If points having affixes z, `-iz` and 1 are collinear, then z lies on

Options

A. a straight line
B. a circle
C. an ellipse
D. a pair of straight lines.

Explanation

To solve the problem where the points with affixes \( z \), \( -iz \), and \( 1 \) are collinear, we can use the determinant condition for collinearity. The points are collinear if the following determinant is equal to zero:

\[ \begin{vmatrix} z & -iz & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

### Step-by-Step Solution:

1. **Set up the determinant**:
We write the determinant of the matrix formed by the points:
\[ \begin{vmatrix} z & -iz & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

2. **Calculate the determinant**:
The determinant can be calculated using the formula for a 3x3 determinant. For our matrix, we can simplify it:
\[ = z \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + iz \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} \]
However, since the second and third rows are identical, the determinant simplifies to:
\[ = 0 \]

3. **Use the property of collinearity**:
Since the determinant equals zero, we can use the property of collinearity:
\[ z - iz + 1 = 0 \]

4. **Rearranging the equation**:
Rearranging gives:
\[ z(1 - i) + 1 = 0 \]
Thus,
\[ z(1 - i) = -1 \]
Therefore,
\[ z = \frac{-1}{1 - i} \]

5. **Multiply by the conjugate**:
To simplify \( z \), multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{-1(1 + i)}{(1 - i)(1 + i)} = \frac{-(1 + i)}{1 + 1} = \frac{-(1 + i)}{2} \]
Hence,
\[ z = -\frac{1}{2} - \frac{1}{2}i \]

6. **Conclusion**:
The point \( z \) lies on the line represented by the equation derived from the collinearity condition. This equation represents a line in the complex plane.