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 given condition and derive the expression for \((z-2)/z\).

### Step 1: Understand the given condition
The condition given is \(|z - 1| = 1\). This means that the complex number \(z\) lies on a circle centered at \(1\) (which is the point \(1 + 0i\) in the complex plane) with a radius of \(1\).

### Step 2: Rewrite the condition
The equation \(|z - 1| = 1\) can be interpreted as:
\[ z - 1 = e^{i\theta} \quad \text{for } \theta \in [0, 2\pi) \]
Thus, we can express \(z\) as:
\[ z = 1 + e^{i\theta} \]

### Step 3: Substitute \(z\) into the expression
We need to find the expression \(\frac{z - 2}{z}\):
\[ \frac{z - 2}{z} = \frac{(1 + e^{i\theta}) - 2}{1 + e^{i\theta}} = \frac{e^{i\theta} - 1}{1 + e^{i\theta}} \]

### Step 4: Simplify the expression
Now we simplify \(\frac{e^{i\theta} - 1}{1 + e^{i\theta}}\):
- The numerator \(e^{i\theta} - 1\) can be expressed as \(e^{i\theta} - e^{0}\).
- The denominator \(1 + e^{i\theta}\) is straightforward.

### Step 5: Analyze the result
To analyze the expression further, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(e^{i\theta} - 1)(1 + e^{-i\theta})}{(1 + e^{i\theta})(1 + e^{-i\theta})} \]
This gives:
\[ = \frac{(e^{i\theta} - 1)(1 + e^{-i\theta})}{2 + e^{i\theta} + e^{-i\theta}} = \frac{(e^{i\theta} - 1)(1 + e^{-i\theta})}{2 + 2\cos(\theta)} \]

### Step 6: Determine the nature of the expression
The numerator \((e^{i\theta} - 1)(1 + e^{-i\theta})\) can be analyzed to show that it is purely imaginary because it represents a rotation in the complex plane. The denominator is real and positive.

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

### Final Answer
The expression \(\frac{z - 2}{z}\) is purely imaginary.

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