Let $ z = \cos\frac{\pi}{24} + \sin\frac{\pi}{24}$.
Compute $ a,b $ such that $ z^8 = a + bi $.
Applying De Moivre's Theorem I get this:
$ z^8 = (\cos\frac{\pi}{24} + i \sin\frac{\pi}{24})^8 $
$ z = \frac12 +\frac {\sqrt{3}} 2i $
I am not sure of the answer, I think it can be done in other ways as well, but I can't think of anything.
How would you do tackle it?
Hint: $$ e^{i\theta} = \cos\theta+i\sin\theta $$