A complex number z is said to be unimodular if `abs(z)=1`. Suppose `z_(1)` and `z_(2)` are complex numbers such that `(z_(1)-2z_(2))/(2-z_(1)z_(2)^_)` is unimodular and `z_(2)` is not unimodular. Then the point `z_(1)` lies on a

Options

A. circle of radius z
B. circle of radius `sqrt(2)`
C. straight line parallel to X-axis
D. straight line parallel to y-axis

Answer

To solve the problem, we need to analyze the given expression and derive the conditions for the complex numbers \( z_1 \) and \( z_2 \).

### Step-by-step Solution:

1. **Understanding the Given Condition**:
We are given that the complex number \( z_1 \) and \( z_2 \) satisfy the condition:
\[ \left| \frac{z_1 - 2z_2}{2 - z_1 \overline{z_2}} \right| = 1 \]
This means that the modulus of the fraction is equal to 1, indicating that the numerator and denominator have the same modulus.

2. **Cross Multiplying**:
From the unimodular condition, we can cross-multiply:
\[ |z_1 - 2z_2| = |2 - z_1 \overline{z_2}| \]

3. **Squaring Both Sides**:
To eliminate the modulus, we square both sides:
\[ |z_1 - 2z_2|^2 = |2 - z_1 \overline{z_2}|^2 \]

4. **Expanding the Modulus**:
Using the property \( |z|^2 = z \overline{z} \), we expand both sides:
- Left Side:
\[ |z_1 - 2z_2|^2 = (z_1 - 2z_2)(\overline{z_1 - 2z_2}) = (z_1 - 2z_2)(\overline{z_1} - 2\overline{z_2}) \]
- Right Side:
\[ |2 - z_1 \overline{z_2}|^2 = (2 - z_1 \overline{z_2})(2 - \overline{z_1} z_2) \]

5. **Simplifying Both Sides**:
After expanding both sides, we have:
\[ |z_1|^2 - 4\text{Re}(z_1 \overline{z_2}) + 4|z_2|^2 = 4 - 2(z_1 \overline{z_2} + \overline{z_1} z_2) + |z_1|^2 |z_2|^2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ |z_1|^2 - |z_1|^2 |z_2|^2 + 4|z_2|^2 - 4 + 2(z_1 \overline{z_2} + \overline{z_1} z_2) = 0 \]

7. **Factoring the Equation**:
Factoring out common terms leads us to:
\[ (1 - |z_2|^2)|z_1|^2 + 4|z_2|^2 - 4 = 0 \]

8. **Finding Conditions**:
This gives us two conditions:
- \( 1 - |z_2|^2 = 0 \) implies \( |z_2|^2 = 1 \) (but \( z_2 \) is not unimodular).
- \( |z_1|^2 = 4 \) implies \( |z_1| = 2 \).

9. **Conclusion**:
Since \( |z_1| = 2 \), we can express \( z_1 \) in terms of its real and imaginary parts:
\[ |z_1| = \sqrt{x^2 + y^2} = 2 \implies x^2 + y^2 = 4 \]
This is the equation of a circle centered at the origin with a radius of 2.

### Final Answer:
The point \( z_1 \) lies on a circle of radius 2.