Let z = x + iy be a complex number,A = {z : | | vedclass

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(A) Given $A = \{z : |z| \leq 2\}$,which implies $\sqrt{x^2 + y^2} \leq 2$,or $x^2 + y^2 \leq 4$. Given $B = \{z : (1-i)z + (1+i)\bar{z} \geq 4\}$. Su

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$\sqrt{3} + \frac{1}{2}i$

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(A) Given $A = \{z : |z| \leq 2\}$,which implies $\sqrt{x^2 + y^2} \leq 2$,or $x^2 + y^2 \leq 4$. Given $B = \{z : (1-i)z + (1+i)\bar{z} \geq 4\}$. Substituting $z = x + iy$ and $\bar{z} = x - iy$: $(1-i)(x+iy) + (1+i)(x-iy) \geq 4$ $(x + iy - ix - i^2y) + (x - iy + ix - i^2y) \geq 4$ $(x + iy - ix + y) + (x - iy + ix + y) \geq 4$ $2x + 2y \geq 4 \implies x + y \geq 2$. Thus,$A \cap B = \{z : x^2 + y^2 \leq 4 	ext{ and } x + y \geq 2\}$. Checking the options: For $A$,$z = \sqrt{3} + \frac{1}{2}i$: $|z|^2 = 3 + \frac{1}{4} = 3.25 \leq 4$ (True). $x+y = \sqrt{3} + 0.5 \approx 1.732 + 0.5 = 2.232 \geq 2$ (True). Thus,$\sqrt{3} + \frac{1}{2}i$ belongs to $A \cap B$.

Let $z = x + iy$ be a complex number,$A = \{z : |z| \leq 2\}$ and $B = \{z : (1-i)z + (1+i)\bar{z} \geq 4\}$. Then which one of the following options belongs to $A \cap B$?

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