I for one have little or no idea about your year 12 course.
But the equation \(z^8=-1\) has eight complex roots.
Now the notation easy to learn \(z=r\exp(\theta {\bf i})=r\cos(\theta)+{\bf i} r\sin(\theta)\)
Thus we write \(-1=\exp(\pi{\bf i})=\cos(\pi)+{\bf i}\sin(\pi)\)
Then one of the eighth roots of \(-1\) is \(\zeta=\exp\left(\dfrac{\pi}{8}{\bf i}\right)\) that is called the principal eighth root.
No we spin that root around the unit circle to get seven more eighth roots.
Here is how we do it. Let say \(\rho=\exp\left(\dfrac{2\pi}{8}{\bf i}\right)=\exp\left(\dfrac{\pi}{4}{\bf i}\right)\)
Now having worked through all that notation here are all eight eighth roots of \(-1\):
\(\zeta\cdot\rho^k,~k=0,1,\cdots 7\).