i Cycle

Where is i^{127}

If you have read $latex i$ Cycle, the Sum of Powers, you have already experienced the powers of i and their hidden cycle. This is a very basic problem, and in fact, it is a prerequisite for understanding i Cycle, the Sum of Powers and everything else about complex numbers (okay, it is a bit of an exaggeration to say so, but you get the point; it is very important).

What is the problem here? 

Notice that I could ask the “same” problem just using a different power. For example, I could ask: Where is $i^{39}$?  You see, the problem here is not about finding this or that power of i, it is about finding the pattern governing these powers. 

The fundamental power in this case is two, as it is the power used in the definition of (i): i^2=-1.

Furthermore, if our question was Where is  i^2, we knew the answer is C. From, this starting point, we can see i^3=i^2\times i=-i is the point D, i^4=i^2\times i^2=1 is on the point A, i^5=i^4\times i=i is on the point B, and i^6=i^4\times i^2=-1 is on the point C, which happens to be the point we started our 90-degree anticlockwise journey. 

It doesn’t matter where we start our rotation. But, maybe, point A is the most natural starting point. Starting from A, the powers are: i^0, i^1, i^2, i^3, i^4, i^5,...

They look different from each other, but we know that there are only four fundamentally different numbers, as the above sequence is in fact the following: 

1, i, -1, -i, 1, i, -1, -i, ....

So, to find the answer to our original question, we only need to put aside the fours that go into 127 and see what remains. 

Starting from point A, it has another interesting advantage. It gives meaning to the very definition of (i). Think of (i) as i \times 1. You might say, Okay, so what? Isn’t it obvious that i \times 1=i. Yes, but notice that the left side is the process of multiplication, and the right side is the result of the process. In this case, the process has an important interpretation: it is rotating 90 degrees anticlockwise. That is to say, when we multiply the point A (that happens to be (1,0) by i it goes to the point B, and likewise, when we multiply the point B by i, it goes to the point C. In other words, i\times i\times 1=-1

Voila: i^2=-1  

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