Let S = S₁ ∩ S₂ ∩ S₃, where S₁ = z ∈ C...

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2013Moderate

Comprehension Passage

Let

S = S_{1} \cap S_{2} \cap S_{3}

, where

S_{1} = {z \in C : ∣ z ∣ < 4}

S_{2} = {z \in C : I m [\frac{z - 1 + 3 i}{1 - 3 i}] > 0}

and

S_{3} = {z \in C : R ez > 0}

Question 1:

Area of $S =$

(A)

f r a c 10 p i 3

(B)

f r a c 20 p i 3

(C)

f r a c 16 p i 3

(D)

f r a c 32 p i 3

Answer:B

Question 2:

$mi n_{z in S} ∣1 - 3 i - z ∣ =$

(A)

f r a c 2 - s q r t 3 2

(B)

f r a c 2 + s q r t 3 2

(C)

f r a c 3 - s q r t 3 2

(D)

f r a c 3 + s q r t 3 2

Answer:C

Visualized Solution (English)

Defining the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S_{1}

Set $S_{1} = {z \in C : ∣ z ∣ < 4}$
This represents the interior of a circle centered at the origin $(0, 0)$ with radius $r = 4$ .
Equation of boundary: $x^{2} + y^{2} = 16$ .

Simplifying the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S_{2}

Set $S_{2} = {z \in C : I m [\frac{z - 1 + 3 i}{1 - 3 i}] > 0}$
Let $z = x + i y$ . Rationalize the fraction by multiplying by $1 + 3 i$ :
$\frac{( x - 1 ) + i ( y + 3 )}{1 - 3 i} \cdot \frac{1 + 3 i}{1 + 3 i} = \frac{[( x - 1 ) - 3 ( y + 3 )] + i [ 3 ( x - 1 ) + ( y + 3 )]}{4}$
$I m [...] = \frac{3 x - 3 + y + 3}{4} = \frac{3 x + y}{4}$
Condition: $3 x + y > 0 ⟹ y > - 3 x$ .

Defining the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S_{3} and Intersection

Set $S_{3} = {z \in C : R e (z) > 0} ⟹ x > 0$ .
Boundary of $S_{2}$ is $y = - 3 x$ , which has a slope of $- 3$ (angle $- 6 0^{\circ}$ ).
Intersection $S = S_{1} \cap S_{2} \cap S_{3}$ is a sector of the circle.

Calculating the Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The sector $S$ spans from the positive y-axis ( $9 0^{\circ}$ ) to the line $y = - 3 x$ ( $- 6 0^{\circ}$ ).
Total angle $θ = 9 0^{\circ} - (- 6 0^{\circ}) = 15 0^{\circ}$ .
Area $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{150}{360} \times π (4^{2})$
Area $= \frac{5}{12} \times 16 π = \frac{20 π}{3}$ .

Finding Minimum Distance

Point $P = 1 - 3 i ⟹ (1, - 3)$ . Check its position:
$∣1 - 3 i ∣ = 10 < 4$ (Inside $S_{1}$ ); $x = 1 > 0$ (Inside $S_{3}$ ).
But $3 (1) + (- 3) = 3 - 3 < 0$ , so $P \in / S_{2}$ .
The minimum distance is the perpendicular distance to the line $3 x + y = 0$ .

Final Calculation and Conclusion

Distance $d = \frac{∣ 3 ( 1 ) + 1 ( - 3 ) ∣}{( 3 ) ^{2} + ( 1 ) ^{2}}$
$d = \frac{∣ 3 - 3∣}{2} = \frac{3 - 3}{2}$
Key Takeaways:
1. $S_{2}$ represents a half-plane defined by a line through the origin.
2. The intersection of these constraints forms a circular sector.
3. Minimum distance to a region often involves perpendiculars to its boundary lines.

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Conceptually Similar Problems

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2008Difficult

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Comprehension Passage

Let

A, B, C

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A = {z : I m z \geq 1}

B = {z : ∣ z - 2 - i ∣ = 3}

C = {z : R e ((1 - i) z) = 2}

Question 1:

The number of elements in the set $A \cap B \cap C$ is

(A)

(B)

(C)

(D)

\infty

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25 and 29

(B)

30 and 34

(C)

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(D)

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Question 3:

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Let $ω = \frac{3 + i}{2}$ and $P = {ω^{n} : n = 1, 2, 3, \dots}$ . Further $H_{1} = {z \in C : R ez > 1/2}$ and $H_{2} = {z \in C : R ez < - 1/2}$ , where $C$ is the set of all complex numbers. If $z_{1} \in P \cap H_{1}, z_{2} \in P \cap H_{2}$ and $O$ represents the origin, then $∠ z_{1} O z_{2} =$

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The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

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(A)

S \geq \frac{1}{e}

(B)

S \geq 1 - \frac{1}{e}

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Match the following:

List-I

(P)

In a triangle

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a, b

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c

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X, Y

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Z

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In a triangle

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(R)

R^{2}

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and

β \hat{i} + (1 - β) \hat{j}

be the position vectors of

X, Y

and

Z

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O

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Z

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O X

with

O Y

\frac{3}{2}

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∣ β ∣

is (are)

(S)

Suppose that

F (α)

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and

y = ∣ α x - 1∣ + ∣ α x - 2∣ + α x

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α \in {0, 1}

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F (α) + \frac{8}{3} 2

, when

α = 0

and

α = 1

, is (are)

List-II

(1)

(2)

(3)

(4)

(5)

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Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

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the circle with radius

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(1/2 a, 0)

for

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(B)

the circle with radius

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the x-axis for

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(D)

the y-axis for

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Match the statements given in Column-I with the intervals/union of intervals given in Column-II.

List-I

(P)

The set

{Re (\frac{2 i z}{1 - z ^{2}}) : z is a complex number, ∣ z ∣ = 1, z \neq = \pm 1}

(Q)

The domain of the function

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(R)

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(S)

f (x) = x^{3/2} (3 x - 10), x \geq 0

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(- \infty, - 1) \cup (1, \infty)

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(- \infty, - 0) \cup (0, \infty)

(3)

[2, \infty)

(4)

(- \infty, - 1] \cup [1, \infty)

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Match the statement in Column-I with the values in Column-II

List-I

(P)

A line from the origin meets the lines

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and

\frac{x - \frac{8}{3}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1}

P

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Q

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(S)

Let

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and

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for

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(1)

-4

(2)

(3)

(4)

(5)

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The complex numbers $z_{1}, z_{2}$ and $z_{3}$ satisfying $\frac{z _{1} - z _{3}}{z _{2} - z _{3}} = \frac{1 - i 3}{2}$ are the vertices of a triangle which is

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right-angled isosceles

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(D)

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If $z = (\frac{3}{2} + \frac{i}{2})^{5} + (\frac{3}{2} - \frac{i}{2})^{5}$ , then

(A)

Re(z) = 0

(B)

Im(z) = 0

(C)

Re(z) > 0, Im(z) > 0

(D)

Re(z) > 0, Im(z) < 0

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area of S=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } minzinS​∣1−3i−z∣=

Visualized Solution (English)

Defining the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Simplifying the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Defining the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S3​ and Intersection

Calculating the Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Finding Minimum Distance

Final Calculation and Conclusion

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Area of $S =$

$mi n_{z in S} ∣1 - 3 i - z ∣ =$

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