Let $S = \{z : 3 \le |2z – 3(1 + i)| \le 7\}$ be a set of complex numbers. Then $\text{Min}_{z \in S} |z + \frac{1}{2}(5 + 3i)|$ is equal to :
A) $\frac{1}{2}$ B) $\frac{5}{2}$ C) $2$ D) $\frac{3}{2}$
A) $\frac{1}{2}$ B) $\frac{5}{2}$ C) $2$ D) $\frac{3}{2}$
✓
Solution Steps
1
Simplify the Set S
The given condition is $3 \le |2z – 3(1 + i)| \le 7$. Divide by 2:
$\frac{3}{2} \le |z – \frac{3}{2}(1 + i)| \le \frac{7}{2}$
This describes an annulus (a ring-shaped region) centered at $C(\frac{3}{2}, \frac{3}{2})$ with inner radius $r_1 = \frac{3}{2}$ and outer radius $r_2 = \frac{7}{2}$.
2
Identify the Target Expression
We need to minimize $|z + \frac{1}{2}(5 + 3i)|$, which can be rewritten as $|z – (-\frac{5}{2} – \frac{3}{2}i)|$.
This represents the distance from a point $z$ in set $S$ to the fixed point $P(-\frac{5}{2}, -\frac{3}{2})$.
3
Calculate Distance between Point P and Center C
Point $P = (-\frac{5}{2}, -\frac{3}{2})$ and Center $C = (\frac{3}{2}, \frac{3}{2})$.
$CP = \sqrt{(\frac{3}{2} – (-\frac{5}{2}))^2 + (\frac{3}{2} – (-\frac{3}{2}))^2}$
$CP = \sqrt{(\frac{8}{2})^2 + (\frac{6}{2})^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.
4
Determine Minimum Distance
Since the distance $CP = 5$ is greater than the outer radius $r_2 = 3.5$, the point $P$ lies outside the annulus.
The minimum distance from $P$ to the region $S$ occurs at the intersection of the line $CP$ and the outer boundary of $S$.
$\text{Min Distance} = CP – r_2 = 5 – \frac{7}{2} = 5 – 3.5 = 1.5$.
5
Final Result
The value is $1.5$, which is $\frac{3}{2}$.
Answer: 3/2 (Option D)
📚
Theory
1. Geometrical Locus of |z – z₀| = r
In the complex plane, the equation $|z – z₀| = r$ represents the locus of a circle centered at the complex number $z₀$ with radius $r$. When expressed as an inequality $r₁ \le |z – z₀| \le r₂$, it defines an annulus or a ring-shaped region between two concentric circles. This concept is vital for visualizing regions in complex analysis problems.
2. Distance Formula in Complex Plane
The modulus of the difference between two complex numbers, $|z₁ – z₂|$, represents the Euclidean distance between them. If $z₁ = x₁ + iy₁$ and $z₂ = x₂ + iy₂$, then $|z₁ – z₂| = \sqrt{(x₁ – x₂)² + (y₁ – y₂)²}$. This allows us to convert complex geometry problems into coordinate geometry problems easily.
3. Minimum and Maximum Distances to a Circle
For a point $P$ and a circle with center $C$ and radius $r$, the minimum distance to the circle is $|CP – r|$ and the maximum distance is $CP + r$. If $P$ is outside the circle, the minimum distance is $CP – r$. If $P$ is inside, it is $r – CP$.
4. Algebra of Modulus
Key properties of modulus used in JEE include $|kz| = |k||z|$ for any scalar $k$. In this problem, factoring out 2 from $|2z – 3(1+i)|$ was the crucial first step to identify the center and radius accurately. Misinterpreting the center or radii is the most common cause of error in these problems.
5. Triangle Inequality Application
The triangle inequality states $|z₁ + z₂| \le |z₁| + |z₂|$ and $|z₁ – z₂| \ge ||z₁| – |z₂||$. While not directly used for the final calculation here, it is often the theoretical basis for finding bounds of complex expressions where geometric visualization is difficult.
6. Parametric Representation of a Circle
A point $z$ on the circle $|z – z₀| = r$ can be represented as $z = z₀ + re^{iθ}$, where $θ \in [0, 2π)$. This representation is sometimes used to minimize expressions using calculus, though the geometric approach used here is almost always faster for JEE Main.
❓
FAQs
1
Why did we divide the entire inequality by 2?
⌄
To isolate ‘z’ inside the modulus. The standard form for a circle is $|z – z₀| = r$. Having a coefficient like 2z complicates identifying the true radius and center.
2
How do you know if the point P is inside or outside the region?
⌄
Compare the distance between the point and the center (CP = 5) with the radii. Since 5 > 3.5 (outer radius), the point is outside the entire region.
3
What would be the maximum distance in this problem?
⌄
The maximum distance would be from point P to the furthest point on the boundary, which is $CP + \text{outer radius} = 5 + 3.5 = 8.5$ or $17/2$.
4
What if CP was less than the inner radius?
⌄
Then the point would be in the “hole” of the annulus. The minimum distance would be $\text{inner radius} – CP$.
5
Can this be solved using $x$ and $y$ coordinates?
⌄
Yes, but it results in much longer calculations involving square roots and inequalities. Geometric interpretation is the preferred “JEE way”.
6
Why did $|z + 1/2(5+3i)|$ become distance from $(-2.5, -1.5)$?
⌄
Because the distance between $z$ and $z₀$ is $|z – z₀|$. We write $|z + (2.5 + 1.5i)|$ as $|z – (-2.5 – 1.5i)|$ to find the target point $z₀$.
7
What is an annulus?
⌄
It is the region between two concentric circles. Think of it as a flat 2D donut shape.
8
Is there any case where the minimum distance is 0?
⌄
Yes, if the point $P$ lies anywhere within the region $S$ (i.e., if $1.5 \le CP \le 3.5$), the minimum distance would be 0.
9
Is the modulus always positive?
⌄
Yes, the modulus represents a distance, and distance is always non-negative by definition.
10
What does $z \in S$ mean?
⌄
It means that the complex number $z$ must satisfy the initial inequality and thus must be a point located within the annulus.
🔗
Related Questions
Q
Let a, b, c be distinct real numbers. Common root of a(x²+1)+bx+c=0 and ax²+bx+c=0… · Options: A) 1, B) 2, C) 1/2, D) 3 · Ans: 1 ✓
Quadratic Equations
→
Q
A rectangle is formed by x=0, y=0, x=3, y=4. Line L bisects area… · Options: A) √10, B) 2√5, C) 2√10, D) 3√10 · Ans: 2√10 ✓
Coordinate Geometry
→
Q
Let the domain of f(x) = log₃ log₅ log₇ (9x – x² – 13) be (m, n)… find b² – a² · Options: A) 5, B) 7, C) 8, D) 10 · Ans: 7 ✓
Functions
→
Q
Let S be the set of first 11 natural numbers… subsets with even product · Options: A) 1024, B) 2048, C) 1979, D) 2000 · Ans: 1979 ✓
Combinatorics
→
Q
Let cos(α + β) = -1/10 and sin(α – β) = 3/8… find r + s · Options: A) 10, B) 20, C) 25, D) 15 · Ans: 20 ✓
Trigonometry
→