Let Papoint denoting a comples number z on the complex plane.
`i.e." "z=Re(z)+i Im(z)," where "i=sqrt(-1)`
`if" "Re(z)=xand Im (z)=y,then z=x+iy`
If Pmovew such that
`|Re(z)|+|Im(z)=a(ainR^(+))`
The locus of P is

Options

A. a parallelogram which is not arhombus
B. a rhombus which is not a square
C. a rectangle which is not a square
D. a square

Answer

To find the locus of the point \( P \) representing the complex number \( z \) such that \( |Re(z)| + |Im(z)| = a \) (where \( a \) is a positive real number), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Complex Number**:
Let \( z = Re(z) + i Im(z) \). We denote \( Re(z) = x \) and \( Im(z) = y \). Thus, we can express \( z \) as:
\[ z = x + iy \]

2. **Set Up the Equation**:
According to the problem, we have:
\[ |Re(z)| + |Im(z)| = a \]
This translates to:
\[ |x| + |y| = a \]

3. **Analyze the Absolute Value Equation**:
The equation \( |x| + |y| = a \) describes a geometric figure in the Cartesian plane. To understand this better, we can consider the different cases based on the signs of \( x \) and \( y \).

4. **Consider Different Cases**:
- **Case 1**: \( x \geq 0 \) and \( y \geq 0 \):
\[ x + y = a \quad \text{(First Quadrant)} \]
- **Case 2**: \( x \geq 0 \) and \( y < 0 \):
\[ x - y = a \quad \text{(Fourth Quadrant)} \]
- **Case 3**: \( x < 0 \) and \( y \geq 0 \):
\[ -x + y = a \quad \text{(Second Quadrant)} \]
- **Case 4**: \( x < 0 \) and \( y < 0 \):
\[ -x - y = a \quad \text{(Third Quadrant)} \]

5. **Identify the Lines**:
From the above cases, we can derive the equations of the lines:
- \( x + y = a \)
- \( x - y = a \)
- \( -x + y = a \)
- \( -x - y = a \)

6. **Graph the Lines**:
Each of these equations represents a line in the Cartesian plane. When graphed, they will form a shape that can be identified as a rhombus.

7. **Find the Vertices**:
The vertices of the rhombus can be found by solving the equations pairwise:
- Intersection of \( x + y = a \) and \( -x + y = a \) gives \( (0, a) \).
- Intersection of \( x + y = a \) and \( x - y = a \) gives \( (a, 0) \).
- Intersection of \( -x - y = a \) and \( x - y = a \) gives \( (0, -a) \).
- Intersection of \( -x - y = a \) and \( -x + y = a \) gives \( (-a, 0) \).

8. **Conclusion**:
The locus of the point \( P \) is a rhombus with vertices at \( (a, 0) \), \( (0, a) \), \( (-a, 0) \), and \( (0, -a) \).

### Final Answer:
The locus of the point \( P \) is a rhombus.

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