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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?

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    $\begingroup$ Did you mean $i\sin(\pi/24)$ instead of $\sin(\pi/24)$? $\endgroup$ – 79037662 Oct 29 at 17:47
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    $\begingroup$ Also, you last line should still contain "$z^8$", not just "$z$". $\endgroup$ – MPW Oct 29 at 17:55
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Hint: $$ e^{i\theta} = \cos\theta+i\sin\theta $$

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  • $\begingroup$ Shouldn't this be a comment? $\endgroup$ – Mason Oct 29 at 20:26

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