2+2x=0 \\ \Rightarrow {\left(x+1\right)}^2+{\left(y-0\right)}^2=1^2 \\ \mathrm{which}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}(-1,0)\mathrm{and}\mathrm{radius}1 \end{gathered}$$

Hence, the correct option is (b).