Let S_{1}, S_{2} and S_{3} be three sets defi | vedclass

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(C) For $S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$, $z$ represents the points on and inside the circle of radius $\sqrt{2}$ with center $(1, 0)$.

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has infinitely many elements

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(C) For $S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$, $z$ represents the points on and inside the circle of radius $\sqrt{2}$ with center $(1, 0)$.For $S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$, let $z = x + iy$.Then $(1 - i)(x + iy) = x + iy - ix - i^2y = (x + y) + i(y - x)$.So, $\operatorname{Re}((1 - i)z) = x + y \geq 1$.For $S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$, we have $y \leq 1$.The intersection $S_{1} \cap S_{2} \cap S_{3}$ represents a region in the complex plane bounded by the circle $(x - 1)^2 + y^2 = 2$, the line $x + y = 1$, and the line $y = 1$.As shown in the figure, this intersection forms a region with a non-zero area, which implies that the set has infinitely many elements.

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:
$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$
$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$
$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$
Then the set $S_{1} \cap S_{2} \cap S_{3}$

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Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$Then the set $S_{1} \cap S_{2} \cap S_{3}$