What is the square root of `1/2-i(sqrt3)/(2)`?

Options

A. `+-((sqrt3)/(2)+(i)/(2))`
B. `+-((sqrt3)/(2)-(i)/(2))`
C. `+-((1)/(2)-(sqrt(3))/(2))`
D. `+-((1)/(2)+i(sqrt(3))/(2))`

Explanation

To find the square root of the complex number \( \frac{1}{2} - i \frac{\sqrt{3}}{2} \), we can follow these steps:

### Step 1: Set up the equation
Let \( z = x + iy \) be the square root of \( \frac{1}{2} - i \frac{\sqrt{3}}{2} \). Therefore, we have:
\[ z^2 = \frac{1}{2} - i \frac{\sqrt{3}}{2} \]

### Step 2: Square the expression
Squaring \( z \) gives us:
\[ (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]
Thus, we can equate the real and imaginary parts:
\[ x^2 - y^2 = \frac{1}{2} \quad \text{(1)} \]
\[ 2xy = -\frac{\sqrt{3}}{2} \quad \text{(2)} \]

### Step 3: Use the equations to find \( x^2 + y^2 \)
We can use the identity:
\[ x^2 + y^2 = (x^2 - y^2) + 2y^2 \]
From equation (1), we know \( x^2 - y^2 = \frac{1}{2} \). We need to find \( x^2 + y^2 \).

### Step 4: Express \( y^2 \) in terms of \( x \)
From equation (2), we can express \( y \) in terms of \( x \):
\[ y = -\frac{\sqrt{3}}{4x} \]
Now substituting \( y \) into \( x^2 + y^2 \):
\[ y^2 = \left(-\frac{\sqrt{3}}{4x}\right)^2 = \frac{3}{16x^2} \]
Thus,
\[ x^2 + y^2 = x^2 + \frac{3}{16x^2} \]

### Step 5: Combine equations
Now we can combine the equations:
\[ x^2 + y^2 = \frac{1}{2} + 2y^2 \]
Substituting \( y^2 \):
\[ x^2 + \frac{3}{16x^2} = \frac{1}{2} + 2\left(\frac{3}{16x^2}\right) \]
This simplifies to:
\[ x^2 + \frac{3}{16x^2} = \frac{1}{2} + \frac{3}{8x^2} \]

### Step 6: Solve for \( x^2 \)
Multiply through by \( 16x^2 \) to eliminate the fractions:
\[ 16x^4 + 3 = 8x^2 + 6 \]
Rearranging gives:
\[ 16x^4 - 8x^2 - 3 = 0 \]
Let \( u = x^2 \):
\[ 16u^2 - 8u - 3 = 0 \]
Using the quadratic formula:
\[ u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 16 \cdot (-3)}}{2 \cdot 16} \]
\[ u = \frac{8 \pm \sqrt{64 + 192}}{32} \]
\[ u = \frac{8 \pm \sqrt{256}}{32} = \frac{8 \pm 16}{32} \]
This gives:
\[ u = \frac{24}{32} = \frac{3}{4} \quad \text{or} \quad u = \frac{-8}{32} \quad \text{(not valid)} \]
Thus, \( x^2 = \frac{3}{4} \) and \( x = \pm \frac{\sqrt{3}}{2} \).

### Step 7: Find \( y \)
Using \( 2xy = -\frac{\sqrt{3}}{2} \):
For \( x = \frac{\sqrt{3}}{2} \):
\[ 2 \cdot \frac{\sqrt{3}}{2} \cdot y = -\frac{\sqrt{3}}{2} \implies y = -\frac{1}{2} \]
For \( x = -\frac{\sqrt{3}}{2} \):
\[ 2 \cdot -\frac{\sqrt{3}}{2} \cdot y = -\frac{\sqrt{3}}{2} \implies y = \frac{1}{2} \]

### Final Answer
Thus, the square roots of the complex number are:
\[ \frac{\sqrt{3}}{2} - \frac{1}{2}i \quad \text{and} \quad -\frac{\sqrt{3}}{2} + \frac{1}{2}i \]