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A variation of the Root of Unity problem.

I want to find all possible answers to this:

$$z^n = i$$

Where $$i^2 = -1$$

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    $\begingroup$ See also math.stackexchange.com/questions/3315/…, which asks for a special case, but whose answers give you everything you need and more. $\endgroup$ – Jonas Meyer Sep 30 '10 at 4:43
  • $\begingroup$ By the way, your title doesn't quite match your question. Any purely imaginary root of unity must be an n-th root where n is a multiple of 4 (for example, (-i)^4 = 1). $\endgroup$ – Weltschmerz Sep 30 '10 at 4:44
  • $\begingroup$ @weltschmerz I don't know what to call it then? $\endgroup$ – Talvi Watia Sep 30 '10 at 4:47
  • $\begingroup$ "All n-th roots of i" would do. $\endgroup$ – Weltschmerz Sep 30 '10 at 4:48
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    $\begingroup$ Or, if you want to emphasize that they're not real, you could refer to "complex roots of i". $\endgroup$ – Michael Lugo Sep 30 '10 at 5:03
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If the polar form of $z$ is $$z=r(\cos\theta + i\sin\theta),$$ there are $n$ distinct solutions to the equation $w^n = z$: $$w=\sqrt[n]{r}(\cos\frac{\theta +2\pi k}{n}+ i \sin\frac{\theta +2\pi k}{n}),$$ where $k=0,1,...,n-1$. In your case, $z=i$, whose polar form is given by $r=1$, $\theta = \pi /2$.

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Generally, the answers would be of the form

$$\sqrt[n]{i}\omega_n^j$$

where $\omega_n=\exp\left(\frac{2i\pi}{n}\right)$ is a root of unity, and $j=0\dots n-1$.

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Also, observe that if $z^n=i$ then $z^{4n}=1$. Thus, the complex numbers you're looking for are particular $4n$-th roots of $1$.

If you know that the $m$-th roots of 1 (any $m$) can be written as powers of a single well-chosen one (a primitive root), it shouldn't be too hardto find exactly which $4n$-th roots have the desired property.

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