If `|z_(1)| = |z_(2)| = 1, z_(1)z_(2) ne -1 and z = (z_(1) + z_(2))/(1+z_(1)z_(2))` then

Options

A. z is a purely real number
B. z is a purely imaginary number
C. |z| = 1
D. None of these

Answer

To solve the problem, we start with the given information:

1. \( |z_1| = |z_2| = 1 \)
2. \( z_1 z_2 \neq -1 \)
3. We need to analyze \( z = \frac{z_1 + z_2}{1 + z_1 z_2} \)

### Step 1: Understanding the Modulus of \( z_1 \) and \( z_2 \)
Since \( |z_1| = 1 \) and \( |z_2| = 1 \), we can express \( z_1 \) and \( z_2 \) in exponential form:
- Let \( z_1 = e^{i\theta_1} \)
- Let \( z_2 = e^{i\theta_2} \)

### Step 2: Express \( z \) in terms of \( z_1 \) and \( z_2 \)
Using the definitions of \( z_1 \) and \( z_2 \):
\[ z = \frac{e^{i\theta_1} + e^{i\theta_2}}{1 + e^{i(\theta_1 + \theta_2)}} \]

### Step 3: Simplifying the Expression
We can simplify \( z \):
\[ z = \frac{e^{i\theta_1} + e^{i\theta_2}}{1 + e^{i(\theta_1 + \theta_2)}} \]
Using the property of complex numbers, we can rewrite the numerator:
\[ z = \frac{2 \cos\left(\frac{\theta_1 + \theta_2}{2}\right) e^{i\left(\frac{\theta_1 + \theta_2}{2}\right)}}{1 + e^{i(\theta_1 + \theta_2)}} \]

### Step 4: Finding the Modulus of \( z \)
To find the modulus \( |z| \):
\[ |z| = \left| \frac{z_1 + z_2}{1 + z_1 z_2} \right| = \frac{|z_1 + z_2|}{|1 + z_1 z_2|} \]
Since \( |z_1| = |z_2| = 1 \):
\[ |z_1 + z_2| = |2 \cos\left(\frac{\theta_1 + \theta_2}{2}\right)| \]
And:
\[ |1 + z_1 z_2| = |1 + e^{i(\theta_1 + \theta_2)}| = \sqrt{(1 + \cos(\theta_1 + \theta_2))^2 + \sin^2(\theta_1 + \theta_2)} \]
This simplifies to:
\[ |1 + z_1 z_2| = \sqrt{2(1 + \cos(\theta_1 + \theta_2))} = 2\cos\left(\frac{\theta_1 + \theta_2}{2}\right) \]
Thus:
\[ |z| = \frac{|2 \cos\left(\frac{\theta_1 + \theta_2}{2}\right)|}{2 \cos\left(\frac{\theta_1 + \theta_2}{2}\right)} = 1 \]

### Step 5: Conclusion
Since \( |z| = 1 \), we conclude that:
- \( z \) is on the unit circle.

### Step 6: Checking for Real or Imaginary
To check if \( z \) is purely real or purely imaginary, we note that:
- If \( z = \frac{z_1 + z_2}{1 + z_1 z_2} \) and \( z_1 z_2 \neq -1 \), the imaginary part must be zero for \( z \) to be real.

Thus, we conclude that \( z \) is purely real.

### Final Answer
The value of \( z \) is purely real.