If `|z_(1)|=|z_(2)|` and arg `(z_(1))+"arg"(z_(2))=0`, then

Options

A. `z_(1)=z_(2)`
B. `z_(1)=barz_(2)`
C. `z_(1)z_(2)=1`
D. `z_(1)barz_(2)=1`

Answer

To solve the problem, we need to analyze the given conditions step by step.

### Step 1: Understand the given conditions
We are given that:
1. \(|z_1| = |z_2|\)
2. \(\arg(z_1) + \arg(z_2) = 0\)

### Step 2: Express \(z_1\) and \(z_2\) in polar form
Let’s express \(z_1\) and \(z_2\) in their polar forms:
- \(z_1 = r e^{i\theta_1}\)
- \(z_2 = r e^{i\theta_2}\)

where \(r = |z_1| = |z_2|\) and \(\theta_1 = \arg(z_1)\), \(\theta_2 = \arg(z_2)\).

### Step 3: Use the first condition
From the first condition, since \(|z_1| = |z_2|\), we have:
\[ r_1 = r_2 = r \]

### Step 4: Use the second condition
From the second condition, we have:
\[ \arg(z_1) + \arg(z_2) = 0 \implies \theta_1 + \theta_2 = 0 \]
This implies:
\[ \theta_2 = -\theta_1 \]

### Step 5: Substitute \(\theta_2\) into the expression for \(z_2\)
Now substituting \(\theta_2\) into the expression for \(z_2\):
\[ z_2 = r e^{i(-\theta_1)} = r e^{-i\theta_1} \]

### Step 6: Find the product \(z_1 \cdot z_2\)
Now, we can find the product \(z_1 \cdot z_2\):
\[ z_1 \cdot z_2 = (r e^{i\theta_1})(r e^{-i\theta_1}) = r^2 e^{i(\theta_1 - \theta_1)} = r^2 e^{0} = r^2 \]

### Step 7: Analyze the result
Since \(r^2\) is a positive real number, the product \(z_1 \cdot z_2\) is equal to \(r^2\), which means that:
\[ z_1 \cdot z_2 = 1 \text{ if } r^2 = 1 \]
Thus, \(z_1\) and \(z_2\) are conjugates of each other.

### Conclusion
The condition implies that \(z_1\) and \(z_2\) are conjugates, which can be expressed as:
\[ z_1 \cdot z_2 = 1 \]

### Final Answer
Thus, the answer is:
\[ \text{Option 3: } z_1 \cdot z_2 = 1 \]

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