Question:
The region represented by $\{z=x+i y \in \mathrm{C}:|z|-\operatorname{Re}(z) \leq 1\}$ is also given by the inequality:
Correct Option: 2,
Solution:
$\because|z|-\operatorname{Re}(z) \leq 1 \quad(\because z=x+i y)$
$\Rightarrow \sqrt{x^{2}+y^{2}}-x \leq 1 \Rightarrow \sqrt{x^{2}+y^{2}} \leq 1+x$
$\Rightarrow x^{2}+y^{2} \leq 1+x^{2}+2 x$
$\Rightarrow y^{2} \leq 1+2 x \Rightarrow y^{2} \leq 2\left(x+\frac{1}{2}\right)$
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