If `z_(1) , z_(2), z_(3)` are in H.P., they lie on a

Options

A. circle
B. sphere
C. straight line
D. none of these

Answer

To solve the problem, we need to show that if \( z_1, z_2, z_3 \) are in Harmonic Progression (H.P.), then they lie on a straight line in the complex plane.

### Step-by-step Solution:

1. **Understanding H.P.**:
- Three complex numbers \( z_1, z_2, z_3 \) are said to be in H.P. if their reciprocals \( \frac{1}{z_1}, \frac{1}{z_2}, \frac{1}{z_3} \) are in Arithmetic Progression (A.P.).
- This means that \( 2/z_2 = 1/z_1 + 1/z_3 \).

2. **Writing the Condition**:
- From the H.P. condition, we can write:
\[ \frac{2}{z_2} = \frac{1}{z_1} + \frac{1}{z_3} \]
- Rearranging gives:
\[ \frac{2}{z_2} = \frac{z_3 + z_1}{z_1 z_3} \]

3. **Cross Multiplying**:
- Cross multiplying yields:
\[ 2 z_1 z_3 = z_2 (z_1 + z_3) \]

4. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ z_2 (z_1 + z_3) - 2 z_1 z_3 = 0 \]

5. **Factoring**:
- This can be factored as:
\[ z_2 = \frac{2 z_1 z_3}{z_1 + z_3} \]

6. **Geometric Interpretation**:
- The equation \( z_2 = \frac{2 z_1 z_3}{z_1 + z_3} \) indicates that \( z_2 \) is a linear combination of \( z_1 \) and \( z_3 \).
- This means that \( z_1, z_2, z_3 \) are collinear in the complex plane, as \( z_2 \) can be expressed as a weighted average of \( z_1 \) and \( z_3 \).

7. **Conclusion**:
- Therefore, if \( z_1, z_2, z_3 \) are in H.P., they lie on a straight line in the complex plane.