The number of elements in the set { z = a + i | vedclass

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(A) The given condition is $1 < |z - (3 - 2i)| < 4$. Let $z = a + ib$,where $a, b \in \mathbb{Z}$.This represents the region between two circles cente

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The given condition is $1 This represents the region between two circles centered at $(3, -2)$ with radii $r_1 = 1$ and $r_2 = 4$.The inequality is $1 Let $x = a - 3$ and $y = b + 2$. Since $a, b \in \mathbb{Z}$,$x, y \in \mathbb{Z}$.We need to find the number of integer pairs $(x, y)$ such that $1 Possible values for $x^2 + y^2$ are $2, 4, 5, 8, 9, 10, 13$.- For $x^2 + y^2 = 2$: $(\pm 1, \pm 1)$ $\rightarrow 4$ points.- For $x^2 + y^2 = 4$: $(\pm 2, 0), (0, \pm 2)$ $\rightarrow 4$ points.- For $x^2 + y^2 = 5$: $(\pm 1, \pm 2), (\pm 2, \pm 1)$ $\rightarrow 8$ points.- For $x^2 + y^2 = 8$: $(\pm 2, \pm 2)$ $\rightarrow 4$ points.- For $x^2 + y^2 = 9$: $(\pm 3, 0), (0, \pm 3)$ $\rightarrow 4$ points.- For $x^2 + y^2 = 10$: $(\pm 1, \pm 3), (\pm 3, \pm 1)$ $\rightarrow 8$ points.- For $x^2 + y^2 = 13$: $(\pm 2, \pm 3), (\pm 3, \pm 2)$ $\rightarrow 8$ points.Total number of points = $4 + 4 + 8 + 4 + 4 + 8 + 8 = 40$.

The number of elements in the set $\{ z = a + ib \in \mathbb{C} : a, b \in \mathbb{Z} \text{ and } 1 < |z - 3 + 2i| < 4 \}$ is:

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