Let C be the set of all complex numbers. Let | vedclass

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(A) Given $S_{1} = \{z \in C : |z-(3+2i)|^{2}=8\}$. Let $z = x+iy$. Then $|(x-3)+i(y-2)|^{2}=8$,which implies $(x-3)^{2}+(y-2)^{2}=8$. This represents

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(A) Given $S_{1} = \{z \in C : |z-(3+2i)|^{2}=8\}$. Let $z = x+iy$. Then $|(x-3)+i(y-2)|^{2}=8$,which implies $(x-3)^{2}+(y-2)^{2}=8$. This represents a circle with center $(3, 2)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.$S_{2} = \{z \in C : x \geq 5\}$.$S_{3} = \{z \in C : |z-\bar{z}| \geq 8\}$. Since $z-\bar{z} = 2iy$,we have $|2iy| = 2|y| \geq 8$,which implies $|y| \geq 4$,so $y \geq 4$ or $y \leq -4$.We need to find the intersection $S_{1} \cap S_{2} \cap S_{3}$.For $S_{1} \cap S_{2}$,we substitute $x=5$ into the circle equation: $(5-3)^{2} + (y-2)^{2} = 8$ $\Rightarrow 4 + (y-2)^{2} = 8$ $\Rightarrow (y-2)^{2} = 4$ $\Rightarrow y-2 = \pm 2$. Thus $y=4$ or $y=0$.For $S_{1} \cap S_{2} \cap S_{3}$,we check the condition $y \geq 4$ or $y \leq -4$ from $S_{3}$.At $x=5$,the points on the circle are $(5, 4)$ and $(5, 0)$.Only the point $(5, 4)$ satisfies $y \geq 4$.Thus,the only point in the intersection is $z = 5+4i$.The number of elements is $1$.

Let $C$ be the set of all complex numbers. Let $S_{1} = \{z \in C : |z-3-2i|^{2}=8\}$,$S_{2} = \{z \in C : \operatorname{Re}(z) \geq 5\}$,and $S_{3} = \{z \in C : |z-\bar{z}| \geq 8\}$. Then the number of elements in $S_{1} \cap S_{2} \cap S_{3}$ is equal to:

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