Let `z!=i` be any complex number such that `(z-i)/(z+i)` is a purely imaginary number. Then `z+ 1/z` is

Options

A. any non-zero real number other than 1.
B. a purely imaginary number.
C. 0
D. any non-zero number

Answer

To solve the problem, we need to analyze the given condition that \((z - i)/(z + i)\) is a purely imaginary number. Let's denote \(z\) as a complex number in the form \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step-by-Step Solution:

1. **Substitute \(z\) in the given expression**:
\[ \frac{z - i}{z + i} = \frac{(x + iy) - i}{(x + iy) + i} = \frac{x + (y - 1)i}{x + (y + 1)i} \]

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

3. **Calculate the denominator**:
The denominator simplifies as follows:
\[ (x + (y + 1)i)(x - (y + 1)i) = x^2 + (y + 1)^2 \]

4. **Calculate the numerator**:
The numerator simplifies as follows:
\[ (x + (y - 1)i)(x - (y + 1)i) = x^2 - (y + 1)(y - 1) + i[(y - 1)x + (y + 1)x] = x^2 - (y^2 - 1) + i[2yx] \]

5. **Combine the results**:
Thus, we have:
\[ \frac{x^2 + 1 - y^2 + i(2yx)}{x^2 + (y + 1)^2} \]

6. **Separate real and imaginary parts**:
The expression can be separated into real and imaginary parts:
\[ \text{Real part: } \frac{x^2 + 1 - y^2}{x^2 + (y + 1)^2}, \quad \text{Imaginary part: } \frac{2yx}{x^2 + (y + 1)^2} \]

7. **Set the real part to zero** (since it is purely imaginary):
\[ x^2 + 1 - y^2 = 0 \implies x^2 + 1 = y^2 \implies y^2 - x^2 = 1 \]

8. **Recognize this as a hyperbola**:
This equation represents a hyperbola in the \(xy\)-plane.

9. **Now, find \(z + \frac{1}{z}\)**:
We have:
\[ z + \frac{1}{z} = (x + iy) + \frac{1}{x + iy} \]
To compute \(\frac{1}{z}\):
\[ \frac{1}{z} = \frac{x - iy}{x^2 + y^2} \]

10. **Combine the two parts**:
\[ z + \frac{1}{z} = \left(x + \frac{x}{x^2 + y^2}\right) + i\left(y - \frac{y}{x^2 + y^2}\right) \]
Simplifying gives:
\[ z + \frac{1}{z} = \left(x + \frac{x}{1}\right) + i\left(y - \frac{y}{1}\right) = 2x \]

11. **Conclusion**:
Since \(x^2 + y^2 = 1\) (from the earlier equation), \(z + \frac{1}{z} = 2x\) is a non-zero real number.

### Final Answer:
\[ z + \frac{1}{z} = 2x \]