If `z ne 1 and (z^(2))/(z-1)` is real, then the point represented by the complex number z lies

Options

A. either on the real axis or on a circle passing through the origin
B. on a circle with centre at the origin
C. either on the real axis or on a circle not passing through the origin
D. on the imaginary axis

Answer

To solve the problem, we need to determine the conditions under which the expression \(\frac{z^2}{z-1}\) is real, given that \(z \neq 1\). We will represent the complex number \(z\) in the form \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step-by-Step Solution:

1. **Substitute \(z\)**:
\[ z = x + iy \]
Substitute this into the expression:
\[ \frac{z^2}{z-1} = \frac{(x + iy)^2}{(x + iy) - 1} \]

2. **Calculate \(z^2\)**:
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]

3. **Calculate \(z - 1\)**:
\[ z - 1 = (x + iy) - 1 = (x - 1) + iy \]

4. **Form the expression**:
\[ \frac{(x^2 - y^2) + 2xyi}{(x - 1) + iy} \]

5. **Multiply numerator and denominator by the conjugate of the denominator**:
\[ \frac{((x^2 - y^2) + 2xyi)((x - 1) - iy)}{((x - 1) + iy)((x - 1) - iy)} \]

6. **Calculate the denominator**:
\[ (x - 1)^2 + y^2 \]

7. **Calculate the numerator**:
\[ \text{Numerator} = (x^2 - y^2)(x - 1) - 2xy^2 + 2xyi(x - 1) + 2xyi(-iy) \]
Simplifying gives:
\[ = (x^2 - y^2)(x - 1) - 2xy^2 + 2xy(x - 1)i + 2xy^2 \]
Combine like terms:
\[ = (x^2 - y^2)(x - 1) + 2xy(x - 1)i \]

8. **Set the imaginary part to zero**:
For the expression to be real, the imaginary part must equal zero:
\[ 2xy(x - 1) = 0 \]

9. **Solve for conditions**:
This gives us two cases:
- Case 1: \(y = 0\) (which means \(z\) lies on the real axis)
- Case 2: \(x = 1\) (which is not allowed since \(z \neq 1\))

10. **Conclusion**:
The only valid case is when \(y = 0\), which means that the locus of the complex number \(z\) is on the real axis.

### Final Answer:
The point represented by the complex number \(z\) lies on the **real axis**.