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I have been struggling on the problem for some time; I would like a gentle tip on answering the question:

Given the equation $z^4 = 1 + \sqrt{3}i $ where $ z \in \mathbb{C}$ , deduce the exact values of $\cos\left(\frac{\pi}{12}\right)$ and $\sin\left(\frac{\pi}{12}\right)$.

So far I am proceeding in the following manner:

I pose $Z' = z^2$

therefore I have $Z'^2 = 1 + \sqrt{3}i $

I calculate that $Z'$ = $\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$

Right now I have to solve for $Z' = z^2 \iff z^2= \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$

After solving this equation I get that $ z = \frac{1}{2}\sqrt{\sqrt{6}+1} + \frac{1}{2}\sqrt{1-\sqrt{6}}i$

If I am taking that $\cos\left(\frac{\pi}{12}\right) = \Re(z)$ and $\sin\left(\frac{\pi}{12}\right) = \Im(z)$, certainly my solution is wrong.

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  • $\begingroup$ use half-angle formulas knowing that $\cos {\pi/ 6}=0.5$ $\endgroup$ – Vasya Oct 31 '18 at 18:32
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    $\begingroup$ Use $$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$$ $\endgroup$ – Crostul Oct 31 '18 at 18:34
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    $\begingroup$ Please read the question fully. $\endgroup$ – AverageMarcin Oct 31 '18 at 18:36
  • $\begingroup$ You have two errors in your expression for $z$. 1) $\sqrt{6} > 1$, so your expression for $z$ is a real number. 2) If we switch the quantity under the radical to $\sqrt{6}-1$, you have $z^2 = (1+i\sqrt{5})/2\ne (\sqrt{6}+i\sqrt{2})/2$. $\endgroup$ – eyeballfrog Oct 31 '18 at 18:38
  • $\begingroup$ Another issue is that, given the defining equation for $z$, the real and imaginary parts of $z$, and hence of the roots of $z$, can't be cosines or sines since $\vert z\vert\ne1$. What must you do to relate $z$ to a complex number with unit modulus? $\endgroup$ – Will Orrick Oct 31 '18 at 18:41
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You start with $z^4=2(\cos\frac\pi3+i\sin\frac\pi3)$, so your final $z$ should be equal to $\sqrt[4]2(\cos\frac\pi{12}+i\sin\frac\pi{12})$. So, to get your cosine and sine divide by $\sqrt[4]2$ in the end. But your $z$ is wrong (square it); I get $\frac12\sqrt{2\sqrt2+\sqrt6}+\frac i2\sqrt{2\sqrt2-\sqrt6}$

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  • $\begingroup$ This can be written without nested radicals using $\sqrt{2\sqrt{2}\pm\sqrt{6}} = (\sqrt{3}\pm1)/2^{1/4}$. $\endgroup$ – eyeballfrog Oct 31 '18 at 18:52
  • $\begingroup$ Yep, you're right. $\endgroup$ – hartkp Oct 31 '18 at 22:13

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