JEE Advanced 2013

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: Comprehension Passage

Q: Why do we rationalize the denominator in $S_2$?

Great question! The expression $Im\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0$ is difficult to interpret directly. By rationalizing, we transform the complex fraction into a form $A + iB$, where $B$ is a real linear expression in $x$ and $y$. This allows us to convert a complex inequality into a simple, familiar coordinate geometry inequality ($y > -\sqrt{3}x$).

Q: Why is the area a sector and not a triangle?

Remember that $S_1$ is $|z| < 4$, which is the interior of a circle. The boundaries of our region are two lines intersecting at the origin and the circular arc of the circle itself. Any region bounded by two radii and an arc of a circle is, by definition, a circular sector.

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 $S =$

Select Answer:

Question 2:

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Select Answer:

Progressive Hints

Hint 1

The region

S

is defined by three constraints.

S_{1}

is a disk of radius

4

centered at the origin.

S_{3}

restricts us to the right half-plane (

x > 0

). The most complex constraint,

S_{2}

, involves a complex fraction; rationalize it by multiplying the numerator and denominator by the conjugate of the denominator (

1 + 3 i

) to reveal a linear inequality in

x

and

y

Hint 2

Once you have the three inequalities, visualize them on the Argand plane. The line from

S_{2}

and the vertical line from

S_{3}

intersect the circle

∣ z ∣ = 4

. Calculate the angle each boundary line makes with the positive x-axis to determine the total angular width of the circular sector

S

Hint 3

For the minimum distance, first check if the point

P (1, - 3)

lies within the region

S

. If it lies outside, the shortest distance to the region is the perpendicular distance to the nearest boundary line. Use the formula

d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

for the line derived from

S_{2}

Visualized Solution

Defining the Region $S_{1}$

Let's analyze the first set: $S_{1} = {z \in C : ∣ z ∣ < 4}$ .
The condition $∣ z ∣ < 4$ represents the interior of a circle.
Center is at the origin $(0, 0)$ and radius is $r = 4$ .

Analyzing the Region $S_{2}$

The second set is $S_{2} = {z \in C : I m [\frac{z - 1 + 3 i}{1 - 3 i}] > 0}$ .
To find the imaginary part, we substitute $z = x + i y$ .
We must rationalize the denominator by multiplying the numerator and denominator by its conjugate, $1 + 3 i$ .

Rationalizing the Expression

Substitute $z = x + i y$ : $\frac{( x - 1 ) + i ( y + 3 )}{1 - 3 i}$
Multiply by $\frac{1 + 3 i}{1 + 3 i}$ :
Denominator becomes $1^{2} + (3)^{2} = 4$ .

Extracting the Imaginary Part

Numerator: $[(x - 1) + i (y + 3)] [1 + 3 i]$
Imaginary part comes from: $(x - 1) (3) + (y + 3) (1)$
$I m = \frac{3 x - 3 + y + 3}{4} = \frac{3 x + y}{4}$

The Boundary Line of $S_{2}$

Condition: $I m > 0 ⟹ \frac{3 x + y}{4} > 0$
This simplifies to $3 x + y > 0$ , or $y > - 3 x$ .
This represents the region above the line $y = - 3 x$ .

Defining the Region $S_{3}$

The third set is $S_{3} = {z \in C : R e (z) > 0}$ .
This means $x > 0$ , which is the right half of the complex plane.
The intersection $S = S_{1} \cap S_{2} \cap S_{3}$ is the region satisfying all three conditions.

Visualizing the Intersection $S$

$S$ is bounded by the circle $∣ z ∣ = 4$ , the positive y-axis ( $x = 0$ ), and the line $y = - 3 x$ .
The line $y = - 3 x$ has a slope of $- 3$ , corresponding to an angle of $- 6 0^{\circ}$ .
The positive y-axis corresponds to an angle of $9 0^{\circ}$ .

Calculating the Area of $S$

The region $S$ is a circular sector.
The total angle subtended at the origin is $θ = 9 0^{\circ} - (- 6 0^{\circ}) = 15 0^{\circ}$ .
Area of a sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2}$ .

Final Area Calculation

Substitute $θ = 15 0^{\circ}$ and $r = 4$ :
Area $= \frac{150}{360} \times π (4)^{2}$
Area $= \frac{5}{12} \times 16 π = \frac{20 π}{3}$ .
This matches option (2) for the first question.

The Minimum Distance Problem

Now, we need to find $min_{z \in S} ∣1 - 3 i - z ∣$ .
This is the minimum distance from the point $P (1, - 3)$ to any point $z$ in the region $S$ .
Let's locate $P (1, - 3)$ on the complex plane.

Locating Point $P$

Check if $P (1, - 3)$ is inside $S$ :
$∣ P ∣ = 1^{2} + (- 3)^{2} = 10 < 4$ (Inside $S_{1}$ ).
$R e (P) = 1 > 0$ (Inside $S_{3}$ ).
$I m (P) = - 3$ . But for $S_{2}$ , we need $y > - 3 x ⟹ - 3 > - 3 (1)$ , which is $- 3 > - 1.732$ (False).

Shortest Distance to the Region

Since $P$ is outside $S$ but inside the circle, the closest point in $S$ lies on the boundary line $3 x + y = 0$ .
The shortest distance from a point $(x_{1}, y_{1})$ to a line $A x + B y + C = 0$ is $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ .

Calculating the Minimum Distance

Substitute $P (1, - 3)$ and line $3 x + y = 0$ :
$d = \frac{∣ 3 ( 1 ) + 1 ( - 3 ) ∣}{( 3 ) ^{2} + ( 1 ) ^{2}}$
$d = \frac{∣ 3 - 3∣}{3 + 1} = \frac{3 - 3}{2}$

Final Conclusion

The minimum distance is $\frac{3 - 3}{2}$ .
This matches option (3) for the second question.
Key Takeaway: Complex number inequalities often translate directly to standard coordinate geometry regions like circles, lines, and half-planes.

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

We define the region

S

as the intersection of three distinct sets in the complex plane. Our first set,

S_{1} = {z \in C : ∣ z ∣ < 4}

, represents an open disk of radius

4

centered at the origin.

The second set,

S_{2}

, is defined by the condition:

I m [\frac{z - 1 + i 3}{1 - i 3}] > 0

To simplify this, we substitute

z = x + i y

and multiply the numerator and denominator by the conjugate

1 + i 3

. The denominator becomes

1^{2} + (3)^{2} = 4

Expanding the numerator, we isolate the imaginary part:

I m [\frac{( x + i y - 1 + i 3 ) ( 1 + i 3 )}{4}] = \frac{3 x + y}{4}

Setting this expression greater than zero yields the linear boundary:

y > - 3 x

Finally,

S_{3} = {z \in C : R e (z) > 0}

restricts our region to the right half-plane, where

x > 0

Visualizing the Sector

The region

S

is the intersection of the disk, the area above the line

y = - 3 x

, and the area to the right of the

y

-axis. The line

y = - 3 x

corresponds to an angle of

- 6 0^{\circ}

(or

30 0^{\circ}

) with the positive

x

-axis.

The

y

-axis (

x = 0

) corresponds to an angle of

9 0^{\circ}

. Thus, the region

S

is a circular sector spanning an angle of:

θ = 9 0^{\circ} - (- 6 0^{\circ}) = 15 0^{\circ}

The area of a circular sector is calculated using the formula

A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}

. Substituting

θ = 15 0^{\circ}

and

r = 4

A = \frac{150}{360} \times π (4)^{2} = \frac{5}{12} \times 16 π = \frac{20 π}{3}

The Final Challenge

Minimum Distance

We now find the minimum distance from the point

P (1, - 3)

to the region

S

. Since

P

lies outside the sector, the shortest path is the perpendicular distance to the nearest boundary.

The nearest boundary is the line

3 x + y = 0

. We apply the point-to-line distance formula:

d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

Substituting

A = 3

B = 1

C = 0

x_{1} = 1

, and

y_{1} = - 3

d = \frac{∣ 3 ( 1 ) + 1 ( - 3 ) ∣}{( 3 ) ^{2} + 1 ^{2}} = \frac{∣ 3 - 3∣}{2}

Since

3 < 3

, the absolute value results in

3 - 3

. The minimum distance is: $d = \frac{3 - 3}{2}$

Common Traps & FAQs

Why do we rationalize the denominator in

S_{2}

Great question! The expression

I m [\frac{z - 1 + 3 i}{1 - 3 i}] > 0

is difficult to interpret directly. By rationalizing, we transform the complex fraction into a form

A + i B

, where

B

is a real linear expression in

x

and

y

. This allows us to convert a complex inequality into a simple, familiar coordinate geometry inequality (

y > - 3 x

How do I know if the point

P (1, - 3)

is inside or outside the region?

To check if a point is in the region, it must satisfy all three inequalities simultaneously. We know

P

is inside the circle (

∣ P ∣ = 10 < 4

) and in the right half-plane (

R e (P) = 1 > 0

). However, when we test

P

in the line inequality

y > - 3 x

, we get

- 3 > - 3 (1)

, which is

- 3 > - 1.732

. This is false! Thus,

P

is outside the region, specifically below the boundary line.

Why is the area a sector and not a triangle?

Remember that

S_{1}

∣ z ∣ < 4

, which is the interior of a circle. The boundaries of our region are two lines intersecting at the origin and the circular arc of the circle itself. Any region bounded by two radii and an arc of a circle is, by definition, a circular sector.

Animated Solution for Mathematics - Complex Numbers: Comprehension Passage

Area of $S =$

Select Answer:

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Select Answer:

Visualized Solution

Defining the Region $S_{1}$

Analyzing the Region $S_{2}$

Rationalizing the Expression

Extracting the Imaginary Part

The Boundary Line of $S_{2}$

Defining the Region $S_{3}$

Visualizing the Intersection $S$

Calculating the Area of $S$

Final Area Calculation

The Minimum Distance Problem

Locating Point $P$

Shortest Distance to the Region

Calculating the Minimum Distance

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Visualizing the Sector

The Final Challenge

Similar Questions

Let $S_{1} = {z \in C : ∣ z ∣ \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - 3 i}{1 - 3 i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0}$ . Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is :

The area (in sq. units) of the region $S = {z \in C : ∣ z - 1∣ \leq 2; (z + \overset{z}{ˉ}) + i (z - \overset{z}{ˉ}) \leq 2, Im (z) \geq 0}$ is

Let $S = {z \in C : ∣ z - 1∣ = 1 and (2 - 1) (z + \overset{z}{ˉ}) - i (z - \overset{z}{ˉ}) = 22}$ . Let $z_{1}, z_{2} \in S$ be such that $∣ z_{1} ∣ = max_{z \in S} ∣ z ∣$ and $∣ z_{2} ∣ = min_{z \in S} ∣ z ∣$ . Then $∣ 2 z_{1} - z_{2} ∣^{2}$ equals :

Comprehension Passage

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

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

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

Let the curve $z (1 + i) + \overline{z} (1 - i) = 4, z \in C$ , divide the region $∣ z - 3∣ \leq 1$ into two parts of areas $α$ and $β$ . Then $∣ α - β ∣$ equals:

Let $C$ be the set of all complex numbers. Let $S_{1} = {z \in C : ∣ z - 2∣ \leq 1}$ and $S_{2} = {z \in C : z (1 + i) + \overset{z}{ˉ} (1 - i) \geq 4}$ . Then, the maximum value of $∣ z - \frac{5}{2} ∣^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to :

Let $C$ be the set of all complex numbers. Let $S_{1} = {z \in C : ∣ z - 3 - 2 i ∣^{2} = 8}$ , $S_{2} = {z \in C : Re (z) \geq 5}$ and $S_{3} = {z \in C : ∣ z - \overset{z}{ˉ} ∣ \geq 8}$ . Then the number of elements in $S_{1} \cap S_{2} \cap S_{3}$ is equal to

Let a complex number be $w = 1 - 3 i$ . Let another complex number $z$ be such that $∣ z w ∣ = 1$ and $ar g (z) - ar g (w) = \frac{π}{2}$ . Then the area of the triangle with vertices origin, $z$ and $w$ is equal to:

Animated Solution for Mathematics - Complex Numbers: Comprehension Passage

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Select Answer:

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Select Answer:

Visualized Solution

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

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

Rationalizing the Expression

Extracting the Imaginary Part

The Boundary Line of .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​

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

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

Final Area Calculation

The Minimum Distance Problem

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Shortest Distance to the Region

Calculating the Minimum Distance

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Visualizing the Sector

The Final Challenge

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S1​={z∈C:∣z∣≤5}, S2​={z∈C:Im(1−3​iz+1−3​i​)≥0} and S3​={z∈C:Re(z)≥0}. Then the area of the region S1​∩S2​∩S3​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region S={z∈C:∣z−1∣≤2;(z+zˉ)+i(z−zˉ)≤2,Im(z)≥0} is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of elements in the set A∩B∩C is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be any point in A∩B∩C. Then, ∣z+1−i∣2+∣z−5−i∣2 lies between

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the curve z(1+i)+z(1−i)=4,z∈C, divide the region ∣z−3∣≤1 into two parts of areas α and β. Then ∣α−β∣ equals:

Area of $S =$

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Defining the Region $S_{1}$

Analyzing the Region $S_{2}$

The Boundary Line of $S_{2}$

Defining the Region $S_{3}$

Visualizing the Intersection $S$

Calculating the Area of $S$

Locating Point $P$

Let $S_{1} = {z \in C : ∣ z ∣ \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - 3 i}{1 - 3 i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0}$ . Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is :

The area (in sq. units) of the region $S = {z \in C : ∣ z - 1∣ \leq 2; (z + \overset{z}{ˉ}) + i (z - \overset{z}{ˉ}) \leq 2, Im (z) \geq 0}$ is

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

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

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

Let the curve $z (1 + i) + \overline{z} (1 - i) = 4, z \in C$ , divide the region $∣ z - 3∣ \leq 1$ into two parts of areas $α$ and $β$ . Then $∣ α - β ∣$ equals: