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

Options

A. `a gt 6`
B. a= 6
C. ` a lt 6 `
D. none of these

Answer

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

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

### Step 2: Interpret the First Condition
The first condition \( |z + \sqrt{2}| = a^2 - 3a - 2 \) describes a circle in the complex plane centered at \( -\sqrt{2} \) with a radius of \( a^2 - 3a - 2 \). For this radius to be valid, it must be non-negative:
\[ a^2 - 3a - 2 \geq 0 \]

### Step 3: Solve the Quadratic Inequality
To solve the inequality \( a^2 - 3a - 2 \geq 0 \), we first find the roots of the equation \( a^2 - 3a - 2 = 0 \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
\[ = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \]
Thus, the roots are:
\[ a_1 = \frac{3 - \sqrt{17}}{2}, \quad a_2 = \frac{3 + \sqrt{17}}{2} \]

### Step 4: Analyze the Quadratic Inequality
The quadratic \( a^2 - 3a - 2 \) opens upwards (since the coefficient of \( a^2 \) is positive). Therefore, the expression is non-negative outside the interval defined by its roots:
\[ a \leq \frac{3 - \sqrt{17}}{2} \quad \text{or} \quad a \geq \frac{3 + \sqrt{17}}{2} \]

### Step 5: Interpret the Second Condition
The second condition \( |z + i\sqrt{2}| < a \) indicates that the distance from \( z \) to \( -i\sqrt{2} \) must be less than \( a \). This describes a circle centered at \( -i\sqrt{2} \) with radius \( a \).

### Step 6: Determine Valid Values for \( a \)
Since \( a \) must be positive (as it represents a radius), we have:
1. From the first condition, we need \( a \geq \frac{3 + \sqrt{17}}{2} \).
2. From the second condition, \( a > 0 \).

### Step 7: Find the Intersection of Conditions
We need to find the intersection of the intervals:
- \( a \geq \frac{3 + \sqrt{17}}{2} \)
- \( a > 0 \)

Since \( \frac{3 + \sqrt{17}}{2} \) is positive, the intersection is simply:
\[ a \geq \frac{3 + \sqrt{17}}{2} \]

### Conclusion
Thus, the final result for the values of \( a \) that satisfy both conditions is:
\[ a \geq \frac{3 + \sqrt{17}}{2} \]