Let Z be a complex number satisfying `|Z-1| <= |Z-3|, |Z-3| <= |Z-5|, |Z+ i| <= |Z- i|, |Z- i| <= |Z- 5i|`. Then area of region in which Z lies is A square units, Where A is equal to :

Answer

To solve the problem, we need to analyze the given inequalities involving the complex number \( Z \). Let's denote \( Z = x + iy \), where \( x \) and \( y \) are real numbers. We will break down the inequalities step by step.

### Step 1: Analyze the first inequality \( |Z - 1| \leq |Z - 3| \)

This can be expressed as:
\[ |Z - 1|^2 \leq |Z - 3|^2 \]
Substituting \( Z = x + iy \):
\[ |(x + iy) - 1|^2 \leq |(x + iy) - 3|^2 \]
This simplifies to:
\[ (x - 1)^2 + y^2 \leq (x - 3)^2 + y^2 \]
Cancelling \( y^2 \) from both sides:
\[ (x - 1)^2 \leq (x - 3)^2 \]
Expanding both sides:
\[ x^2 - 2x + 1 \leq x^2 - 6x + 9 \]
Cancelling \( x^2 \):
\[ -2x + 1 \leq -6x + 9 \]
Rearranging gives:
\[ 4x \leq 8 \quad \Rightarrow \quad x \leq 2 \]

### Step 2: Analyze the second inequality \( |Z - 3| \leq |Z - 5| \)

Similarly, we have:
\[ |Z - 3|^2 \leq |Z - 5|^2 \]
Substituting \( Z = x + iy \):
\[ |(x + iy) - 3|^2 \leq |(x + iy) - 5|^2 \]
This simplifies to:
\[ (x - 3)^2 + y^2 \leq (x - 5)^2 + y^2 \]
Cancelling \( y^2 \):
\[ (x - 3)^2 \leq (x - 5)^2 \]
Expanding both sides:
\[ x^2 - 6x + 9 \leq x^2 - 10x + 25 \]
Cancelling \( x^2 \):
\[ -6x + 9 \leq -10x + 25 \]
Rearranging gives:
\[ 4x \leq 16 \quad \Rightarrow \quad x \leq 4 \]

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

This can be expressed as:
\[ |Z + i|^2 \leq |Z - i|^2 \]
Substituting \( Z = x + iy \):
\[ |(x + (y + 1)i)|^2 \leq |(x + (y - 1)i)|^2 \]
This simplifies to:
\[ x^2 + (y + 1)^2 \leq x^2 + (y - 1)^2 \]
Cancelling \( x^2 \):
\[ (y + 1)^2 \leq (y - 1)^2 \]
Expanding both sides:
\[ y^2 + 2y + 1 \leq y^2 - 2y + 1 \]
Cancelling \( y^2 + 1 \):
\[ 2y \leq -2y \]
Rearranging gives:
\[ 4y \leq 0 \quad \Rightarrow \quad y \leq 0 \]

### Step 4: Analyze the fourth inequality \( |Z - i| \leq |Z - 5i| \)

This can be expressed as:
\[ |Z - i|^2 \leq |Z - 5i|^2 \]
Substituting \( Z = x + iy \):
\[ |(x + (y - 1)i)|^2 \leq |(x + (y - 5)i)|^2 \]
This simplifies to:
\[ x^2 + (y - 1)^2 \leq x^2 + (y - 5)^2 \]
Cancelling \( x^2 \):
\[ (y - 1)^2 \leq (y - 5)^2 \]
Expanding both sides:
\[ y^2 - 2y + 1 \leq y^2 - 10y + 25 \]
Cancelling \( y^2 \):
\[ -2y + 1 \leq -10y + 25 \]
Rearranging gives:
\[ 8y \leq 24 \quad \Rightarrow \quad y \leq 3 \]

### Step 5: Combine the results

From the inequalities, we have:
1. \( x \leq 2 \)
2. \( x \leq 4 \) (but the more restrictive condition is \( x \leq 2 \))
3. \( y \leq 0 \)
4. \( y \leq 3 \)

Thus, the region where \( Z \) lies is bounded by:
- \( x \) can range from \( -\infty \) to \( 2 \)
- \( y \) can range from \( -\infty \) to \( 0 \) (but also must be less than or equal to \( 3 \))

### Step 6: Determine the area of the region

The area of the rectangle formed by these boundaries is:
- Width (in the x-direction) = \( 2 - (-\infty) \) (but we will consider the relevant area)
- Height (in the y-direction) = \( 0 - (-3) = 3 \)

The area \( A \) of the region is:
\[ A = (2 - 0)(3 - 0) = 2 \times 3 = 6 \, \text{square units} \]

### Final Answer
The area \( A \) is equal to \( 6 \) square units.