The square roots of `- 2 + 2 sqrt(3)i` are :

Options

A. `pm (1 + sqrt(3)i)`
B. `pm (1 - sqrt(3)i)`
C. `pm (-1 + sqrt(3)i)`
D. None of these

Answer

To find the square roots of the complex number \(-2 + 2\sqrt{3}i\), we can express it in the form \(a + bi\) where \(a\) and \(b\) are real numbers. We will find \(x\) and \(y\) such that:

\[ x^2 - y^2 = -2 \quad (1) \]
\[ 2xy = 2\sqrt{3} \quad (2) \]

### Step 1: Solve for \(xy\)
From equation (2), we can simplify it to:

\[ xy = \sqrt{3} \quad (3) \]

### Step 2: Express \(y\) in terms of \(x\)
From equation (3), we can express \(y\) in terms of \(x\):

\[ y = \frac{\sqrt{3}}{x} \quad (4) \]

### Step 3: Substitute \(y\) in equation (1)
Now, substitute equation (4) into equation (1):

\[ x^2 - \left(\frac{\sqrt{3}}{x}\right)^2 = -2 \]

This simplifies to:

\[ x^2 - \frac{3}{x^2} = -2 \]

### Step 4: Multiply through by \(x^2\)
To eliminate the fraction, multiply through by \(x^2\):

\[ x^4 + 2x^2 - 3 = 0 \]

### Step 5: Let \(u = x^2\)
Let \(u = x^2\). Then we have a quadratic equation:

\[ u^2 + 2u - 3 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]
\[ u = \frac{-2 \pm \sqrt{4 + 12}}{2} \]
\[ u = \frac{-2 \pm \sqrt{16}}{2} \]
\[ u = \frac{-2 \pm 4}{2} \]

This gives us:

\[ u = 1 \quad \text{or} \quad u = -3 \]

Since \(u = x^2\), we discard \(u = -3\) (as \(x^2\) cannot be negative).

### Step 7: Find \(x\)
Thus, we have:

\[ x^2 = 1 \implies x = \pm 1 \]

### Step 8: Find \(y\)
Substituting \(x\) back into equation (4):

For \(x = 1\):

\[ y = \frac{\sqrt{3}}{1} = \sqrt{3} \]

For \(x = -1\):

\[ y = \frac{\sqrt{3}}{-1} = -\sqrt{3} \]

### Step 9: Write the square roots
Thus, the square roots of \(-2 + 2\sqrt{3}i\) are:

\[ 1 + \sqrt{3}i \quad \text{and} \quad -1 - \sqrt{3}i \]

### Final Answer
The square roots of \(-2 + 2\sqrt{3}i\) are:

\[ \boxed{1 + \sqrt{3}i \quad \text{and} \quad -1 - \sqrt{3}i} \]