Let complex number z satisfy ` |z - 2/z| =1 " then " |z|` can take all values except

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the equation given by the complex number \( z \) satisfying \( |z - \frac{2}{z}| = 1 \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ |z - \frac{2}{z}| = 1 \]

2. **Multiply both sides by \( |z| \) (assuming \( z \neq 0 \)):**
\[ |z| \cdot |z - \frac{2}{z}| = |z| \]
This simplifies to:
\[ |z|^2 - 2 = |z| \]

3. **Let \( |z| = r \):**
\[ r^2 - 2 = r \]

4. **Rearranging the equation gives us a quadratic equation:**
\[ r^2 - r - 2 = 0 \]

5. **Factor the quadratic equation:**
\[ (r - 2)(r + 1) = 0 \]

6. **Solve for \( r \):**
\[ r - 2 = 0 \quad \text{or} \quad r + 1 = 0 \]
This gives:
\[ r = 2 \quad \text{or} \quad r = -1 \]

7. **Since \( r = |z| \) must be non-negative, we discard \( r = -1 \):**
\[ |z| = 2 \]

8. **Now, we need to find the values that \( |z| \) can take. The equation \( |z - \frac{2}{z}| = 1 \) leads us to conclude that \( |z| \) can take all values except for \( 1 \) and \( 2 \).**

### Conclusion:
Thus, the modulus \( |z| \) can take all values except \( 1 \) and \( 2 \).