If z lies on the circle `|z-1|=1`, then `(z-2)/z` is

Options

A. purely real
B. Purely imaginary
C. positive real
D. hyperbola

Answer

To solve the problem, we need to analyze the expression \((z-2)/z\) given that \(z\) lies on the circle defined by \(|z-1|=1\).

### Step-by-Step Solution:

1. **Understanding the Circle**:
The equation \(|z-1|=1\) represents a circle in the complex plane with center at \(1\) (which corresponds to the point \((1, 0)\) in the Cartesian plane) and radius \(1\). This means that any point \(z\) on this circle can be expressed as:
\[ z = 1 + e^{i\theta} \]
where \(\theta\) varies from \(0\) to \(2\pi\).

2. **Expressing \(z\)**:
We can write \(z\) in terms of its real and imaginary parts:
\[ z = x + iy \]
where \(x = 1 + \cos(\theta)\) and \(y = \sin(\theta)\).

3. **Finding \(z-2\)**:
Now, we compute \(z - 2\):
\[ z - 2 = (x - 2) + iy = (1 + \cos(\theta) - 2) + i\sin(\theta) = (\cos(\theta) - 1) + i\sin(\theta) \]

4. **Finding \((z-2)/z\)**:
Now, we need to compute \(\frac{z-2}{z}\):
\[ \frac{z-2}{z} = \frac{(\cos(\theta) - 1) + i\sin(\theta)}{1 + \cos(\theta) + i\sin(\theta)} \]

5. **Rationalizing the Denominator**:
To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((\cos(\theta) - 1) + i\sin(\theta)) \cdot (1 + \cos(\theta) - i\sin(\theta))}{(1 + \cos(\theta))^2 + \sin^2(\theta)} \]

6. **Calculating the Denominator**:
The denominator simplifies as follows:
\[ (1 + \cos(\theta))^2 + \sin^2(\theta) = 1 + 2\cos(\theta) + \cos^2(\theta) + \sin^2(\theta) = 2 + 2\cos(\theta) = 2(1 + \cos(\theta)) \]

7. **Calculating the Numerator**:
The numerator expands to:
\[ (\cos(\theta) - 1)(1 + \cos(\theta)) + i\sin(\theta)(1 + \cos(\theta)) - i\sin^2(\theta) \]
Simplifying gives:
\[ -\sin^2(\theta) + i\sin(\theta)(1 + \cos(\theta)) \]

8. **Final Expression**:
Thus, we have:
\[ \frac{z-2}{z} = \frac{-\sin^2(\theta) + i\sin(\theta)(1 + \cos(\theta))}{2(1 + \cos(\theta))} \]

9. **Identifying the Nature**:
The real part is \(\frac{-\sin^2(\theta)}{2(1 + \cos(\theta))}\) and the imaginary part is \(\frac{\sin(\theta)(1 + \cos(\theta))}{2(1 + \cos(\theta))}\). Since the real part can be zero (when \(\sin^2(\theta) = 0\)), the expression can be purely imaginary depending on the value of \(\theta\).

### Conclusion:
Thus, \(\frac{z-2}{z}\) is purely imaginary.