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

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Solution

## $\left|z+1\right|=1\phantom{\rule{0ex}{0ex}}⇒{\left|z+1\right|}^{2}={1}^{2}\phantom{\rule{0ex}{0ex}}⇒\left(z+1\right)\overline{\left(z+1\right)}=1\phantom{\rule{0ex}{0ex}}⇒\left(z+1\right)\left(\overline{z}+1\right)=1\phantom{\rule{0ex}{0ex}}⇒z\overline{z}+z+\overline{z}+1=1\phantom{\rule{0ex}{0ex}}⇒z\overline{z}+z+\overline{z}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since},z=x+iy\phantom{\rule{0ex}{0ex}}\therefore z\overline{z}+z+\overline{z}=0\phantom{\rule{0ex}{0ex}}⇒\left(x+iy\right)\left(x-iy\right)+x+iy+x-iy=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+2x=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(y-0\right)}^{2}={1}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\left(-1,0\right)\mathrm{and}\mathrm{radius}1$ Hence, the correct option is (b).

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