Consider the region `R` in the Argand plane described by the complex number. `Z` satisfying the inequalities `|Z-2| le |Z-4|`, `|Z-3| le |Z+3|`, `|Z-i| le |Z-3i|`, `|Z+i| le |Z+3i|`
Answer the followin questions :
The maximum value of `|Z|` for any `Z` in `R` is

Options

A. `5`
B. `3`
C. `1`
D. `sqrt(13)`

Explanation

To solve the problem, we need to analyze the inequalities given for the complex number \( Z \) in the Argand plane. Let's denote \( Z = x + iy \), where \( x \) and \( y \) are the real and imaginary parts of \( Z \), respectively.

### Step 1: Analyze the first inequality \( |Z - 2| \leq |Z - 4| \)

This inequality can be rewritten as:
\[ |Z - 2|^2 \leq |Z - 4|^2 \]
Substituting \( Z = x + iy \):
\[ |(x - 2) + iy|^2 \leq |(x - 4) + iy|^2 \]
This expands to:
\[ (x - 2)^2 + y^2 \leq (x - 4)^2 + y^2 \]
Cancelling \( y^2 \) from both sides:
\[ (x - 2)^2 \leq (x - 4)^2 \]
Expanding both sides:
\[ x^2 - 4x + 4 \leq x^2 - 8x + 16 \]
Simplifying gives:
\[ 4x \leq 12 \implies x \leq 3 \]

### Step 2: Analyze the second inequality \( |Z - 3| \leq |Z + 3| \)

Rewriting this:
\[ |Z - 3|^2 \leq |Z + 3|^2 \]
Substituting \( Z = x + iy \):
\[ |(x - 3) + iy|^2 \leq |(x + 3) + iy|^2 \]
This expands to:
\[ (x - 3)^2 + y^2 \leq (x + 3)^2 + y^2 \]
Cancelling \( y^2 \) gives:
\[ (x - 3)^2 \leq (x + 3)^2 \]
Expanding both sides:
\[ x^2 - 6x + 9 \leq x^2 + 6x + 9 \]
Simplifying gives:
\[ -12x \leq 0 \implies x \geq 0 \]

### Step 3: Analyze the third inequality \( |Z - i| \leq |Z - 3i| \)

Rewriting this:
\[ |Z - i|^2 \leq |Z - 3i|^2 \]
Substituting \( Z = x + iy \):
\[ |(x) + (y - 1)i|^2 \leq |(x) + (y - 3)i|^2 \]
This expands to:
\[ x^2 + (y - 1)^2 \leq x^2 + (y - 3)^2 \]
Cancelling \( x^2 \) gives:
\[ (y - 1)^2 \leq (y - 3)^2 \]
Expanding both sides:
\[ y^2 - 2y + 1 \leq y^2 - 6y + 9 \]
Simplifying gives:
\[ 4y \leq 8 \implies y \leq 2 \]

### Step 4: Analyze the fourth inequality \( |Z + i| \leq |Z + 3i| \)

Rewriting this:
\[ |Z + i|^2 \leq |Z + 3i|^2 \]
Substituting \( Z = x + iy \):
\[ |(x) + (y + 1)i|^2 \leq |(x) + (y + 3)i|^2 \]
This expands to:
\[ x^2 + (y + 1)^2 \leq x^2 + (y + 3)^2 \]
Cancelling \( x^2 \) gives:
\[ (y + 1)^2 \leq (y + 3)^2 \]
Expanding both sides:
\[ y^2 + 2y + 1 \leq y^2 + 6y + 9 \]
Simplifying gives:
\[ -4y \leq 8 \implies y \geq -2 \]

### Step 5: Combine the inequalities

From the inequalities derived:
1. \( 0 \leq x \leq 3 \)
2. \( -2 \leq y \leq 2 \)

This describes a rectangle in the Argand plane with vertices at \( (0, -2) \), \( (0, 2) \), \( (3, -2) \), and \( (3, 2) \).

### Step 6: Find the maximum value of \( |Z| \)

The maximum modulus \( |Z| \) occurs at the vertex farthest from the origin, which is \( (3, 2) \):
\[ |Z| = \sqrt{x^2 + y^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \]

### Final Answer:
The maximum value of \( |Z| \) for any \( Z \) in \( R \) is \( \sqrt{13} \).

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