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That $e^{i\theta}$ traces a circle $\theta \in R$ has been well discussed elsewhere.

However, I was always curious with it's relation to the following property of $i$:

$i^0$ = 1

$i^1$ = i

$i^2$ = -1

$i^3$ = -i

$i^4$ = 1

These properties seem like they should be related, but I can't map them too eachother.

Joseph
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    $i = e^{i \pi/2} \implies i^k = (e^{i \pi/2})^k=e^{i k \pi/2}$ therefore, just take $\theta=k \pi/2$ in $e^{i \theta}$. – Jean Marie Feb 14 '24 at 11:25

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Since $i=e^{i\pi/2}$, you see that taking powers of $i$ is equivalent to making successive rotations of $\pi/2$ around the unit circle. This gives exactly the periodic sequence of $4$ elements you describe.

GSofer
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