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Complex Numbers question

2022 · 28 Jun · Shift 1 · Q42
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Complex Numbers question

2022 · 28 Jun · Shift 1 · Q42

JEE MainMathematicsComplex NumbersNumerical+4 / −1
The number of elements in the set {z = a + ib ∈\in∈ C : a, b ∈\in∈ Z and 1 < | z −-− 3 + 2i | < 4} is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 40

  1. Interpret the condition carefully

The set is

{z=a+ib∈C:a,b∈Z and 1<∣z−3+2i∣<4}.\{z=a+ib\in \mathbb C: a,b\in \mathbb Z \text{ and } 1<|z-3+2i|<4\}.{z=a+ib∈C:a,b∈Z and 1<∣z−3+2i∣<4}.

Since

z=a+ib,z=a+ib,z=a+ib,

we have

z−3+2i=(a−3)+i(b+2).z-3+2i=(a-3)+i(b+2).z−3+2i=(a−3)+i(b+2).

So,

∣z−3+2i∣=(a−3)2+(b+2)2.|z-3+2i|=\sqrt{(a-3)^2+(b+2)^2}.∣z−3+2i∣=(a−3)2+(b+2)2​.

Thus we need integer lattice points (a,b)(a,b)(a,b) satisfying

1<(a−3)2+(b+2)2<4.1<\sqrt{(a-3)^2+(b+2)^2}<4.1<(a−3)2+(b+2)2​<4.
  1. Square the inequality

Since all quantities are positive,

1<(a−3)2+(b+2)2<16.1<(a-3)^2+(b+2)^2<16.1<(a−3)2+(b+2)2<16.

More precisely,

1<(a−3)2+(b+2)2<4iff1<(a−3)2+(b+2)2<161<\sqrt{(a-3)^2+(b+2)^2}<4 iff 1<(a-3)^2+(b+2)^2<161<(a−3)2+(b+2)2​<4iff1<(a−3)2+(b+2)2<16

is incorrect directly; we must square properly:

1<(a−3)2+(b+2)2<41<\sqrt{(a-3)^2+(b+2)^2}<41<(a−3)2+(b+2)2​<4

becomes

1<(a−3)2+(b+2)2<16.1<(a-3)^2+(b+2)^2<16.1<(a−3)2+(b+2)2<16.

This is correct because squaring gives

12<(a−3)2+(b+2)2<42.1^2<(a-3)^2+(b+2)^2<4^2.12<(a−3)2+(b+2)2<42.

Let

x=a−3,y=b+2.x=a-3,\qquad y=b+2.x=a−3,y=b+2.

Then x,y∈Zx,y\in\mathbb Zx,y∈Z, and we need

1<x2+y2<16.1<x^2+y^2<16.1<x2+y2<16.

So we are counting integer lattice points (x,y)(x,y)(x,y) whose squared distance from the origin is one of

2,4,5,8,9,10,13.2,4,5,8,9,10,13.2,4,5,8,9,10,13.

(We exclude 0,1,160,1,160,1,16.)

  1. Count solutions for each possible value of x2+y2x^2+y^2x2+y2

(i) x2+y2=2x^2+y^2=2x2+y2=2

Possible points:

(±1,±1)(\pm1,\pm1)(±1,±1)

Count: 444.

(ii) x2+y2=4x^2+y^2=4x2+y2=4

Possible points:

(±2,0),(0,±2)(\pm2,0),(0,\pm2)(±2,0),(0,±2)

Count: 444.

(iii) x2+y2=5x^2+y^2=5x2+y2=5

Possible points:

(±1,±2),(±2,±1)(\pm1,\pm2),(\pm2,\pm1)(±1,±2),(±2,±1)

Count: 888.

(iv) x2+y2=8x^2+y^2=8x2+y2=8

Possible points:

(±2,±2)(\pm2,\pm2)(±2,±2)

Count: 444.

(v) x2+y2=9x^2+y^2=9x2+y2=9

Possible points:

(±3,0),(0,±3)(\pm3,0),(0,\pm3)(±3,0),(0,±3)

Count: 444.

(vi) x2+y2=10x^2+y^2=10x2+y2=10

Possible points:

(±1,±3),(±3,±1)(\pm1,\pm3),(\pm3,\pm1)(±1,±3),(±3,±1)

Count: 888.

(vii) x2+y2=13x^2+y^2=13x2+y2=13

Possible points:

(±2,±3),(±3,±2)(\pm2,\pm3),(\pm3,\pm2)(±2,±3),(±3,±2)

Count: 888.

  1. Add all counts

Total number of points is

4+4+8+4+4+8+8=40.4+4+8+4+4+8+8=40.4+4+8+4+4+8+8=40.
  1. Final answer

Hence, the number of elements in the set is

40.\boxed{40}.40​.
  1. Comparison with stored answer

Stored correct answer = 404040.

Our derived answer also equals 404040, so they agree.

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