If at least one value of the complex number `z=x+iy` satisfies the condition `|z+sqrt(2)|=sqrt(a^(2)-3a+2)` and the inequality `|z+isqrt(2)| lt a`, then

Options

A. `a gt 2`
B. `a=2`
C. `a lt 2`
D. `a gt 1`

Answer

To solve the problem, we need to analyze the given conditions involving the complex number \( z = x + iy \).

### Step 1: Understand the conditions
We have two conditions:
1. \( |z + \sqrt{2}| = \sqrt{a^2 - 3a + 2} \)
2. \( |z + i\sqrt{2}| < a \)

### Step 2: Rewrite the first condition
The first condition can be interpreted geometrically. The expression \( |z + \sqrt{2}| \) represents the distance from the point \( z \) to the point \( -\sqrt{2} \) in the complex plane. Thus, we can write:
\[ |z + \sqrt{2}| = |(x + iy) + \sqrt{2}| = |(x + \sqrt{2}) + iy| \]
This means the center of the circle is at \( (-\sqrt{2}, 0) \) and the radius is \( \sqrt{a^2 - 3a + 2} \).

### Step 3: Ensure the radius is positive
For the radius to be valid, we need:
\[ a^2 - 3a + 2 > 0 \]
Factoring this quadratic inequality gives:
\[ (a - 1)(a - 2) > 0 \]

### Step 4: Solve the inequality
To find the intervals where this inequality holds, we find the roots:
- \( a = 1 \)
- \( a = 2 \)

We can test intervals around these points:
- For \( a < 1 \): Choose \( a = 0 \) → \( (0 - 1)(0 - 2) = 2 > 0 \) (True)
- For \( 1 < a < 2 \): Choose \( a = 1.5 \) → \( (1.5 - 1)(1.5 - 2) = -0.25 < 0 \) (False)
- For \( a > 2 \): Choose \( a = 3 \) → \( (3 - 1)(3 - 2) = 2 > 0 \) (True)

Thus, the solution for this inequality is:
\[ a < 1 \quad \text{or} \quad a > 2 \]

### Step 5: Analyze the second condition
The second condition is:
\[ |z + i\sqrt{2}| < a \]
This represents the distance from \( z \) to the point \( -i\sqrt{2} \). For this distance to be less than \( a \), we need \( a > 0 \).

### Step 6: Combine the conditions
From the first condition, we have:
1. \( a < 1 \) or \( a > 2 \)
From the second condition, we have:
2. \( a > 0 \)

### Step 7: Determine the valid intervals
Combining these conditions:
- From \( a < 1 \) and \( a > 0 \), we get \( 0 < a < 1 \).
- From \( a > 2 \), we have \( a > 2 \).

Thus, the final solution is:
\[ a \in (0, 1) \cup (2, \infty) \]

### Step 8: Conclusion
The values of \( a \) that satisfy both conditions are:
\[ \text{At least one value of } a \text{ must be in } (0, 1) \cup (2, \infty) \]