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Let's suppose t be in the inverval $(-\pi, \pi]$ and that $n$ is a natural number. What is $(\cos t + i\sin t )^{\frac 1n}$? Using Euler's formula would give us the following:

$(\cos t + i\sin t )^{\frac 1n}=$

$(e^{it})^{\frac 1n}=$

$e^{it\times \frac 1n} = $

$e^{\frac{it}{n}} = $

$\cos \frac 1n + i\sin \frac 1n$.

However, this would be problematic if we're taking on odd root of $-1$. When we're just dealing with the real numbers, any odd root of $-1$ is $-1$. However, if $n$ is odd, then based on formula above, since the argument of $-1$ is $\pi$, $(-1)^{\frac 1n} = \cos \frac{\pi}{n}+ i\sin \frac{\pi}{n}$. So $(-1)^{\frac 13}$ would be equal to $ \frac 12 + i \frac{\sqrt 3}{2}$, even though it would just be $-1$ if we were only dealing with the real numbers. It doesn't make sense that the same operation would yield a different result just because we've extended the number-system we're working with. So, I was wondering if this in fact, is the correct formula for determining the principal root of a unti complex number.

$(\cos t + i\sin t )^{\frac 1n}= \cos \frac 1n + i\sin \frac 1n$ if $-\pi<t<\pi$ or $n$ is even

$=-1$ if $t=\pi$ and $n$ is odd

If this isn't the correct formula, then what is the correct formula?

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This is the reason why one should never talk about "the" root of a number when dealing with complex numbers -1 has 3 cube roots, $\frac{1}{2}+i\frac{\sqrt{3}}{2},-1$ and $\frac{1}{2}-i\frac{\sqrt{3}}{2}$. Your method only gives the root with the smallest argument which is $\frac{1}{2}+i\frac{\sqrt{3}}{2}$.

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I could be wrong, but I don't think there is a universally accepted definition of "principal root" in the context of complex analysis. Since we are working over an algebraically closed field, $\mathbb{C},$ the equation $z^n-z_0=0$ for $z_0$ having unit radius (or any modulus for that matter) has $n$ solutions.

The selection of one of these $n$-th roots as being special somehow is more arbitrary that (e.g.) the selection of a square root of a positive number in $\mathbb{R}.$ since there aren't really notions of positive and negative in the same way.

One could do something by forcing all the angles of the roots between $[0,2\pi)$ and then select the root with the smallest angle

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For the n-th root there are n complex solutions, to visualize it see the picture.enter image description here

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  • $\begingroup$ I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to. $\endgroup$ – Wire Bowl Jan 31 at 19:40

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