2. \( \sqrt{-1+\sqrt{3} i} \)

To find the square root of the complex number \( z = -1 + \sqrt{3} i \), we can express it in polar form and then apply the square root formula for complex numbers. 1. **Convert to polar form**: - First, we find the modulus \( r \) of \( z \): \[ r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2. \] - Next, we find the argument \( \theta \): \[ \theta = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right). \] Since the real part is negative and the imaginary part is positive, \( z \) is in the second quadrant. The reference angle is \( \frac{\pi}{3} \), so: \[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}. \] - Thus, we can express \( z \) in polar form: \[ z = 2 \left( \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) \right). \] 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: \[ \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right). \] Here, \( r = 2 \) and \( \theta = \frac{2\pi}{3} \). - Calculate \( \sqrt{r} \): \[ \sqrt{r} = \sqrt{2}. \] - Calculate \( \frac{\theta}{2} \): \[ \frac{\theta}{2} = \frac{2\pi/3}{2} = \frac{\pi}{3}. \] - Now, substitute into the formula: \[ \sqrt{-1 + \sqrt{3} i} = \sqrt{2} \left( \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) \right). \] - Evaluate \( \cos\left(\frac{\pi}{3}\right) \) and \( \sin\left(\frac{\pi}{3}\right) \): \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}. \] - Therefore: \[ \sqrt{-1 + \sqrt{3} i} = \sqrt{2} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2}. \] 3. **Final result**: The square root of \( -1 + \sqrt{3} i \) is: \[ \sqrt{-1 + \sqrt{3} i} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2}. \]