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I am trying to solve $z^6 = 1$ where $z\in\mathbb{C}$. So What I have so far is : $$z^6 = 1 \rightarrow r^6\operatorname{cis}(6\theta) = 1\operatorname{cis}(0 + \pi k)$$ $$r = 1,\ \theta = \frac{\pi k}{6}$$ $$k=0: z=\operatorname{cis}(0)=1$$ $$k=1: z=\operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3} + i}{2}$$ $$k=2: z=\operatorname{cis}\left(\frac{\pi}{3}\right)=\frac{1 + \sqrt{3}i}{2}$$ $$k=3: z=\operatorname{cis}\left(\frac{\pi}{2}\right)=i$$ According to my book I have a mistake since non of the roots starts with $\frac{\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots. Can anyone please tell me where my mistake is? Thanks!

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    So what is written in your book?2012-02-06
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    $cis=cos+i\cdot sin$?2012-02-06
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    yes its is. why do i have to write 12 chars?2012-02-06
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    You want $\theta=\frac{2\pi k}6 = \frac{\pi k}3$.2012-02-06
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    What ever "start with ...", I think there is a 2 missing. Shouldn't it be ${\rm cis}(x)=e^{ix}=e^{i(x+2k\pi)}$? This would also drop the solution that "starts with $\sqrt{3}/2$", which is a root of -1 anyhow.2012-02-06
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    @yotamoo While some books write $cis$, not all trained mathematicians have seen that notation, so he was just asking for clarification. Usually, a mathematician would just write $e^{i\theta}$.2012-02-06
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    @yotamoo: I cannot answer your question because I don't agree with the premise (that you have to write twelve characters).2012-02-08

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