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

Options

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

Explanation

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

### Step 1: Identify the values of \( \omega \) and \( \omega^2 \)
The cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \]
\[ \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \]

### Step 2: Substitute \( \omega^2 \) into the expression
We substitute \( \omega^2 \) into the expression for \( z \):
\[ z = 2\omega^2 + 3i = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) + 3i \]

### Step 3: Simplify the expression
Now, simplify the expression:
\[ z = 2\left(-\frac{1}{2}\right) + 2\left(-\frac{\sqrt{3}}{2} i\right) + 3i \]
\[ z = -1 - \sqrt{3} i + 3i \]
Combine the imaginary parts:
\[ z = -1 + (3 - \sqrt{3}) i \]

### Step 4: Find the conjugate of \( z \)
The conjugate of a complex number \( a + bi \) is \( a - bi \). Thus, the conjugate of \( z \) is:
\[ \overline{z} = -1 - (3 - \sqrt{3}) i \]
Simplifying this gives:
\[ \overline{z} = -1 - 3i + \sqrt{3} i \]
\[ \overline{z} = -1 + (\sqrt{3} - 3)i \]

### Final Answer
The conjugate of \( 2\omega^2 + 3i \) is:
\[ \overline{z} = -1 + (\sqrt{3} - 3)i \]

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