How to rewrite $i^3$ I learned that: $\sqrt[3]i=(e^{\frac{\pi}{2}i+2k\pi i})^\frac{1}{3}=e^{\frac{\pi i}{6}+\frac{2}{3}k\pi i}$ for k={0, 1, 2} 
Now how about this case:
$i^3=(e^{\frac{\pi}{2}i+2k\pi i})^3=e^{\frac{3\pi i}{2}+6k\pi i}$ for k={???}
Why would it be $+6k\pi i$? Why not $+2k\pi i$? It seems that for example $e^{\frac{3\pi i}{2}+2\pi i}$ should also be a solution?
And, for what $k$ is it valid?
 A: Well, $i^3$ is just $-i$, no multiple values here.
Why? Because: $i^3 = i^2 \cdot i = -1 \cdot i$ 
A: The $+6k i \pi$ follows from exponent laws.
Due to the periodicity of $\exp$,
$$ e^{3i\pi/2 + 6ki\pi} = e^{3i\pi/2 + 2i\pi} = e^{3i\pi/2} = -i
$$
so all the solutions are the same.
A: All $k\in\mathbb{Z}$ "count", but we only list a subset that between them obtain all possible values for the final result. In the first calculation, $k\mapsto k+3$ does nothing, so we may as well take $k\in\{ 0,\,1,\,2\}$. In the second, $k\mapsto k+1$ does nothing, so we may as well take $k=0$.
A: Actually, for any case, the correct and full set of values of $k$ should be
$$k\in\mathbb Z$$
However, if the power $n\in\mathbb Z$, then any single $k\in\mathbb Z$ will do, but if the power is $n\in\mathbb Q$, where $n$ is in the form of $\frac ab$ and $\text{gcf}(a,b)=1$, then any consecutive sequence of $b$ values from $k\in\mathbb Z$ will do.
In the case that $n\notin\mathbb Q$, $k\in\mathbb Z$ must be used to have all solutions, there is no simplifications for this one.
A: When we say  $z = re^{ti + 2k\pi i}$ the $2k\pi i$ is just the period and is always there much as "plus a constant"
So really  $\sqrt[3]i=(e^{\frac{\pi}{2}i+2k\pi i})^\frac{1}{3}=e^{\frac{\pi i}{6}+\frac{2}{3}k\pi i + 2k'\pi i}$  But $\frac{2}{3}k\pi i$ is now significant and means more that simply "give or take a few periods".  But $2k'\pi i$ is redundant.  $e^{\frac{\pi i}{6}+\frac{2}{3}k\pi i + 2k'\pi i}= e^{\frac{\pi i}{6}+\frac{2k+6k'}{3}\pi i }$ so we simply don't need to write the period term.  So we don't.
To go the other way should really be $i^3=(e^{\frac{\pi}{2}i+2k\pi i})^3=e^{\frac{3\pi i}{2}+6k\pi i+ 2j\pi i}$ to get the "give or take some periods" back in.  But that's now excessive.  If we set $j = -2k$ we just get back to $6k\pi i+ 2j\pi i=2k\pi i$.  
tl;dr
If the tacked on $\gamma \pi i$ is such that $\gamma$ is not an integer multiple of $2$, then $\gamma \pi i$  is significant.  If $\gamma$ is  an integer multiple of $2$, then $\gamma \pi i$  is not significant.
