Suppose `a in R and z in C. "If" |z| = 2 and (z-alpha)/(z+alpha)` is purely imaginary, then 1.25 `|alpha|` is equal to ____________.

Answer

To solve the problem step by step, we will analyze the given conditions and derive the required value.

### Step 1: Understand the given conditions
We have:
- \( |z| = 2 \) (which means the modulus of the complex number \( z \) is 2)
- \( \frac{z - \alpha}{z + \alpha} \) is purely imaginary.

### Step 2: Represent \( z \) in terms of its components
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The modulus condition gives us:
\[ |z| = \sqrt{x^2 + y^2} = 2 \]
Squaring both sides, we get:
\[ x^2 + y^2 = 4 \]

### Step 3: Analyze the expression \( \frac{z - \alpha}{z + \alpha} \)
We can rewrite the expression:
\[ \frac{z - \alpha}{z + \alpha} = \frac{(x - \alpha) + iy}{(x + \alpha) + iy} \]

### Step 4: Multiply by the conjugate of the denominator
To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((x - \alpha) + iy)((x + \alpha) - iy)}{((x + \alpha) + iy)((x + \alpha) - iy)} \]
The denominator simplifies to:
\[ (x + \alpha)^2 + y^2 \]
The numerator simplifies to:
\[ (x - \alpha)(x + \alpha) - i y(x - \alpha) + i y(x + \alpha) + y^2 \]
This gives:
\[ (x^2 - \alpha^2 + y^2) + i(2y\alpha) \]

### Step 5: Set the real part to zero
Since the expression is purely imaginary, the real part must be zero:
\[ x^2 - \alpha^2 + y^2 = 0 \]
Substituting \( x^2 + y^2 = 4 \) into this equation, we have:
\[ 4 - \alpha^2 = 0 \]
Thus,
\[ \alpha^2 = 4 \implies |\alpha| = 2 \]

### Step 6: Calculate \( 1.25 | \alpha | \)
Now we can calculate:
\[ 1.25 |\alpha| = 1.25 \times 2 = 2.5 \]

### Final Answer
Therefore, the value of \( 1.25 |\alpha| \) is:
\[ \boxed{2.5} \]