If an ant jumps from one vertex to another of a regular decagon, based on a condition, which vertex will he be on after his workout?

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QUESTION: Brilli the ant stands on vertex $1$ of a regular decagon..

He starts by hopping $1$ space at a time (from $1$ to $2$, then from $2$ to $3$, and so on). He performs $10$ hops in this way.

He then hops $2$ spaces at a time (from $1$ to $3$, then from $3$ to $5$, and so on). He performs $10$ hops in this way.

He continues to increase the hop distance every $10$ hops: hopping $3$ spaces $10$ times, then hopping $4$ spaces $10$ times, and so on.

After Brilli has hopped $10$ spaces $10$ times, he ends his workout.

When Brilli has completed his workout, which vertex will he be standing on?

MY ANSWER: This is an easy problem. I have figured out a simple number theoretic approach to this one.. It's trivial to see that the first round brings the ant back to vertex $1$. Then, he basically jumps $2 \times 10$ spaces. That results in vertex $20$, which is nothing but vertex $1$ $(\because 10 \mid 20 \text{ and every vertex is } \equiv 0 \pmod{10})$. Then for the third round we get $30$ spaces, which brings him back to vertex $1$ again. Continuing in this fashion, we get that eventually after the workout, Brilli will be back to where he started from, i.e. vertex $1$.

Now, my question is a bit different. Lets look at this same problem from the complex geometry side.. We know that the vertices are nothing but the $10^{\text{th}}$ roots of unity. Now, at an official solution, it was written that --

Peforming the $10$ "hops" around the decagon is just like raising a $10^\text{th}$ root of unity to the $10^\text{th}$ power.

I did not understand this statement... Why is that? The ant is just hopping around from one root to another (that is to say, the vertices)... How does "raising the $10^{\text{th}}$ root (which is $1$) to the $10^{\text{th}}$ power help?"

Can anyone please clarify this geometry a bit clearly. Thank you so much..

Source: Brilliant.

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$$\underbrace{ \left( e^{(2 \pi i)/10} \right)^{10}}_{=1} \underbrace{ \left( e^{2 (2 \pi i)/10} \right)^{10}}_{=1} \cdots \underbrace{ \left( e^{ 10 (2 \pi i)/10} \right)^{10}}_{=1} = 1$$

where the power is because you take $10$ steps of each size.