Animated Solution for Mathematics - Complex Numbers: For n∈N, let Sn={z∈C:∣z−(3+2i)∣=4n} and Tn={z∈C:∣z−(2+3i)∣=n1}. Then the number of elements in the set {n∈N:Sn∩Tn=∅} is:
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Visualized Solution
Locus of Sn
Sn={z∈C:∣z−(3+2i)∣=4n}
This represents a circle in the complex plane.
Center: C1(3,2)
Radius: r1=4n
Locus of Tn
Tn={z∈C:∣z−(2+3i)∣=n1}
This represents another circle.
Center: C2(2,3)
Radius: r2=n1
Distance Between Centers
Distance d=(3−2)2+(2−3)2
d=12+(−1)2
d=2≈1.414
Condition for Sn∩Tn=∅
The problem requires Sn∩Tn=∅.
This means the two circles must not intersect.
Case 1: Circles are completely outside each other (d>r1+r2).
Case 2: One circle is completely inside the other (d<∣r1−r2∣).
Case 1: Circles Outside Each Other
Condition: d>r1+r2
Substitute values: 2>4n+n1
Multiply by 4n (since n>0): 42n>n2+4
Rearrange: n2−42n+4<0
Solving Case 1 Inequality
Find roots of n2−42n+4=0
n=242±32−16=22±2
Roots are approximately 2(1.414)−2=0.828 and 2(1.414)+2=4.828
Inequality holds for n∈(0.828,4.828)
Valid Natural Numbers for Case 1
We need n∈N such that 0.828<n<4.828
The possible natural numbers are n∈{1,2,3,4}
So, Case 1 gives 4 valid values for n.
Case 2: One Circle Inside the Other
Condition: d<∣r1−r2∣
Substitute values: 2<∣4n−n1∣
For n≥2, 4n>n1, so the modulus opens positively.
2<4n−n1
Solving Case 2 Inequality
Multiply by 4n: 42n<n2−4
Rearrange: n2−42n−4>0
Roots of equation: n=242±32+16=22±23
Valid Natural Numbers for Case 2
Approximate the positive root: 2(1.414)+2(1.732)=2.828+3.464=6.292
The inequality n2−42n−4>0 holds for n>6.292
Since n∈N, the valid values are n∈{7,8,9,…}
Final Conclusion
Combining both cases, the valid values for n are:
n∈{1,2,3,4}∪{7,8,9,10,…}
The set contains an Infinite number of elements.
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The Sigma Insight: Geometrical Applications of Complex Numbers
Solution Diagram
Analyzing the Setup
Welcome, fellow traveler of the complex plane! Today, we are going to explore a beautiful problem that bridges the gap between algebra and geometry.
We are looking at two sets, Sn and Tn, defined by the modulus of complex numbers. At first glance, these might look like abstract equations, but they are actually the heartbeat of geometry: circles.
Visualizing the Locus
Let Sn={z∈C:∣z−(3+2i)∣=4n}. This is the definition of a circle in the complex plane with center C1(3,2) and radius r1=4n.
Now, consider Tn={z∈C:∣z−(2+3i)∣=n1}. This is a second circle centered at C2(2,3) with radius r2=n1.
We have a dynamic system where one circle expands and the other contracts as n increases. Our goal is to find the values of n for which these circles never touch.
The Dance of the Centers
To understand their interaction, we calculate the distance d between their centers. Using the distance formula between (3,2) and (2,3):
d=(3−2)2+(2−3)2=12+(−1)2=2≈1.414
This distance is our anchor; it remains constant regardless of the value of n.
The Geometry of Separation
For two circles to have an empty intersection, they must be separated. This occurs in two distinct geometric configurations:
1. The circles are completely outside each other: d>r1+r2.
2. One circle is entirely contained within the other: d<∣r1−r2∣.
Solving the Inequalities
For Case 1 (d>r1+r2), we have:
2>4n+n1
Multiplying by 4n (given n>0), we obtain 42n>n2+4, or:
n2−42n+4<0
The roots of n2−42n+4=0 are 22±2. Approximating these, we find n∈(0.828,4.828). Since n is a natural number, n∈{1,2,3,4}.
For Case 2 (d<∣r1−r2∣), we have:
2<4n−n1
For n≥2, this simplifies to 2<4n−n1, which rearranges to n2−42n−4>0. The positive root is 22+23≈6.292.
Thus, n>6.292, giving us the set n∈{7,8,9,…}.
Final Conclusion
Combining these results, the valid values for n are {1,2,3,4,7,8,9,…}.
Because the second set extends to infinity, the total number of elements is infinite. You have successfully navigated the complex plane and mastered the geometry of circles!