What is the square root of `(1)/(2) - i(sqrt(3))/(2)` ?

Options

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

Answer

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

### Step 1: Convert to Polar Form
First, we need to express the complex number in polar form. A complex number \( z = x + iy \) can be represented as:
\[ z = r (\cos \theta + i \sin \theta) \]
where \( r = |z| = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \).

For our \( z \):
- \( x = \frac{1}{2} \)
- \( y = -\frac{\sqrt{3}}{2} \)

Calculating \( r \):
\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

Calculating \( \theta \):
\[ \theta = \tan^{-1} \left( \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) = \tan^{-1}(-\sqrt{3}) \]
This corresponds to an angle of \( -60^\circ \) or \( 300^\circ \) (in standard position).

Thus, we can write:
\[ z = 1 \left( \cos 300^\circ + i \sin 300^\circ \right) \]

### Step 2: Find the Square Root
To find the square root of \( z \), we use the formula:
\[ \sqrt{z} = \sqrt{r} \left( \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) \]
Since \( r = 1 \):
\[ \sqrt{z} = 1 \left( \cos \frac{300^\circ}{2} + i \sin \frac{300^\circ}{2} \right) = \cos 150^\circ + i \sin 150^\circ \]

### Step 3: Evaluate \( \cos 150^\circ \) and \( \sin 150^\circ \)
Now we calculate:
\[ \cos 150^\circ = -\frac{\sqrt{3}}{2}, \quad \sin 150^\circ = \frac{1}{2} \]
Thus,
\[ \sqrt{z} = -\frac{\sqrt{3}}{2} + i \frac{1}{2} \]

### Step 4: Consider Both Roots
Since we are looking for the square root, we must consider both the positive and negative roots:
\[ \sqrt{z} = \pm \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]

### Final Answer
The square root of \( z = \frac{1}{2} - i \frac{\sqrt{3}}{2} \) is:
\[ \sqrt{z} = \pm \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]