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 (Hindi)

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

View in:English Hindi

Comprehension Passage

Let

A, B, C

be three sets of complex numbers as defined below

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

Answer:B

Question 2:

Let $z$ be any point in $A \cap B \cap C$ . Then, $∣ z + 1 - i ∣^{2} + ∣ z - 5 - i ∣^{2}$ lies between

(A)

25 and 29

(B)

30 and 34

(C)

35 and 39

(D)

40 and 44

Answer:C

Question 3:

Let $z$ be any point $A \cap B \cap C$ and let $w$ be any point satisfying $∣ w - 2 i ∣ < 3$ . Then, $∣ z ∣ - ∣ w ∣ + 3$ lies between

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-6 and 3

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(C)

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MathematicsArgand Plane and Polar RepresentationJEE Advanced 2013Moderate

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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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2 π /3

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MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2005SModerate

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

(A)

z : ∣ z + 1∣ > 2

and

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(B)

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∣ a r g (z - 1) ∣ < π /4

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z : ∣ z + 1∣ < 2

and

∣ a r g (z + 1) ∣ < π /2

(D)

z : ∣ z - 1∣ < 2

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∣ a r g (z + 1) ∣ < π /2

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MathematicsArea Bounded by CurvesJEE Advanced 2012Moderate

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Let $S$ be the area of the region enclosed by $y = e^{- x^{2}}, y = 0, x = 0$ and $x = 1$ ; then

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(A)

S \geq \frac{1}{e}

(B)

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

(C)

S \leq \frac{1}{4} (1 + \frac{1}{e})

(D)

S \leq \frac{1}{2} + \frac{1}{e} (1 - \frac{1}{2})

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MathematicsComponents of a VectorJEE Advanced 2015Difficult

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Match the following:

List-I

(P)

In a triangle

Δ X Y Z

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a, b

, and

c

be the lengths of the sides opposite to the angles

X, Y

and

Z

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2 (a^{2} - b^{2}) = c^{2}

and

λ = \frac{s i n ( X - Y )}{s i n Z}

, then possible values of

n

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is (are)

(Q)

In a triangle

Δ X Y Z

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a, b

and

c

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X, Y

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Z

respectively. If

1 + cos 2 X - 2 cos 2 Y = 2 sin X sin Y

, then possible value(s) of

\frac{a}{b}

is (are)

(R)

R^{2}

, let

3 \hat{i} + \hat{j}, \hat{i} + 3 \hat{j}

and

β \hat{i} + (1 - β) \hat{j}

be the position vectors of

X, Y

and

Z

with respect to the origin

O

, respectively. If the distance of

Z

from the bisector of the acute angle of

O X

with

O Y

\frac{3}{2}

, then possible value(s) of

∣ β ∣

is (are)

(S)

Suppose that

F (α)

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x = 0, x = 2, y^{2} = 4 x

and

y = ∣ α x - 1∣ + ∣ α x - 2∣ + α x

, where

α \in {0, 1}

. Then the value(s) of

F (α) + \frac{8}{3} 2

, when

α = 0

and

α = 1

, is (are)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → NaN, Q → 1, P → 2, S → 5

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2016Moderate

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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

* Multiple Correct Options

(A)

the circle with radius

1/2 a

and centre

(1/2 a, 0)

for

a > 0, b \neq = 0

(B)

the circle with radius

- 1/2 a

and centre

(- 1/2 a, 0)

for

a < 0, b \neq = 0

(C)

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

f (x) = sin^{- 1} (\frac{8 ( 3 ) ^{x - 2}}{1 - 3 ^{2 (x - 1)}})

(R)

f (θ) = 1 - tan θ - 1 tan θ 1 - tan θ 1 tan θ 1

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{f (θ) : 0 \leq θ < \frac{π}{2}}

(S)

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

then

f (x)

is increasing in

List-II

(1)

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

(2)

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

(3)

[2, \infty)

(4)

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

(5)

(0, 0] \cup [2, \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

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

and

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

P

and

Q

respectively. If length

P Q = d

, then

d^{2}

(Q)

The values of

x

satisfying

tan^{- 1} (x + 3) - tan^{- 1} (x - 3) = sin^{- 1} (\frac{3}{5})

are

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Non-zero vectors

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and

c

satisfy

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(b - a) \cdot (b + c) = 0

and

2∣ b + c ∣ = ∣ b - a ∣

. If

a = μ b + 4 c

, then the possible values of

μ

are

(S)

Let

f

be the function on

[- π, π]

given by

f (0) = 9

and

f (x) = \frac{s i n ( \frac{9 x}{2} )}{s i n ( \frac{x}{2} )}

for

x \neq = 0

. The value of

\frac{2}{π} \int_{- π}^{π} f (x) d x

List-II

(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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(B)

right-angled isosceles

(C)

equilateral

(D)

obtuse-angled isosceles

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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

Answer:B

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 (Hindi)

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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Calculating the Area of $S$

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