Let A-z- z-z-2i2 - 2z- z- - where i-1 and B-z-z-2 - 5- Then number of points with integral real and imaginary parts of z lying in A - B is

Let z=x+iy nbsp; Then A:{(x,y): y^2 ≤ 4x} and B:{(x,y): x^2+y^2 ≤ 5} A ∩ B is the shaded region. Points in A ∩ B with integral coordinates ={(0,0), (1,0), ( ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of a Complex Number

Let A:z: z-z...

Question

Let $A : {z : (\frac{z - ¯ z}{2 i})^{2} \leq 2 (z + ¯ z)}$ , where $i = \sqrt{- 1} and B : {z : | z |^{2} \leq 5}$ . Then number of points with integral real and imaginary parts of $z$ lying in $A \cap B$ is

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Solution

The correct option is D $9$
$Let z = x + i y$
$Then A : {(x, y) : y^{2} \leq 4 x}$
$and B : {(x, y) : x^{2} + y^{2} \leq 5}$

$A \cap B$ is the shaded region.
Points in $A \cap B$ with integral coordinates $= {(0, 0), (1, 0), (2, 0), (1, 1), (1, - 1), (1, 2), (1, - 2), (2, 1), (2, - 1)}$
Hence, there are 9 points.

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Let A:{z:(z−¯z2i)2≤2(z+¯z)}, where i=√−1 and B:{z:|z|2≤5}. Then number of points with integral real and imaginary parts of z lying in A∩B is

The correct option is D 9 Let z=x+iy Then A:{(x,y):y2≤4x} and B:{(x,y):x2+y2≤5} A∩B is the shaded region. Points in A∩B with integral coordinates ={(0,0),(1,0),(2,0),(1,1), (1,−1),(1,2),(1,−2),(2,1),(2,−1)} Hence, there are 9 points.

Let $A : {z : (\frac{z - ¯ z}{2 i})^{2} \leq 2 (z + ¯ z)}$ , where $i = \sqrt{- 1} and B : {z : | z |^{2} \leq 5}$ . Then number of points with integral real and imaginary parts of $z$ lying in $A \cap B$ is