What is the square root of 1/2 − i√3/2?

• A: ±(√3/2 + i/2)
• B: ±(√3/2 - i/2)
• C: ±(1/2 − i√3/2)
• D: ±(1/2 + i√3/2)

The expression you provided involves finding the square root of a complex number. To solve for the square root of \( \frac{1}{2} - i\frac{\sqrt{3}}{2} \), we can express it in polar form and then apply the square root formula for complex numbers.

Step 1: Convert to Polar Form

The complex number can be represented in polar form as:

• Magnitude: \( r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = 1 \)
• Argument: \( \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = -\frac{\pi}{3} \)

Step 2: Apply the Square Root Formula

The square root of a complex number in polar form \( r(\cos \theta + i \sin \theta) \) is given by:

Square root: \( \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) \)

For our case:

• Magnitude: \( \sqrt{1} = 1 \)
• Argument: \( \frac{-\frac{\pi}{3}}{2} = -\frac{\pi}{6} \)

Step 3: Calculate the Result

Now, substituting back into the formula:

Result: \( 1 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) \)

This simplifies to:

• Real part: \( \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
• Imaginary part: \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \)

Final Answer

The square roots are:

±\( \left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) \)

Thus, the correct answer is:

B: ±(√3/2 - i/2)