The area of the region S = z C: |z - 1|^2 4, z + z 2, Im(z) 0 is: — Solution | JEE Mathematics

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The area of the region S = { $z \in C$ : $|z - 1|^2 \le 4$ , $z + \bar{z} \ge 2$ , $\operatorname{Im}(z) \ge 0$ } is:

A
$\pi/3$
B
$3\pi/2$
C
$17\pi/8$
D
$7\pi/4$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The region is $S = \{z \in C : |z-1|^2 \le 4,\; z+\bar z \ge 2,\; \operatorname{Im}(z) \ge 0\}$ .

Find: The area of this region.

The solution is unrelated to this question and discusses a different problem about a parabola and a line. Therefore, the working for this complex-number region could not be extracted from the solution.

Using the answer key as fallback, the correct option is B.

Common mistakes

Interpreting $z+\bar z$ incorrectly. Since $z = x+iy$ , we have $z+\bar z = 2x$ , not $2y$ . So $z+\bar z \ge 2$ means $x \ge 1$ .
Misreading $|z-1|^2 \le 4$ as a line or as $|z-1| \le 4$ . It represents the disk centered at $(1,0)$ with radius $2$ , because $|z-1| \le 2$ .
Ignoring the condition $\operatorname{Im}(z) \ge 0$ . This restricts the region to the upper half-plane, so only the upper part of the disk should be considered.

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Answer

Step-by-step solution

Common mistakes

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