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?
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 $\frac{1}{2}+i\frac{\sqrt{3}}{2}$.