If `a gt 0` and the equation `|z-a^(2)|+|z-2a|=3`, represents an ellipse, then 'a' belongs to the interval

Options

A. (1,3)
B. `(sqrt(2),sqrt(3))`
C. (0,3)
D. `(1,sqrt(3))`

Answer

To solve the problem, we need to determine the interval for \( a \) such that the equation \( |z - a^2| + |z - 2a| = 3 \) represents an ellipse.

### Step-by-Step Solution:

1. **Understanding the Equation**:
The given equation \( |z - a^2| + |z - 2a| = 3 \) can be interpreted geometrically. It represents the sum of distances from a point \( z \) in the complex plane to two fixed points \( a^2 \) and \( 2a \).

2. **Identifying Fixed Points**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The fixed points in the complex plane are \( a^2 \) (which is \( a^2 + 0i \)) and \( 2a \) (which is \( 2a + 0i \)).

3. **Condition for an Ellipse**:
The equation \( |z - z_1| + |z - z_2| = d \) represents an ellipse if \( d \) is greater than the distance between the two points \( z_1 \) and \( z_2 \). Here, the distance between the points \( a^2 \) and \( 2a \) is given by:
\[ |a^2 - 2a| = |a(a - 2)| \]

4. **Setting Up the Inequality**:
For the equation to represent an ellipse, we need:
\[ 3 > |a(a - 2)| \]
This leads to two cases based on the absolute value.

5. **Case Analysis**:
- **Case 1**: \( a(a - 2) \geq 0 \)
\[ a(a - 2) < 3 \]
- **Case 2**: \( a(a - 2) < 0 \)
\[ -a(a - 2) < 3 \implies a(a - 2) > -3 \]

6. **Solving Case 1**:
For \( a(a - 2) \geq 0 \):
- This occurs when \( a \leq 0 \) or \( a \geq 2 \). Since \( a > 0 \), we only consider \( a \geq 2 \).
- The inequality becomes:
\[ a^2 - 2a < 3 \implies a^2 - 2a - 3 < 0 \]
Factoring gives:
\[ (a - 3)(a + 1) < 0 \]
This implies \( -1 < a < 3 \). Since \( a > 0 \), we have \( 0 < a < 3 \).

7. **Solving Case 2**:
For \( a(a - 2) < 0 \):
- This occurs when \( 0 < a < 2 \).
- The inequality becomes:
\[ -a(a - 2) < 3 \implies a^2 - 2a + 3 > 0 \]
The quadratic \( a^2 - 2a + 3 \) has no real roots (discriminant \( < 0 \)), hence it is always positive.

8. **Combining Results**:
From both cases, we find that the valid interval for \( a \) is:
\[ 0 < a < 3 \]

### Final Answer:
Thus, \( a \) belongs to the interval \( (0, 3) \).