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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5$\begingroup$ 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 MeyerSep 30, 2010 at 4:43
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$\begingroup$ 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). $\endgroup$– WeltschmerzSep 30, 2010 at 4:44
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$\begingroup$ @weltschmerz I don't know what to call it then? $\endgroup$– Talvi WatiaSep 30, 2010 at 4:47
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$\begingroup$ "All n-th roots of i" would do. $\endgroup$– WeltschmerzSep 30, 2010 at 4:48
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2$\begingroup$ Or, if you want to emphasize that they're not real, you could refer to "complex roots of i". $\endgroup$– Michael LugoSep 30, 2010 at 5:03
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3 Answers
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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 $j=0\dots n-1$.
answered Sep 30, 2010 at 4:41
J. M. ain't a mathematicianJ. M. ain't a mathematician
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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.
answered Sep 30, 2010 at 9:45