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 \(i\omega\) and \(i\omega^2\), where \(\omega\) is a complex cube root of unity. The cube roots of unity are given by:

\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \]
\[ \omega^2 = e^{4\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \]

### Step 1: Find \(\arg(i\omega)\)

We start with \(i\omega\):
\[ i\omega = i\left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) = -\frac{1}{2}i - \frac{\sqrt{3}}{2} \]

In Cartesian coordinates, this can be represented as:
\[ \text{Re} = -\frac{\sqrt{3}}{2}, \quad \text{Im} = -\frac{1}{2} \]

To find the argument, we use:
\[ \arg(i\omega) = \tan^{-1}\left(\frac{\text{Im}}{\text{Re}}\right) = \tan^{-1}\left(\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

This gives us:
\[ \arg(i\omega) = \frac{\pi}{6} \]

Since both the real and imaginary parts are negative, \(i\omega\) lies in the third quadrant, so we adjust the argument:
\[ \arg(i\omega) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \]

### Step 2: Find \(\arg(i\omega^2)\)

Next, we calculate \(i\omega^2\):
\[ i\omega^2 = i\left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) = -\frac{1}{2}i + \frac{\sqrt{3}}{2} \]

In Cartesian coordinates, this can be represented as:
\[ \text{Re} = \frac{\sqrt{3}}{2}, \quad \text{Im} = -\frac{1}{2} \]

To find the argument:
\[ \arg(i\omega^2) = \tan^{-1}\left(\frac{\text{Im}}{\text{Re}}\right) = \tan^{-1}\left(\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \]

This gives us:
\[ \arg(i\omega^2) = -\frac{\pi}{6} \]

Since the real part is positive and the imaginary part is negative, \(i\omega^2\) lies in the fourth quadrant, so the argument remains:
\[ \arg(i\omega^2) = -\frac{\pi}{6} \]

### Step 3: Sum the arguments

Now we sum the two arguments:
\[ \arg(i\omega) + \arg(i\omega^2) = \frac{7\pi}{6} - \frac{\pi}{6} = \frac{6\pi}{6} = \pi \]

### Final Answer

Thus, the final answer is:
\[ \arg(i\omega) + \arg(i\omega^2) = \pi \]

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