If `a gt 0 and z=(1+i)^2/(a-i)` has magnitude `sqrt(2/5) "then " bar z is ` equal to

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the expression for \( z \)
Given:
\[ z = \frac{(1 + i)^2}{a - i} \]
We will first simplify the numerator.

### Step 2: Simplify the numerator
Using the formula for the square of a binomial:
\[ (1 + i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i \]
Thus, we can rewrite \( z \):
\[ z = \frac{2i}{a - i} \]

### Step 3: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{2i(a + i)}{(a - i)(a + i)} = \frac{2i(a + i)}{a^2 + 1} \]
Now, simplify the numerator:
\[ z = \frac{2ai + 2i^2}{a^2 + 1} = \frac{2ai - 2}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1} \]

### Step 4: Separate real and imaginary parts
From the expression:
\[ z = \frac{-2}{a^2 + 1} + \frac{2a}{a^2 + 1}i \]
Let \( x = \frac{-2}{a^2 + 1} \) and \( y = \frac{2a}{a^2 + 1} \).

### Step 5: Find the magnitude of \( z \)
The magnitude of \( z \) is given by:
\[ |z| = \sqrt{x^2 + y^2} \]
Substituting \( x \) and \( y \):
\[ |z| = \sqrt{\left(\frac{-2}{a^2 + 1}\right)^2 + \left(\frac{2a}{a^2 + 1}\right)^2} \]
This simplifies to:
\[ |z| = \sqrt{\frac{4 + 4a^2}{(a^2 + 1)^2}} = \frac{2\sqrt{1 + a^2}}{a^2 + 1} \]

### Step 6: Set the magnitude equal to \( \sqrt{\frac{2}{5}} \)
We know:
\[ \frac{2\sqrt{1 + a^2}}{a^2 + 1} = \sqrt{\frac{2}{5}} \]
Squaring both sides:
\[ \frac{4(1 + a^2)}{(a^2 + 1)^2} = \frac{2}{5} \]
Cross-multiplying gives:
\[ 20(1 + a^2) = 2(a^2 + 1)^2 \]

### Step 7: Expand and simplify
Expanding the right side:
\[ 20 + 20a^2 = 2(a^4 + 2a^2 + 1) \]
This simplifies to:
\[ 20 + 20a^2 = 2a^4 + 4a^2 + 2 \]
Rearranging gives:
\[ 2a^4 - 16a^2 - 18 = 0 \]
Dividing by 2:
\[ a^4 - 8a^2 - 9 = 0 \]

### Step 8: Let \( u = a^2 \)
Let \( u = a^2 \):
\[ u^2 - 8u - 9 = 0 \]
Using the quadratic formula:
\[ u = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm 10}{2} \]
This gives:
\[ u = 9 \quad \text{or} \quad u = -1 \]
Since \( u = a^2 \) and must be positive, we have:
\[ a^2 = 9 \implies a = 3 \]

### Step 9: Substitute \( a \) back into \( z \)
Substituting \( a = 3 \) back into \( z \):
\[ z = \frac{-2 + 2(3)i}{3^2 + 1} = \frac{-2 + 6i}{10} = -\frac{1}{5} + \frac{3}{5}i \]

### Step 10: Find the conjugate of \( z \)
The conjugate \( \bar{z} \) is:
\[ \bar{z} = -\frac{1}{5} - \frac{3}{5}i \]

### Final Answer
Thus, the value of \( \bar{z} \) is:
\[ \bar{z} = -\frac{1}{5} - \frac{3}{5}i \]