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If the Complex Number Z = X + I Y Satisfies the Condition | Z + 1 | = 1 , Then Z Lies on

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प्रश्न

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

विकल्प

  • x−axis

  • circle with centre (−1, 0) and radius 1

  • y−axis

  • none of these

MCQ
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उत्तर

\[\left| z + 1 \right| = 1\]

\[ \Rightarrow \left| z + 1 \right|^2 = 1^2 \]

\[ \Rightarrow \left( z + 1 \right) \bar{\left( z + 1 \right)} = 1\]

\[ \Rightarrow \left( z + 1 \right)\left( \bar{z} + 1 \right) = 1\]

\[ \Rightarrow z \bar{z} + z + \bar{z} + 1 = 1\]

\[ \Rightarrow z \bar{z} + z + \bar{z} = 0\]

\[\text { Since }, z = x + iy\]

\[ \therefore z \bar{z} + z + \bar{z} = 0\]

\[ \Rightarrow \left( x + iy \right)\left( x - iy \right) + x + iy + x - iy = 0\]

\[ \Rightarrow x^2 + y^2 + 2x = 0\]

\[ \Rightarrow \left( x + 1 \right)^2 + \left( y - 0 \right)^2 = 1^2 \]

\[\text { which is the equation of a circle with centre } ( - 1, 0) \text { and radius }1\]

Hence, the correct option is (b).

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अध्याय 13: Complex Numbers - Exercise 13.6 [पृष्ठ ६६]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 13 Complex Numbers
Exercise 13.6 | Q 43 | पृष्ठ ६६

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