U can have tons of roots of unity.......1.......can u have roots of i...................nestled fractions............???

How can you find the complex roots of i?

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A variation of the Root of Unity problem. I want to find all possible answers to this: zn=i$$z^n=i$$ Where  i2=−1$$i^2=-1$$ complex-numbers shareciteimprove this question edited Jul 24 '11 at 16:44 Willie Wong 50.8k890186 asked Sep 30 '10 at 4:28 Talvi Watia 12817

5   See also math.stackexchange.com/questions/3315/…, which asks for a special case, but whose answers give you everything you need and more. – Jonas Meyer Sep 30 '10 at 4:43 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). – Weltschmerz Sep 30 '10 at 4:44 @weltschmerz I don't know what to call it then? – Talvi Watia Sep 30 '10 at 4:47 "All n-th roots of i" would do. – Weltschmerz Sep 30 '10 at 4:48 2   Or, if you want to emphasize that they're not real, you could refer to "complex roots of i". – Michael Lugo Sep 30 '10 at 5:03 show 1 more comment

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If the polar form of z$z$ is  z=r(cosθ+isinθ),$$z=r(\cos \theta +i\sin \theta ),$$ there are n$n$ distinct solutions to the equation wn=z$w^n=z$:  w=r√n(cosθ+2πkn+isinθ+2πkn),$$w=\sqrt[n]{r}(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}),$$ where k=0,1,...,n−1$k=0,1,...,n-1$. In your case, z=i$z=i$, whose polar form is given by r=1$r=1$, θ=π/2$\theta =\pi /2$. shareciteimprove this answer edited Dec 21 '11 at 3:45 Jack 25.6k1463155 answered Sep 30 '10 at 4:38 Weltschmerz 3,5971536

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up vote4down vote	Generally, the answers would be of the form i√nωjn$$\sqrt[n]{i}{\omega }_n^j$$ where ωn=exp(2iπn)${\omega }_n=\exp \left(\frac{2i\pi }{n}\right)$ is a root of unity, and j=0…n−1$j=0\dots n-1$. shareciteimprove this answer answered Sep 30 '10 at 4:41 J. M. is not a mathematician 56.2k5131266
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Also, observe that if zn=i$z^n=i$ then z4n=1$z^{4n}=1$. Thus, the complex numbers you're looking for are particular 4n$4n$-th roots of 1$1$. If you know that the m$m$-th roots of 1 (any m$m$) can be written as powers of a single well-chosen one (a primitive root), it shouldn't be too hardto find exactly which 4n$4n$-th roots have the desired property.