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Complex Numbers question

2022 · 25 Jul · Shift 1 · Q23
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Complex Numbers question

2022 · 25 Jul · Shift 1 · Q23

JEE MainMathematicsComplex NumbersMCQ+4 / −1
For n∈N\mathrm{n} \in \mathbf{N}n∈N, let Sn={z∈C:∣z−3+2i∣=n4}\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}Sn​={z∈C:∣z−3+2i∣=4n​} and Tn={z∈C:∣z−2+3i∣=1n}\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}Tn​={z∈C:∣z−2+3i∣=n1​}. Then the number of elements in the set {n∈N:Sn∩Tn=ϕ}\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\}{n∈N:Sn​∩Tn​=ϕ} is :
  1. A
    0
  2. B
    2
  3. C
    3
  4. D
    4
View written solutionFree

Correct answer: D

  1. Interpret the sets geometrically

    Sn={z∈C:∣z−(3−2i)∣=n4}S_n=\{z\in \mathbb C:|z-(3-2i)|=\tfrac n4\}Sn​={z∈C:∣z−(3−2i)∣=4n​} is a circle with center C1=3−2i⇒(3,−2)C_1=3-2i\quad \Rightarrow \quad (3,-2)C1​=3−2i⇒(3,−2) and radius r1=n4.r_1=\frac n4.r1​=4n​.

    Similarly, Tn={z∈C:∣z−(2−3i)∣=1n}T_n=\{z\in \mathbb C:|z-(2-3i)|=\tfrac1n\}Tn​={z∈C:∣z−(2−3i)∣=n1​} is a circle with center C2=2−3i⇒(2,−3)C_2=2-3i\quad \Rightarrow \quad (2,-3)C2​=2−3i⇒(2,−3) and radius r2=1n.r_2=\frac1n.r2​=n1​.

  2. Find distance between centers

    d=∣(3−2i)−(2−3i)∣=∣1+i∣=2.d=| (3-2i)-(2-3i)|=|1+i|=\sqrt{2}.d=∣(3−2i)−(2−3i)∣=∣1+i∣=2​.

  3. Condition for two circles to have empty intersection

    Two circles do not intersect if either

    • they are externally disjoint: d>r1+r2,d>r_1+r_2,d>r1​+r2​,
    • or one lies completely inside the other without touching: d<∣r1−r2∣.d<|r_1-r_2|.d<∣r1​−r2​∣.

    So we check both cases.

  4. Check the possibility d<∣r1−r2∣d<|r_1-r_2|d<∣r1​−r2​∣

    Here ∣r1−r2∣=∣n4−1n∣.|r_1-r_2|=\left|\frac n4-\frac1n\right|.∣r1​−r2​∣=​4n​−n1​​.

    For small and moderate natural nnn, this is much smaller than 2\sqrt22​ except possibly larger nnn. Let us see directly:

    • For n=1n=1n=1: ∣r1−r2∣=∣14−1∣=34<2.|r_1-r_2|=\left|\frac14-1\right|=\frac34<\sqrt2.∣r1​−r2​∣=​41​−1​=43​<2​.
    • For n≥2n\ge 2n≥2, n4−1n\frac n4-\frac1n4n​−n1​ increases, but if this exceeds 2\sqrt22​, then certainly also r1+r2>2r_1+r_2>\sqrt2r1​+r2​>2​ and the circles will overlap or contain one another depending on exact values. We only need values where intersection is empty.

    The more relevant disjoint condition here is external disjointness, since radii are positive and center distance is fixed.

  5. Use external disjointness condition

    We need 2>n4+1n.\sqrt2>\frac n4+\frac1n.2​>4n​+n1​.

    Multiply by 4n>04n>04n>0: 42 n>n2+4.4\sqrt2\,n>n^2+4.42​n>n2+4.

    Rearranging, n2−42 n+4<0.n^2-4\sqrt2\,n+4<0.n2−42​n+4<0.

    Solve the quadratic equation n2−42 n+4=0.n^2-4\sqrt2\,n+4=0.n2−42​n+4=0.

    Its roots are

    =\frac{4\sqrt2\pm\sqrt{32-16}}{2} =\frac{4\sqrt2\pm4}{2} =2\sqrt2\pm2.$$ Therefore, $$2\sqrt2-2<n<2\sqrt2+2.$$ Since $$2\sqrt2\approx 2.828,$$ we get $$0.828<n<4.828.$$ So the natural numbers satisfying this are $$n=1,2,3,4.$$
  6. Check these values

    • n=1n=1n=1: r1+r2=14+1=1.25<2,r_1+r_2=\frac14+1=1.25<\sqrt2,r1​+r2​=41​+1=1.25<2​, so disjoint.
    • n=2n=2n=2: r1+r2=12+12=1<2,r_1+r_2=\frac12+\frac12=1<\sqrt2,r1​+r2​=21​+21​=1<2​, so disjoint.
    • n=3n=3n=3: r1+r2=34+13=1312≈1.083<2,r_1+r_2=\frac34+\frac13=\frac{13}{12}\approx1.083<\sqrt2,r1​+r2​=43​+31​=1213​≈1.083<2​, so disjoint.
    • n=4n=4n=4: r1+r2=1+14=1.25<2,r_1+r_2=1+\frac14=1.25<\sqrt2,r1​+r2​=1+41​=1.25<2​, so disjoint.

    For n≥5n\ge 5n≥5, n4+1n≥54+15=1.45>2,\frac n4+\frac1n\ge \frac54+\frac15=1.45>\sqrt2,4n​+n1​≥45​+51​=1.45>2​, so they are not externally disjoint.

    Also, checking internal disjointness for n≥5n\ge5n≥5: ∣n4−1n∣<2\left|\frac n4-\frac1n\right|<\sqrt2​4n​−n1​​<2​ fails to make them disjoint in the required range of natural numbers here, so no extra values arise.

  7. Count the values

    The set is {1,2,3,4},\{1,2,3,4\},{1,2,3,4}, so the number of elements is 4.4.4.

  8. Option matching

    Correct option is: D\boxed{\text{D}}D​

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