JEE Main 2022 (29 June Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: Let $S = {z \in C : ∣ z - 2∣ \leq 1, z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2}$ . Let $∣ z - 4 i ∣$ attains minimum and maximum values, respectively, at $z_{1} \in S$ and $z_{2} \in S$ . If $5 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}) = α + β 5$ , where $α$ and $β$ are integers, then the value of $α + β$ is equal to \_\_\_\_\_.

Q: Wait, why is the second condition $z(1 + i) + \bar{z}(1 - i) \le 2$ equivalent to $y \ge x - 1$?

Great question! Let's break it down. If $z = x + iy$, then $\bar{z} = x - iy$. Substituting these into the expression: $(x + iy)(1 + i) + (x - iy)(1 - i) = (x + ix + iy - y) + (x - ix - iy - y) = 2x - 2y$. The inequality becomes $2x - 2y \le 2$, which simplifies to $x - y \le 1$, or $y \ge x - 1$. It's just a matter of careful algebraic expansion!

Q: How do I know the minimum distance point $z_1$ lies on the line segment $PC$?

The shortest path from any point $P$ to a circle is along the radial line that passes through the center $C$. If you draw a line from $P(0, 4)$ to $C(2, 0)$, it will intersect the circle at two points. The point closer to $P$ is the minimum distance point $z_1$. Since $z_1$ must be in $S$, we verify it lies within our feasible region.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the complex plane. The first condition

∣ z - 2∣ \leq 1

defines a solid disk centered at

(2, 0)

with radius

1

. The second condition

z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2

is a linear inequality. Substitute

z = x + i y

to find the boundary line

y = x - 1

. The region

S

is the intersection of the disk and the half-plane

y \geq x - 1

Hint 2

Hint 2: To find the minimum distance from

P (0, 4)

to the region

S

, recall that the shortest distance from an external point to a circle lies along the normal line passing through the center. For the maximum distance, since the region is convex, check the boundary points (vertices) of the intersection.

Hint 3

Hint 3: The line

P C

connecting

P (0, 4)

and

C (2, 0)

has the equation

y = - 2 x + 4

. Find the intersection of this line with the circle

(x - 2)^{2} + y^{2} = 1

to locate

z_{1}

. Then, calculate

∣ z_{1} ∣^{2}

and

∣ z_{2} ∣^{2}

where

z_{2}

is the vertex

(1, 0)

that maximizes the distance.

Visualized Solution

The First Constraint: The Circular Disk

The first condition is $∣ z - 2∣ \leq 1$ .
In the complex plane, $∣ z - z_{0} ∣ \leq r$ represents a solid disk.
The center is $C (2, 0)$ and the radius is $r = 1$ .

The Second Constraint: Algebraic Setup

The second condition is $z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2$ .
To interpret this, we convert to Cartesian coordinates.
Substitute $z = x + i y$ and $\overset{z}{ˉ} = x - i y$ .

Simplifying the Linear Inequality

Expand: $(x + i y) (1 + i) + (x - i y) (1 - i) \leq 2$ .
The imaginary parts cancel out: $2 x - 2 y \leq 2$ .
Dividing by 2 gives $x - y \leq 1$ , which means $y \geq x - 1$ .

Defining the Feasible Region $S$

Region $S$ is the intersection of the disk and the half-plane.
The line $y = x - 1$ cuts the circle at $(1, 0)$ and $(2, 1)$ .
Therefore, $S$ is the upper segment of the circular disk.

The Objective: Distance from $P (0, 4)$

We need to optimize $∣ z - 4 i ∣$ .
Geometrically, this is the distance from any point $z$ in $S$ to $P (0, 4)$ .
We must find the minimum distance ( $z_{1}$ ) and maximum distance ( $z_{2}$ ).

Concept: Minimum Distance to a Circle

The shortest distance from an external point to a circle lies along the normal.
The normal must pass through the center $C (2, 0)$ .
The point $z_{1}$ will be where the line segment $P C$ intersects the region $S$ .

Calculating the Minimum Point $z_{1}$

Equation of line $P C$ : $y - 0 = \frac{4 - 0}{0 - 2} (x - 2) ⟹ y = - 2 x + 4$ .
Substitute into circle: $(x - 2)^{2} + (- 2 x + 4)^{2} = 1$ .
Solving gives $5 (x - 2)^{2} = 1 ⟹ x = 2 - \frac{1}{5}$ and $y = \frac{2}{5}$ .

Evaluating $∣ z_{1} ∣^{2}$

We need the square of the distance from the origin: $∣ z_{1} ∣^{2} = x^{2} + y^{2}$ .
$∣ z_{1} ∣^{2} = (2 - \frac{1}{5})^{2} + (\frac{2}{5})^{2}$ .
Expanding: $4 + \frac{1}{5} - \frac{4}{5} + \frac{4}{5} = 5 - \frac{4}{5}$ .

Concept: Maximum Distance to a Region

The maximum distance from a point to a convex region occurs at the boundary vertices.
For region $S$ , the extreme vertices are $(1, 0)$ and $(2, 1)$ .
We must check the distance from $P (0, 4)$ to both these points.

Calculating the Maximum Point $z_{2}$

Distance squared to $(1, 0)$ : $1^{2} + 4^{2} = 17$ .
Distance squared to $(2, 1)$ : $2^{2} + 3^{2} = 13$ .
Since $17 > 13$ , the maximum distance is at $z_{2} = (1, 0)$ .
Therefore, $∣ z_{2} ∣^{2} = 1^{2} + 0^{2} = 1$ .

Setting Up the Final Expression

We need to evaluate $5 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$ .
Substitute the values: $5 (5 - \frac{4}{5} + 1)$ .
Combine the integers inside the bracket: $5 (6 - \frac{4}{5})$ .

The Final Answer

Distribute the 5: $30 - \frac{20}{5} = 30 - 45$ .
Compare with $α + β 5$ to get $α = 30$ and $β = - 4$ .
The required value is $α + β = 30 - 4 = 26$ .

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

The region

S

is defined by two constraints in the complex plane. The first constraint,

∣ z - 2∣ \leq 1

, represents a solid disk centered at

C (2, 0)

with radius

r = 1

The second constraint is given by

z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2

. By substituting

z = x + i y

and

\overset{z}{ˉ} = x - i y

, we expand the expression:

(x + i y) (1 + i) + (x - i y) (1 - i) \leq 2

Simplifying this, the imaginary components cancel out to yield

2 x - 2 y \leq 2

, which simplifies further to the linear inequality:

y \geq x - 1

The Intersection

The region

S

is the intersection of the disk

(x - 2)^{2} + y^{2} \leq 1

and the half-plane

y \geq x - 1

By solving the system of equations, we find that the line

y = x - 1

intersects the boundary of the disk at the points

(1, 0)

and

(2, 1)

. Thus,

S

is the upper circular segment bounded by these points.

The Optimization

We aim to optimize the distance

∣ z - 4 i ∣

, which represents the distance from any point

z \in S

to the fixed point

P (0, 4)

For the minimum distance, we consider the line segment connecting

P (0, 4)

and the center

C (2, 0)

. The equation of this line is

y = - 2 x + 4

The intersection of this line with the circle

(x - 2)^{2} + y^{2} = 1

provides the point

z_{1}

closest to

P

. Substituting

y = - 2 x + 4

into the circle equation:

(x - 2)^{2} + (- 2 x + 4)^{2} = 1

(x - 2)^{2} + 4 (x - 2)^{2} = 1

5 (x - 2)^{2} = 1 \Rightarrow x = 2 - \frac{1}{5}

Calculating the squared distance

∣ z_{1} - 4 i ∣^{2}

(noting the distance from

P

C

2^{2} + 4^{2} = 20

), the minimum distance is

20 - 1

. Thus, the squared minimum distance is:

∣ z_{1} - 4 i ∣^{2} = (20 - 1)^{2} = 20 + 1 - 220 = 21 - 45

Final Calculation

For the maximum distance, we evaluate the distance from

P (0, 4)

to the boundary vertices of

S

, which are

(1, 0)

and

(2, 1)

The squared distance to

(1, 0)

1^{2} + (4 - 0)^{2} = 17

. The squared distance to

(2, 1)

2^{2} + (4 - 1)^{2} = 4 + 9 = 13

The maximum squared distance is

17

. We are asked to evaluate the sum of the minimum and maximum squared distances:

(21 - 45) + 17 = 38 - 45

Comparing this to the form

α + β 5

, we identify

α = 38

and

β = - 4

. The final result is:

α + β = 38 - 4 = 34

Common Traps & FAQs

Wait, why is the second condition

z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2

equivalent to

y \geq x - 1

Great question! Let's break it down. If

z = x + i y

, then

\overset{z}{ˉ} = x - i y

. Substituting these into the expression:

(x + i y) (1 + i) + (x - i y) (1 - i) = (x + i x + i y - y) + (x - i x - i y - y) = 2 x - 2 y

. The inequality becomes

2 x - 2 y \leq 2

, which simplifies to

x - y \leq 1

, or

y \geq x - 1

. It's just a matter of careful algebraic expansion!

Why do we check the boundary points for the maximum distance?

In optimization problems involving convex regions, the extreme values of a distance function from an external point always occur at the boundary. Since our region

S

is a circular segment, the maximum distance must occur at one of the 'corners' where the line

y = x - 1

intersects the circle. We tested both

(1, 0)

and

(2, 1)

and found the maximum.

How do I know the minimum distance point

z_{1}

lies on the line segment

P C

The shortest path from any point

P

to a circle is along the radial line that passes through the center

C

. If you draw a line from

P (0, 4)

C (2, 0)

, it will intersect the circle at two points. The point closer to

P

is the minimum distance point

z_{1}

. Since

z_{1}

must be in

S

, we verify it lies within our feasible region.

Enter Numerical Value:

Visualized Solution

The First Constraint: The Circular Disk

The Second Constraint: Algebraic Setup

Simplifying the Linear Inequality

Defining the Feasible Region $S$

The Objective: Distance from $P (0, 4)$

Concept: Minimum Distance to a Circle

Calculating the Minimum Point $z_{1}$

Evaluating $∣ z_{1} ∣^{2}$

Concept: Maximum Distance to a Region

Calculating the Maximum Point $z_{2}$

Setting Up the Final Expression

The Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Intersection

The Optimization

Final Calculation

Similar Questions

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 :

Let z be those complex number which satisfy $∣ z + 5∣ \leq 4$ and $z (1 + i) + \overset{z}{ˉ} (1 - i) \geq - 10$ , $i = - 1$ If the maximum value of $∣ z + 1 ∣^{2}$ is $α + β 2$ , then the value of $(α + β)$ is

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 $S = {z \in C : ∣ z - 3∣ \leq 1 and z (4 + 3 i) + \overset{z}{ˉ} (4 - 3 i) \leq 24}$ . If $α + i β$ is the point in $S$ which is closest to $4 i$ , then $25 (α + β)$ is equal to \_\_\_\_\_.

Let complex numbers $α$ and $1/ \overset{α}{ˉ}$ lie on circles $(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$ and $(x - x_{0})^{2} + (y - y_{0})^{2} = 4 r^{2}$ respectively. If $z_{0} = x_{0} + i y_{0}$ satisfies the equation $2∣ z_{0} ∣^{2} = r^{2} + 2$ , then $∣ α ∣ =$

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:

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

Let $A = {z \in C : ∣ z - 2∣ \leq 4}$ and $B = {z \in C : ∣ z - 2∣ + ∣ z + 2∣ = 5}$ . Then the $max {∣ z_{1} - z_{2} ∣ : z_{1} \in A and z_{2} \in B}$ is :

Let $∣ z_{1} - 8 - 2 i ∣ \leq 1$ and $∣ z_{2} - 2 + 6 i ∣ \leq 2, z_{1}, z_{2} \in C$ . Then the minimum value of $∣ z_{1} - z_{2} ∣$ is :

Enter Numerical Value:

Visualized Solution

The First Constraint: The Circular Disk

The Second Constraint: Algebraic Setup

Simplifying the Linear Inequality

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

The Objective: Distance from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(0,4)

Concept: Minimum Distance to a Circle

Calculating the Minimum Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z1​∣2

Concept: Maximum Distance to a Region

Calculating the Maximum Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​

Setting Up the Final Expression

The Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Intersection

The Optimization

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z∈C:∣z−3∣≤1 and z(4+3i)+zˉ(4−3i)≤24}. If α+iβ is the point in S which is closest to 4i, then 25(α+β) is equal to \_\_\_\_\_.

.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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for the complex numbers z satisfying ∣z−2−2i∣≤1, the maximum value of ∣3iz+6∣ is attained at a+ib, then a+b is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={z∈C:∣z−2∣≤4} and B={z∈C:∣z−2∣+∣z+2∣=5}. Then the max{∣z1​−z2​∣:z1​∈A and z2​∈B} is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∣z1​−8−2i∣≤1 and ∣z2​−2+6i∣≤2,z1​,z2​∈C. Then the minimum value of ∣z1​−z2​∣ is :

Defining the Feasible Region $S$

The Objective: Distance from $P (0, 4)$

Calculating the Minimum Point $z_{1}$

Evaluating $∣ z_{1} ∣^{2}$

Calculating the Maximum Point $z_{2}$

Let $S = {z \in C : ∣ z - 3∣ \leq 1 and z (4 + 3 i) + \overset{z}{ˉ} (4 - 3 i) \leq 24}$ . If $α + i β$ is the point in $S$ which is closest to $4 i$ , then $25 (α + β)$ is equal to \_\_\_\_\_.

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:

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

Let $A = {z \in C : ∣ z - 2∣ \leq 4}$ and $B = {z \in C : ∣ z - 2∣ + ∣ z + 2∣ = 5}$ . Then the $max {∣ z_{1} - z_{2} ∣ : z_{1} \in A and z_{2} \in B}$ is :

Let $∣ z_{1} - 8 - 2 i ∣ \leq 1$ and $∣ z_{2} - 2 + 6 i ∣ \leq 2, z_{1}, z_{2} \in C$ . Then the minimum value of $∣ z_{1} - z_{2} ∣$ is :