Raising a Complex Number to a Power

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?

• Did you mean $i\sin(\pi/24)$ instead of $\sin(\pi/24)$? – 79037662 Oct 29 at 17:47
• Also, you last line should still contain "$z^8$", not just "$z$". – MPW Oct 29 at 17:55

Hint: $$e^{i\theta} = \cos\theta+i\sin\theta$$