If `omega` is a complex cube root of unity, then arg `(iomega) + "arg" (iomega^(2))=`

Options

A. 0
B. `pi//2`
C. `pi`
D. `pi//4`

Explanation

To solve the problem, we need to find the sum of the arguments of the complex numbers \( i\omega \) and \( i\omega^2 \), where \( \omega \) is a complex cube root of unity.

### Step-by-Step Solution:

1. **Identify the Cube Roots of Unity**:
The complex cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3} \quad \text{and} \quad \omega^2 = e^{4\pi i / 3} \]
The third root is \( 1 \), but we will focus on \( \omega \) and \( \omega^2 \).

2. **Find the Argument of \( i\omega \)**:
We know that \( i = e^{i\pi/2} \). Thus, we can express \( i\omega \) as:
\[ i\omega = e^{i\pi/2} \cdot e^{2\pi i / 3} = e^{i(\pi/2 + 2\pi/3)} \]
Now, we calculate the argument:
\[ \text{arg}(i\omega) = \frac{\pi}{2} + \frac{2\pi}{3} \]
To add these, we need a common denominator:
\[ \frac{\pi}{2} = \frac{3\pi}{6}, \quad \frac{2\pi}{3} = \frac{4\pi}{6} \]
Therefore:
\[ \text{arg}(i\omega) = \frac{3\pi}{6} + \frac{4\pi}{6} = \frac{7\pi}{6} \]

3. **Find the Argument of \( i\omega^2 \)**:
Similarly, for \( i\omega^2 \):
\[ i\omega^2 = e^{i\pi/2} \cdot e^{4\pi i / 3} = e^{i(\pi/2 + 4\pi/3)} \]
Again, we calculate the argument:
\[ \text{arg}(i\omega^2) = \frac{\pi}{2} + \frac{4\pi}{3} \]
Using the common denominator:
\[ \frac{\pi}{2} = \frac{3\pi}{6}, \quad \frac{4\pi}{3} = \frac{8\pi}{6} \]
Thus:
\[ \text{arg}(i\omega^2) = \frac{3\pi}{6} + \frac{8\pi}{6} = \frac{11\pi}{6} \]

4. **Sum the Arguments**:
Now we can find the sum of the two arguments:
\[ \text{arg}(i\omega) + \text{arg}(i\omega^2) = \frac{7\pi}{6} + \frac{11\pi}{6} = \frac{18\pi}{6} = 3\pi \]

5. **Simplify the Result**:
Since the argument is periodic with period \( 2\pi \):
\[ 3\pi \equiv \pi \quad (\text{mod } 2\pi) \]

### Final Answer:
Thus, the final answer is:
\[ \text{arg}(i\omega) + \text{arg}(i\omega^2) = \pi \]