If `omega` is the cube root of unity , then what is the conjugate of `2 omega^(2) + 3i` ?

Options

A. `2 omega - 3i`
B. ` 3 omega+ 2i`
C. ` 2 omega + 3i`
D. ` 3 omega-2i`

Answer

To find the conjugate of the expression \(2\omega^2 + 3i\), where \(\omega\) is a cube root of unity, we can follow these steps:

### Step 1: Understand the cube roots of unity
The cube roots of unity are the solutions to the equation \(x^3 = 1\). They are given by:
- \(1\)
- \(\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\)
- \(\omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\)

### Step 2: Substitute \(\omega^2\)
We know that:
\[ \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \]
Now, we can substitute this into the expression \(2\omega^2 + 3i\).

### Step 3: Calculate \(2\omega^2\)
Substituting \(\omega^2\):
\[ 2\omega^2 = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -1 - \sqrt{3}i \]

### Step 4: Combine with \(3i\)
Now, we add \(3i\) to \(2\omega^2\):
\[ 2\omega^2 + 3i = (-1 - \sqrt{3}i) + 3i = -1 + (3 - \sqrt{3})i \]

### Step 5: Find the conjugate
The conjugate of a complex number \(a + bi\) is \(a - bi\). Therefore, the conjugate of \(-1 + (3 - \sqrt{3})i\) is:
\[ -1 - (3 - \sqrt{3})i = -1 - 3i + \sqrt{3}i = -1 - 3i + \sqrt{3}i \]

### Final Result
Thus, the conjugate of \(2\omega^2 + 3i\) is:
\[ -1 + (\sqrt{3} - 3)i \]