Question:
The region represented by
\(\{z=x+iy\in C : |z|-\mathrm{Re}(z)\le 1\}\) is also given by the inequality :
\(\{z=x+iy\in C : |z|-\mathrm{Re}(z)\le 1\}\)
The region represented by
\(\{z=x+iy\in C : |z|-\mathrm{Re}(z)\le 1\}\) is also given by the inequality :
\(\{z=x+iy\in C : |z|-\mathrm{Re}(z)\le 1\}\)
\(\{z=x+iy\in C : |z|-\mathrm{Re}(z)\le 1\}\) is also given by the inequality :
\(\{z=x+iy\in C : |z|-\mathrm{Re}(z)\le 1\}\)
Show Hint
Write \(\sqrt{x^2+y^2}\le 1+x\) and square it (note \(1+x\ge 0\) is then automatic).
Updated On: Nov 5, 2026
- \(y^2\le 2\left(x+\frac12\right)\)
- \(y^2\le x+\frac12\)
- \(y^2\ge 2(x+1)\)
- \(y^2\ge x+1\)
Show Solution
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The Correct Option is A
Solution and Explanation
\(|z|-\mathrm{Re}(z)\le 1\Rightarrow\sqrt{x^2+y^2}\le 1+x\).
This needs \(1+x\ge 0\); squaring: \(x^2+y^2\le 1+2x+x^2\Rightarrow y^2\le 2x+1=2\left(x+\tfrac12\right)\).
Any point with \(y^2\le 2x+1\) has \(x\ge-\tfrac12\), so \(1+x\ge 0\) holds automatically.
Hence the region is \(y^2\le 2\left(x+\tfrac12\right)\).
Answer: (1) \(y^2\le 2\left(x+\frac12\right)\)
This needs \(1+x\ge 0\); squaring: \(x^2+y^2\le 1+2x+x^2\Rightarrow y^2\le 2x+1=2\left(x+\tfrac12\right)\).
Any point with \(y^2\le 2x+1\) has \(x\ge-\tfrac12\), so \(1+x\ge 0\) holds automatically.
Hence the region is \(y^2\le 2\left(x+\tfrac12\right)\).
Answer: (1) \(y^2\le 2\left(x+\frac12\right)\)
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