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 given problem, we need to analyze the inequalities involving the complex number \( Z \). We will interpret these inequalities geometrically in the complex plane.

### Step 1: Interpret the inequalities

The inequalities given are:
1. \( |Z - 1| \leq |Z - 3| \)
2. \( |Z - 3| \leq |Z - 5| \)
3. \( |Z + i| \leq |Z - i| \)
4. \( |Z - i| \leq |Z - 5i| \)

### Step 2: Analyze the first inequality \( |Z - 1| \leq |Z - 3| \)

This inequality means that the point \( Z \) is closer to \( 1 \) than to \( 3 \). The line that represents this inequality is the perpendicular bisector of the segment joining \( 1 \) and \( 3 \).

- The midpoint of \( 1 \) and \( 3 \) is \( 2 \).
- The perpendicular bisector is the vertical line \( x = 2 \).

Thus, the region satisfying this inequality is to the left of the line \( x = 2 \).

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

Similarly, this inequality indicates that \( Z \) is closer to \( 3 \) than to \( 5 \).

- The midpoint of \( 3 \) and \( 5 \) is \( 4 \).
- The perpendicular bisector is the vertical line \( x = 4 \).

Thus, the region satisfying this inequality is to the left of the line \( x = 4 \).

### Step 4: Combine the first two inequalities

From the first two inequalities, we have:
- \( Z \) is to the left of \( x = 2 \) and \( x = 4 \).

This means that \( 2 \leq x \leq 4 \).

### Step 5: Analyze the third inequality \( |Z + i| \leq |Z - i| \)

This inequality means that \( Z \) is closer to \( -i \) than to \( i \).

- The midpoint of \( -i \) and \( i \) is \( 0 \).
- The perpendicular bisector is the horizontal line \( y = 0 \).

Thus, the region satisfying this inequality is above the line \( y = 0 \).

### Step 6: Analyze the fourth inequality \( |Z - i| \leq |Z - 5i| \)

This inequality indicates that \( Z \) is closer to \( i \) than to \( 5i \).

- The midpoint of \( i \) and \( 5i \) is \( 3i \).
- The perpendicular bisector is the horizontal line \( y = 3 \).

Thus, the region satisfying this inequality is below the line \( y = 3 \).

### Step 7: Combine the last two inequalities

From the last two inequalities, we have:
- \( Z \) is above the line \( y = 0 \) and below the line \( y = 3 \).

This means that \( 0 \leq y \leq 3 \).

### Step 8: Determine the area of the region

Now we have the region defined by:
- \( 2 \leq x \leq 4 \)
- \( 0 \leq y \leq 3 \)

This region is a rectangle with:
- Width = \( 4 - 2 = 2 \)
- Height = \( 3 - 0 = 3 \)

The area \( A \) of this rectangle is given by:
\[ A = \text{Width} \times \text{Height} = 2 \times 3 = 6 \text{ square units} \]

### Final Answer

Thus, the area \( A \) is equal to \( 6 \) square units.

---