If `|z_1|`= `|z_2|`=1 and amp `z_1`+amp`z_2`=0 then

Options

A. `z_1z_2=1`
B. `z_1+z_2`=0
C. `z_1=barz_2`
D. none of these

Answer

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

1. \( |z_1| = |z_2| = 1 \)
2. \( \text{arg}(z_1) + \text{arg}(z_2) = 0 \)

### Step 1: Express the complex numbers in exponential form
Since the magnitudes of both complex numbers are 1, we can express them in the form:
\[ z_1 = e^{i\theta_1} \]
\[ z_2 = e^{i\theta_2} \]

### Step 2: Use the argument condition
From the condition \( \text{arg}(z_1) + \text{arg}(z_2) = 0 \), we have:
\[ \theta_1 + \theta_2 = 0 \]
This implies:
\[ \theta_2 = -\theta_1 \]

### Step 3: Substitute \( \theta_2 \) into the expression for \( z_2 \)
Now substituting \( \theta_2 \) into the expression for \( z_2 \):
\[ z_2 = e^{i(-\theta_1)} = e^{-i\theta_1} \]

### Step 4: Check the product \( z_1 z_2 \)
Now, we can find the product \( z_1 z_2 \):
\[ z_1 z_2 = e^{i\theta_1} \cdot e^{-i\theta_1} = e^{0} = 1 \]

### Step 5: Check the sum \( z_1 + z_2 \)
Next, we check the sum \( z_1 + z_2 \):
\[ z_1 + z_2 = e^{i\theta_1} + e^{-i\theta_1} \]
Using Euler's formula, we can rewrite this as:
\[ z_1 + z_2 = \cos(\theta_1) + i\sin(\theta_1) + \cos(-\theta_1) + i\sin(-\theta_1) \]
Since \( \sin(-\theta_1) = -\sin(\theta_1) \), we have:
\[ z_1 + z_2 = \cos(\theta_1) + \cos(\theta_1) + i(\sin(\theta_1) - \sin(\theta_1)) = 2\cos(\theta_1) \]

### Step 6: Check the conjugate relationship
The conjugate of \( z_2 \) is:
\[ \overline{z_2} = e^{-i\theta_2} = e^{i\theta_1} = z_1 \]
Thus, we find that:
\[ z_1 = \overline{z_2} \]

### Conclusion
From the above steps, we conclude:
1. \( z_1 z_2 = 1 \) (Option A is correct)
2. \( z_1 + z_2 = 2\cos(\theta_1) \) (This is not always zero)
3. \( z_1 = \overline{z_2} \) (Option C is correct)

Thus, the correct options are A and C.