Which complex number is a 6th root of i? A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2018 - Paper 1

According to De Moivre's Theorem, the n-th roots of a complex number can be found as follows:
[ z_k = r^{1/n} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i \sin\left( \frac{\theta + 2k\pi}{n} \right) \right) ]
For our case: ( r = 1, \theta = \frac{\pi}{2}, n = 6 ). This gives us:
[ z_k = 1^{1/6} \left( \cos\left( \frac{\frac{\pi}{2} + 2k\pi}{6} \right) + i \sin\left( \frac{\frac{\pi}{2} + 2k\pi}{6} \right) \right) ]

Calculating for values of ( k ):

• For ( k = 0 ):
  [ z_0 = \cos\left( \frac{\pi/12} \right) + i \sin\left( \frac{\pi/12} \right) ]
• For ( k = 1 ):
  [ z_1 = \cos\left( \frac{5\pi/12} \right) + i \sin\left( \frac{5\pi/12} \right) ]
• For ( k = 2 ):
  [ z_2 = \cos\left( \frac{3\pi/4} \right) + i \sin\left( \frac{3\pi/4} \right) ]
  Continuing this process will yield all 6 roots.

After evaluating the roots, we can identify that Option A corresponds to one of the computed roots, specifically: ( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i ). Therefore, the answer is A.