(B) The given inequality is $|z - (2 + 2i)| \leq 1$. Let $z = x + iy$,where $x, y \in \mathbb{R}$. Substituting $z$ into the inequality,we get $|(x -

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a closed circular disc with center at $(2, 2)$ and radius $1$

(B) The given inequality is $|z - (2 + 2i)| \leq 1$. Let $z = x + iy$,where $x, y \in \mathbb{R}$. Substituting $z$ into the inequality,we get $|(x - 2) + i(y - 2)| \leq 1$. Squaring both sides,we obtain $(x - 2)^2 + (y - 2)^2 \leq 1^2$. This represents a closed circular disc in the complex plane with center at $(2, 2)$ and radius $r = 1$. Therefore,option $B$ is correct.

Class 11 Mathematics 4-1.Complex numbers Geometry of complex numbers

Medium

Geometrically,the set $\{z \in \mathbb{C} : |z - 2 - 2i| \leq 1\}$ represents

AP EAMCET 2020 All AP EAMCET

A
a closed circular disc with center at $(-2, -2)$ and radius $1$
B
a closed circular disc with center at $(2, 2)$ and radius $1$
C
a closed circular disc with center at $(1, 1)$ and radius $0.5$
D
a closed circular disc with center at $(-1, -1)$ and radius $0.5$

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4-1.Complex numbers

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Let $C$ be the set of all complex numbers. Let $S_{1}=\{z \in C:|z-2| \leq 1\}$ and $S_{2}=\{z \in C: z(1+i)+\overline{z}(1-i) \geq 4\}$. Then,the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to:

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If $|z-2|=|z-1|$,where $z$ is a complex number,then the locus of $z$ is a straight line:

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Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

DifficultJEE MAIN 2022

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If $A = \{z : |\frac{z - 2}{z + 2}| = 3, z \in C\}$ and $z_1, z_2, z_3, z_4 \in A$ are $4$ complex numbers representing points $P, Q, R, S$ respectively on the complex plane such that $z_1 - z_2 = z_4 - z_3$,then the maximum value of the area of quadrilateral $PQRS$ is:

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The triangle formed by the complex numbers $z_1$,$z_2$,and $-\omega z_1 - \omega^2 z_2$ on the Argand plane is:

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Geometrically,the set $\{z \in \mathbb{C} : |z - 2 - 2i| \leq 1\}$ represents

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Let $C$ be the set of all complex numbers. Let $S_{1}=\{z \in C:|z-2| \leq 1\}$ and $S_{2}=\{z \in C: z(1+i)+\overline{z}(1-i) \geq 4\}$. Then,the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to:

If $|z-2|=|z-1|$,where $z$ is a complex number,then the locus of $z$ is a straight line:

Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

If $A = \{z : |\frac{z - 2}{z + 2}| = 3, z \in C\}$ and $z_1, z_2, z_3, z_4 \in A$ are $4$ complex numbers representing points $P, Q, R, S$ respectively on the complex plane such that $z_1 - z_2 = z_4 - z_3$,then the maximum value of the area of quadrilateral $PQRS$ is:

The triangle formed by the complex numbers $z_1$,$z_2$,and $-\omega z_1 - \omega^2 z_2$ on the Argand plane is:

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