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

The 6th roots of $i$ can be found using the formula:

$z_k = r^{1/n} e^{i(\frac{\theta + 2k\pi}{n})}$

For our case:

• $r = 1$, $n = 6$, and $\theta = \frac{\pi}{2}$.
• Thus:

$z_k = 1^{1/6} e^{i(\frac{\frac{\pi}{2} + 2k\pi}{6})} = e^{i(\frac{\pi}{12} + \frac{k\pi}{3})}$

where $k = 0, 1, 2, 3, 4, 5$.

Calculating each root:

• For $k = 0$: $z_0 = e^{i\frac{\pi}{12}}$
• For $k = 1$: $z_1 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ (this is option A)
• Continue calculating for $k = 2, 3, 4, 5$: which will yield the other roots, but we can see that option A is indeed one of the 6th roots.

Thus, the complex number that is a 6th root of $i$ is: A. $-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$.