Let A = {z in mathbb{C} : |z - 2 - i| = 3}, B | vedclass

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(C) Let $z = x + iy$. For set $A$, $|(x - 2) + i(y - 1)| = 3$, which implies $(x - 2)^2 + (y - 1)^2 = 9$. For set $B$, $\operatorname{Re}((x + iy) - i

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$22$

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(C) Let $z = x + iy$. For set $A$, $|(x - 2) + i(y - 1)| = 3$, which implies $(x - 2)^2 + (y - 1)^2 = 9$. For set $B$, $\operatorname{Re}((x + iy) - i(x + iy)) = \operatorname{Re}((x + y) + i(y - x)) = x + y = 2$. Substituting $y = 2 - x$ into the equation for $A$: $(x - 2)^2 + (2 - x - 1)^2 = 9$ $(x - 2)^2 + (1 - x)^2 = 9$ $x^2 - 4x + 4 + 1 - 2x + x^2 = 9$ $2x^2 - 6x - 4 = 0 \implies x^2 - 3x - 2 = 0$. The roots are $x = \frac{3 \pm \sqrt{17}}{2}$. Since $y = 2 - x$, the points are $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$. $|z|^2 = x^2 + y^2 = x^2 + (2 - x)^2 = 2x^2 - 4x + 4$. Using $x^2 = 3x + 2$, we get $|z|^2 = 2(3x + 2) - 4x + 4 = 2x + 8$. Sum $= (2x_1 + 8) + (2x_2 + 8) = 2(x_1 + x_2) + 16$. Since $x_1 + x_2 = 3$, Sum $= 2(3) + 16 = 22$.

Let $A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$, $B = \{z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2\}$ and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to . . . . . . .

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