# Category: Conceptual Understanding

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## i Cycle

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- Post author By Amir
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One of the many interesting aspects of imaginary numbers is that we can use them to find out “real” facts (facts about real numbers).

Perhaps the most used examples are the derivation of the trigonometric identities for \(sin 2\theta\) and \(cos 2\theta\)*.

This post offers something more exciting and less-known: the radical forms of \(cos \frac{\pi}{12}\) and \(sin \frac{\pi}{12}\).

Consider \(\frac{\frac{1}{2}+\frac{\sqrt{3}}{2}i}{\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}\)

**Algebraically**, simplify it and write it in the form of \(a+bi\).

\[\frac{\sqrt6+\sqrt2}{4}+\frac{\sqrt6-\sqrt2}{4} i\]

We know the answer. But as a wise man once said, mathematics is all about seeing the same thing from different angles (or something like this). So let us, calculate the same thing in some other way.

We write \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and \(\frac {\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) in the form \(r e^{i\alpha}\). And then evaluate the original fraction , that is going to be:

\[\frac{e^{i \frac{\pi}{3}}}{e^{i \frac{\pi}{4}}}\]

That can be simplified as \[e^{i \frac{\pi}{12}}\] That, can be written as \[cos \frac{\pi}{12}+i sin\frac{\pi}{12}\] Compare this result with the one we had in the form of \(a+bi\) above, where \(a\) and \(b\) were in a radical form. They are** the same things, represented in two different forms.** Thus, their real parts should be equal with each other, and also their imaginary part should be equal with each other. So, we have:

\[cos\frac{\pi}{12} = \frac{\sqrt6+\sqrt2}{4}\]

\[sin\frac{\pi}{12} = \frac{\sqrt6-\sqrt2}{4}\]

\[cos\frac{\pi}{12} = \frac{\sqrt6+\sqrt2}{4}\]

\[sin\frac{\pi}{12} = \frac{\sqrt6-\sqrt2}{4}\]

Design a problem to find \(cos \frac{\pi}{8}\) and \(sin \frac{\pi}{8}\) in the radical form.

* We know that (a post will be written for this):

\[(cos \alpha + i sin \alpha)^2=cos (2\alpha) + i sin(2\alpha)\]

Remember what the wise man once said and find \((cos\alpha+i sin\alpha)^{2}\) by actually finding the power. Then we have:

\[(cos \alpha + i sin \alpha)^2=(cos^2\alpha -sin^2\alpha)+ i (2 sin\alpha cos\alpha)\] The right sides of the above two equality are the same represented in two different forms (both are equal to the same left side). So the real parts of the two should be equal, and the imaginary parts of the two should be also equal.

\[cos 2\alpha=cos^2 \alpha -sin^2 \alpha\]\[sin 2\alpha=2 sin \alpha cos\alpha\]

**Design** a problem to find \(cos 3\alpha\) and \(sin 3\alpha\) .

Where is \(i^{127}\)?

If you have read i Cycle, the Sum of Powers, you have already experienced the powers of i, and their hidden cycle. This is a much 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 exaggeration to say so, but you got 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, So if our question was Where is \(i^2\), we knew the answer is C. From, this starting point, we can see \(i^3=i^2*i=-i\) is the point D, \(i^4=i^2*i^2=1\) is on the point A, \(i^5=i^4*i=i\) is on the point B, and \(i^6=i^4*i^2=-1\) is on the point C, that happens to be the point we started our 90 degrees anticlockwise journey.

It doesn’t matter where we start our rotation. But, maybe, the 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 number, as the above sequence is in fact the following sequence:

\[1, i, -1, -i, 1, i, -1, -i, …\]

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

Starting from the point A, has another interesting advantage. It gives meaning the 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, that 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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- Post author By Amir
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