Let z and omega be complex numbers such that `|z|=|omega|` and arg (z) dentoe the principal of z.
Statement-1: If argz+ arg `omega=pi`, then `z=-baromega`
Statement -2: `|z|=|omega|` implies arg z-arg `baromega=pi`, then `z=-baromega`

Options

A. Statement-1 is True, Statement-2 is True: Statement-2 is a correct exp,anation for statement-1.
B. Statement-1 is true, statement -2 is true, Statement-2 is not a correct explanation for statement-1.
C. Statement-1 is True, statement-2 is false,
D. statement-1 is False, Statement-2 is true.

Answer

To solve the problem, we will analyze both statements step by step.

### Step-by-Step Solution

**Given:**
- Let \( z \) and \( \omega \) be complex numbers such that \( |z| = |\omega| \).
- \( \arg(z) \) denotes the principal argument of \( z \).

### Statement 1: If \( \arg(z) + \arg(\omega) = \pi \), then \( z = -\overline{\omega} \).

1. **Express \( z \) and \( \omega \) in polar form:**
\[ z = |z|(\cos(\arg(z)) + i \sin(\arg(z))) = |z| e^{i \arg(z)} \]
\[ \omega = |\omega|(\cos(\arg(\omega)) + i \sin(\arg(\omega))) = |\omega| e^{i \arg(\omega)} \]

2. **Using the condition \( |z| = |\omega| \):**
Let \( r = |z| = |\omega| \). Thus, we can write:
\[ z = r e^{i \theta} \quad \text{and} \quad \omega = r e^{i \phi} \]
where \( \theta = \arg(z) \) and \( \phi = \arg(\omega) \).

3. **Given \( \arg(z) + \arg(\omega) = \pi \):**
This implies:
\[ \theta + \phi = \pi \implies \phi = \pi - \theta \]

4. **Substituting \( \phi \) into the expression for \( \omega \):**
\[ \omega = r e^{i(\pi - \theta)} = r(\cos(\pi - \theta) + i \sin(\pi - \theta)) = r(-\cos(\theta) + i \sin(\theta)) \]
\[ \omega = -r(\cos(\theta) - i \sin(\theta)) = -\overline{z} \]

5. **Thus, we have:**
\[ z = -\overline{\omega} \]
Therefore, Statement 1 is **True**.

### Statement 2: If \( |z| = |\omega| \) implies \( \arg(z) - \arg(\overline{\omega}) = \pi \), then \( z = -\overline{\omega} \).

1. **Recall that \( \overline{\omega} \) has the same modulus as \( \omega \):**
\[ |\overline{\omega}| = |\omega| = r \]

2. **Express \( \overline{\omega} \) in polar form:**
\[ \overline{\omega} = r e^{-i \phi} \]

3. **Using the condition \( \arg(z) - \arg(\overline{\omega}) = \pi \):**
Let \( \theta = \arg(z) \) and \( \phi = \arg(\overline{\omega}) \). Then:
\[ \theta - (-\phi) = \pi \implies \theta + \phi = \pi \]

4. **This is the same condition as in Statement 1:**
Therefore, we can conclude:
\[ z = -\overline{\omega} \]
Thus, Statement 2 is also **True**.

### Conclusion
Both statements lead to the conclusion that \( z = -\overline{\omega} \). However, the implication in Statement 2 is not a direct consequence of the condition given; hence, it is misleading.

### Final Answer
- Statement 1: True
- Statement 2: False