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Step-by-Step Solution
Step 1: Interpret the Given Sets
We have two sets defined for each natural number $n$:
$S_{n} = \{\,z \in \mathbf{C} : |z - (3 - 2i)| = \frac{n}{4}\}\,$, which is a circle in the complex plane with center $C_{1}(3, -2)$ and radius $r_1 = \frac{n}{4}$.
$T_{n} = \{\,z \in \mathbf{C} : |z - (2 - 3i)| = \frac{1}{n}\}\,$, which is another circle in the complex plane with center $C_{2}(2, -3)$ and radius $r_2 = \frac{1}{n}$.
Step 2: Distance Between the Centers
The distance between the centers $C_{1}(3, -2)$ and $C_{2}(2, -3)$ is computed using the distance formula:
$C_{1} C_{2} = \sqrt{(3 - 2)^{2} + (-2 - (-3))^{2}}
= \sqrt{1^{2} + 1^{2}}
= \sqrt{2}.$
Step 3: Condition for Non-intersecting Circles
Two circles with radii $r_1$ and $r_2$, and distance $d$ between their centers, do not intersect if either of the following is true:
$d > r_1 + r_2$ (the distance between centers is greater than the sum of radii)
$d < |\,r_1 - r_2\,|$ (the distance between centers is less than the absolute difference of the radii)
In our case, $d = \sqrt{2}$. Hence the conditions for $S_{n} \cap T_{n} = \varnothing$ become:
1) $ \sqrt{2} > \frac{n}{4} + \frac{1}{n}, $
2) $ \sqrt{2} < \Bigl|\frac{n}{4} - \frac{1}{n}\Bigr|. $
Step 4: Analyze Each Inequality Separately
First inequality: $ \sqrt{2} > \frac{n}{4} + \frac{1}{n} $
Second inequality: $ \sqrt{2} < \Bigl|\frac{n}{4} - \frac{1}{n}\Bigr| $
We need to find all natural numbers $n$ that satisfy either the first inequality (1) or the second inequality (2), because satisfying either condition implies non-intersection.
Step 5: Check Small Values of n
For n = 1:
Check (1): $\frac{1}{4} + \frac{1}{1} = 1.25 \,$. Since $\sqrt{2} \approx 1.414 > 1.25$, inequality (1) is true.
Check (2): $ \Bigl|\frac{1}{4} - 1\Bigr| = 0.75 \,$. $\sqrt{2} \approx 1.414 \nless 0.75$, so inequality (2) is false.
Because (1) is satisfied, $S_{1} \cap T_{1} = \varnothing$.
For n = 2:
Check (1): $\frac{2}{4} + \frac{1}{2} = 0.5 + 0.5 = 1.0 \,$. $\sqrt{2} \approx 1.414 > 1.0$, so (1) is true.
Check (2): $ \Bigl|\frac{2}{4} - \frac{1}{2}\Bigr| = 0 $, so $\sqrt{2} \approx 1.414 \nless 0$, thus (2) is false.
Because (1) is satisfied, $S_{2} \cap T_{2} = \varnothing$.
For n = 3:
Check (1): $\frac{3}{4} + \frac{1}{3} = 0.75 + 0.3333 \approx 1.0833 \,$. $\sqrt{2} \approx 1.414 > 1.0833$, so (1) is true.
Check (2): $ \Bigl|\frac{3}{4} - \frac{1}{3}\Bigr| \approx 0.4167 \,$, and $\sqrt{2} \approx 1.414 \nless 0.4167$, so (2) is false.
Because (1) is satisfied, $S_{3} \cap T_{3} = \varnothing$.
For n = 4:
Check (1): $\frac{4}{4} + \frac{1}{4} = 1 + 0.25 = 1.25 \,$. Since $\sqrt{2} \approx 1.414 > 1.25$, inequality (1) is true.
Check (2): $ \Bigl|\frac{4}{4} - \frac{1}{4}\Bigr| = 0.75 \,$, and $\sqrt{2} \approx 1.414 \nless 0.75$, so (2) is false.
Because (1) is satisfied, $S_{4} \cap T_{4} = \varnothing$.
For n = 5:
Check (1): $\frac{5}{4} + \frac{1}{5} = 1.25 + 0.2 = 1.45 \,$, and $\sqrt{2} \approx 1.414 \nless 1.45$, so (1) is false.
Check (2): $ \Bigl|\frac{5}{4} - \frac{1}{5}\Bigr| = 1.25 - 0.2 = 1.05 \,$, and $\sqrt{2} \approx 1.414 \nless 1.05$, so (2) is false.
Neither condition holds for $n = 5$, so the circles do intersect. Thus $n = 5$ is not included in the set.
For $n \ge 5$, one can check that $\frac{n}{4} + \frac{1}{n}$ grows sufficiently quickly that $\sqrt{2}$ is no longer greater than the sum of the radii. Meanwhile, the absolute difference of the radii remains too small to be greater than $\sqrt{2}$. Hence no further $n$ values will satisfy either inequality.
Step 6: Count the Suitable n Values
The set of natural numbers $n$ for which $S_{n} \cap T_{n} = \varnothing$ is $\{1,2,3,4\}$. The number of elements in this set is therefore $4$.
Final Answer
The number of elements in the set $\{n \in \mathbf{N} : S_{n} \cap T_{n} = \varnothing\}$ is 4.
