JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: 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 :

Select Answer:

Visualized Solution

The Argand Plane Setup

We are given two regions in the complex plane.
Region 1: $∣ z_{1} - (8 + 2 i) ∣ \leq 1$
Region 2: $∣ z_{2} - (2 - 6 i) ∣ \leq 2$
Goal: Find the minimum value of $∣ z_{1} - z_{2} ∣$ .

Decoding Region 1

The standard equation of a disk in the complex plane is $∣ z - z_{0} ∣ \leq r$ .
Here, $z_{0}$ is the center and $r$ is the radius.
Let's rewrite Region 1: $∣ z_{1} - (8 + 2 i) ∣ \leq 1$ .

Center and Radius of Region 1

Comparing $∣ z_{1} - (8 + 2 i) ∣ \leq 1$ with $∣ z - z_{0} ∣ \leq r$ :
Center $A = 8 + 2 i \equiv (8, 2)$
Radius $r_{1} = 1$

Decoding Region 2

Now consider Region 2: $∣ z_{2} - 2 + 6 i ∣ \leq 2$ .
We must rewrite this in the standard form $∣ z - z_{0} ∣ \leq r$ .
Factor out the negative sign: $∣ z_{2} - (2 - 6 i) ∣ \leq 2$ .

Center and Radius of Region 2

Comparing $∣ z_{2} - (2 - 6 i) ∣ \leq 2$ with $∣ z - z_{0} ∣ \leq r$ :
Center $B = 2 - 6 i \equiv (2, - 6)$
Radius $r_{2} = 2$

The Shortest Path

We need the minimum value of $∣ z_{1} - z_{2} ∣$ .
Geometrically, this is the shortest distance between the two disks.
The shortest distance always lies along the line joining their centers.

Distance Between Centers

First, we must find the total distance between center $A (8, 2)$ and center $B (2, - 6)$ .
We use the standard distance formula:
$A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

Substituting Coordinates

Substitute $A (8, 2)$ and $B (2, - 6)$ into the formula:
$A B = (8 - 2)^{2} + (2 - (- 6))^{2}$

Calculating the Differences

Simplify the terms inside the brackets:
$8 - 2 = 6$
$2 - (- 6) = 2 + 6 = 8$
$A B = 6^{2} + 8^{2}$

Final Distance $A B$

Square the numbers: $6^{2} = 36$ and $8^{2} = 64$ .
Add them up: $36 + 64 = 100$ .
$A B = 100 = 10$

Formula for Minimum Distance

The minimum distance $d_{min}$ between two disjoint circles is:
$d_{min} = A B - r_{1} - r_{2}$
This removes the radii portions from the total center-to-center distance.

Substituting Values for $d_{min}$

We know $A B = 10$ .
Radius $r_{1} = 1$ .
Radius $r_{2} = 2$ .
$d_{min} = 10 - 1 - 2$

The Final Answer

$d_{min} = 10 - 3 = 7$
The minimum value of $∣ z_{1} - z_{2} ∣$ is $7$ .

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

The expression

∣ z - z_{0} ∣ \leq r

represents a closed disk in the complex plane, where

z_{0}

is the center and

r

is the radius.

For the first region,

∣ z_{1} - (8 + 2 i) ∣ \leq 1

, the center is

A = (8, 2)

and the radius is

r_{1} = 1

For the second region,

∣ z_{2} - (2 - 6 i) ∣ \leq 2

, the center is

B = (2, - 6)

and the radius is

r_{2} = 2

The Intuition of the Shortest Path

Geometrically, the shortest distance between two non-overlapping disks is found along the line segment connecting their centers.

d

is the distance between the centers

A

and

B

, the minimum distance between the disks is given by:

d_{min} = d - r_{1} - r_{2}

The Pythagorean Calculation

We calculate the distance

d

between

A (8, 2)

and

B (2, - 6)

using the distance formula:

d = (8 - 2)^{2} + (2 - (- 6))^{2}

Simplifying the terms inside the square root:

d = 6^{2} + 8^{2}

d = 36 + 64 = 100 = 10

Final Calculation

With the distance between centers

d = 10

and the radii

r_{1} = 1

and

r_{2} = 2

, we subtract the radii from the total distance to find the gap:

d_{min} = 10 - 1 - 2

d_{min} = 7

The minimum distance between any point

z_{1}

in the first region and any point

z_{2}

in the second region is 7.

Common Traps & FAQs

Wait, why do we subtract the radii? Shouldn't we add them to get the distance?

That is a fantastic question! Think of it this way: the distance between the centers

A

and

B

is the total gap between the two points. Since the disks have radii

r_{1}

and

r_{2}

, the 'inner' edges of the disks are closer to each other than the centers are. To find the minimum distance between the boundaries of the disks, we must remove the 'buffer' provided by the radii from the total distance. Hence,

d_{min} = A B - r_{1} - r_{2}

Can I just pick any point in the disks to find the distance?

No, that would only give you the distance between two specific points. The problem asks for the minimum value of

∣ z_{1} - z_{2} ∣

, which implies we are looking for the absolute minimum distance possible between *any* point in the first disk and *any* point in the second. Geometry tells us this minimum occurs along the line segment connecting the centers.

What if the disks overlap? Does the formula still work?

If the disks overlap, the distance between them is zero. Our formula

A B - r_{1} - r_{2}

would yield a negative number in that case. Always check if

A B < r_{1} + r_{2}

first. If it is, the minimum distance is zero because the regions intersect!

Animated Solution for Mathematics - Complex Numbers: 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 :

Select Answer:

Visualized Solution

The Argand Plane Setup

Decoding Region 1

Center and Radius of Region 1

Decoding Region 2

Center and Radius of Region 2

The Shortest Path

Distance Between Centers

Substituting Coordinates

Calculating the Differences

Final Distance $A B$

Formula for Minimum Distance

Substituting Values for $d_{min}$

The Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Intuition of the Shortest Path

The Pythagorean Calculation

Final Calculation

Similar Questions

For all complex numbers $z_{1}, z_{2}$ satisfying $∣ z_{1} ∣ = 12$ and $∣ z_{2} - 3 - 4 i ∣ = 5$ , the minimum value of $∣ z_{1} - z_{2} ∣$ is

Let $S_{1} = {z_{1} \in C : ∣ z_{1} - 3∣ = \frac{1}{2}}$ and $S_{2} = {z_{2} \in C : ∣ z_{2} - ∣ z_{2} + 1∣∣ = ∣ z_{2} + ∣ z_{2} - 1∣∣}$ . Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$ , the least value of $∣ z_{2} - z_{1} ∣$ is:

Let $Z_{1}$ and $Z_{2}$ be two complex numbers satisfying $∣ Z_{1} ∣ = 9$ and $∣ Z_{2} - 3 - 4 i ∣ = 4$ . Then the minimum value of $∣ Z_{1} - Z_{2} ∣$ is :

If $z$ is any complex number satisfying $∣ z - 3 - 2 i ∣ \leq 2$ , then the minimum value of $∣2 z - 6 + 5 i ∣$ is .........

If $∣ z - 3 + 2 i ∣ \leq 4$ then the difference between the greatest value and the least value of $∣ z ∣$ is :-

For $z \in C$ if the minimum value of $(∣ z - 32 ∣ + ∣ z - p 2 i ∣)$ is $52$ , then a value of $p$ is

If $z$ is a complex number such that $∣ z ∣ \geq 1$ , then the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ is:

Let $S = {z : 3 \leq ∣2 z - 3 (1 + i) ∣ \leq 7}$ be a set of complex numbers. Then $min_{z \in S} z + \frac{1}{2} (5 + 3 i)$ is equal to :

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 :

Select Answer:

Visualized Solution

The Argand Plane Setup

Decoding Region 1

Center and Radius of Region 1

Decoding Region 2

Center and Radius of Region 2

The Shortest Path

Distance Between Centers

Substituting Coordinates

Calculating the Differences

Final Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Formula for Minimum Distance

Substituting Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dmin​

The Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Intuition of the Shortest Path

The Pythagorean Calculation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For all complex numbers z1​,z2​ satisfying ∣z1​∣=12 and ∣z2​−3−4i∣=5, the minimum value of ∣z1​−z2​∣ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Z1​ and Z2​ be two complex numbers satisfying ∣Z1​∣=9 and ∣Z2​−3−4i∣=4. Then the minimum value of ∣Z1​−Z2​∣ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z is any complex number satisfying ∣z−3−2i∣≤2, then the minimum value of ∣2z−6+5i∣ is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∣z−3+2i∣≤4 then the difference between the greatest value and the least value of ∣z∣ is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For z∈C if the minimum value of (∣z−32​∣+∣z−p2​i∣) is 52​, then a value of p is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z is a complex number such that ∣z∣≥1, then the minimum value of ∣z+21​(3+4i)∣ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z:3≤∣2z−3(1+i)∣≤7} be a set of complex numbers. Then minz∈S​​z+21​(5+3i)​ is equal to :

Final Distance $A B$

Substituting Values for $d_{min}$

For all complex numbers $z_{1}, z_{2}$ satisfying $∣ z_{1} ∣ = 12$ and $∣ z_{2} - 3 - 4 i ∣ = 5$ , the minimum value of $∣ z_{1} - z_{2} ∣$ is

Let $Z_{1}$ and $Z_{2}$ be two complex numbers satisfying $∣ Z_{1} ∣ = 9$ and $∣ Z_{2} - 3 - 4 i ∣ = 4$ . Then the minimum value of $∣ Z_{1} - Z_{2} ∣$ is :

If $z$ is any complex number satisfying $∣ z - 3 - 2 i ∣ \leq 2$ , then the minimum value of $∣2 z - 6 + 5 i ∣$ is .........

If $∣ z - 3 + 2 i ∣ \leq 4$ then the difference between the greatest value and the least value of $∣ z ∣$ is :-

For $z \in C$ if the minimum value of $(∣ z - 32 ∣ + ∣ z - p 2 i ∣)$ is $52$ , then a value of $p$ is

If $z$ is a complex number such that $∣ z ∣ \geq 1$ , then the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ is:

Let $S = {z : 3 \leq ∣2 z - 3 (1 + i) ∣ \leq 7}$ be a set of complex numbers. Then $min_{z \in S} z + \frac{1}{2} (5 + 3 i)$ is equal to :