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 determine the locus of the point \( z \) such that the points with affixes \( z \), \( -iz \), and \( 1 \) are collinear, we can follow these steps:

### Step 1: Set up the condition for collinearity
The points \( z \), \( -iz \), and \( 1 \) are collinear if the area of the triangle formed by these points is zero. We can use the determinant method to express this condition.

### Step 2: Formulate the determinant
We can express the condition for collinearity using the determinant of the following matrix:

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

### Step 3: Calculate the determinant
Expanding the determinant, we have:

\[ \begin{vmatrix} z & -iz & 1 \\ \overline{z} & -i\overline{z} & 1 \\ 1 & 1 & 1 \end{vmatrix} = z(-i\overline{z} - 1) - (-iz)(\overline{z} - 1) + 1(\overline{z}(-iz) - z) \]

Calculating this determinant step by step:

1. The first term: \( z(-i\overline{z} - 1) = -z(i\overline{z} + z) \)
2. The second term: \( -(-iz)(\overline{z} - 1) = iz(\overline{z} - 1) = iz\overline{z} - iz \)
3. The third term: \( 1(\overline{z}(-iz) - z) = -iz\overline{z} - z \)

Combining these terms gives:

\[ -z(i\overline{z} + z) + iz\overline{z} - iz - iz\overline{z} - z = 0 \]

### Step 4: Simplify the expression
Now, simplifying the expression:

\[ -z(i\overline{z} + z) - iz = 0 \]

### Step 5: Rearranging
Rearranging gives:

\[ z(i\overline{z} + z) + iz = 0 \]

### Step 6: Factor out common terms
Factoring out \( z \):

\[ z(i\overline{z} + z + i) = 0 \]

### Step 7: Solve for \( z \)
Since \( z \) cannot be zero (as it represents a point), we set:

\[ i\overline{z} + z + i = 0 \]

### Step 8: Rearranging to find the locus
Rearranging gives:

\[ i\overline{z} + z = -i \]

### Step 9: Substitute \( z = x + iy \)
Let \( z = x + iy \) where \( x \) and \( y \) are real numbers. Then:

\[ i(x - iy) + (x + iy) = -i \]

This simplifies to:

\[ ix + y + x + iy = -i \]

Separating real and imaginary parts gives:

1. Real part: \( x + y = 0 \)
2. Imaginary part: \( x = -1 \)

### Step 10: Conclusion
From \( x + y = 0 \), we can express \( y \) in terms of \( x \):

\[ y = -x \]

Thus, the locus of \( z \) is a line given by \( y = -x \).