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Question
If the complex number \[z = x + iy\] satisfies the condition \[\left| z + 1 \right| = 1\], then z lies on
Options
x−axis
circle with centre (−1, 0) and radius 1
y−axis
none of these
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Solution
\[\left| z + 1 \right| = 1\]
\[ \Rightarrow \left| z + 1 \right|^2 = 1^2 \]
\[ \Rightarrow \left( z + 1 \right) \bar{\left( z + 1 \right)} = 1\]
\[ \Rightarrow \left( z + 1 \right)\left( \bar{z} + 1 \right) = 1\]
\[ \Rightarrow z \bar{z} + z + \bar{z} + 1 = 1\]
\[ \Rightarrow z \bar{z} + z + \bar{z} = 0\]
\[\text { Since }, z = x + iy\]
\[ \therefore z \bar{z} + z + \bar{z} = 0\]
\[ \Rightarrow \left( x + iy \right)\left( x - iy \right) + x + iy + x - iy = 0\]
\[ \Rightarrow x^2 + y^2 + 2x = 0\]
\[ \Rightarrow \left( x + 1 \right)^2 + \left( y - 0 \right)^2 = 1^2 \]
\[\text { which is the equation of a circle with centre } ( - 1, 0) \text { and radius }1\]
Hence, the correct option is (b).
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