## Circular View: Rotate and Out
Yet *another* way to create a kind of arithmetics of coordinates
is by looking at the 2D plane geometrically as *circles*.
Imagine a circle centered at the origin, with radius of 1.
That's the **standard unit circle**.
Note that every point on that unit circle forms an angle
with the **rt** ruler unit (from above) placed at the origin.
That angle measurement is called an
**angle in standard position**, but we measure it in radians (*rad*)
instead of degrees.
What is a [radian](https://en.wikipedia.org/wiki/Radian)?
Here's a quick and dirty explanation.
First, I assume you know what a *degree* is. Well, that's
just a way to measure
an angle by first evenly dividing up a *full turn* (turning all
the way around a circle) as 360°, so you know how far to
turn --- how far to rotate an angle --- for each degree.
What if, instead of dividing up a full turn by 360°, we
divide it up by approximately 6.28 (i.e. about \\(2 \\times 3.14\\) )
--- or to be more precise, divide a full turn by \\(2\\pi\\).
Well, we'd get angles measured in **radians** instead of degrees.
Technically radians isn't a unit the way degrees is, but whatever,
physical intuition wins here so we're going to abbreviate it as
a unit as measured in **rad**s.
(*Fun note: mathematicians like measuring angles with radians
because for some reasons, \\(\\pi\\) and \\(2\\pi\\) comes up
a lot of times in nature whenever circles and angles come up.
So much so that there's
[some people](http://tauday.com/tau-manifesto "The Tau Manifesto")
who thinks
students should be taught to think in terms of \\(2\\pi\\)
instead of just \\(\\pi\\). How? By calling \\(2\\pi\\)
a \\(\\tau\\), easily translated into English as a "turn"...
but anyway*).
With angles in mind, now every point on the plane can be
found as if by shooting a laser from the origin.
Just say how big of an angle (in standard position) to rotate
the **rt** arrow placed starting from the origin,
then say how far to shoot out the laser in the direction
of the arrow.
(I freely admit a more accurate analogy might be extending a ladder
out a certain distance after rotating it a certain angle, but laser
seems more fun.)
Anyway, let's make this more like "arithmetic"
like we did with the rectangular view before.
We'll define a function **rot** to mean the rotation
described above of the **rt** arrow.
Then every point can be addressed by
starting with the **rt** arrow from the origin,
and rotating it counter-clockwise by a certain angle
(in standard position, measured in radians),
then shoot out by a certain distance eequal to
some multiple of the rotated **rt** arrow.
This means:
$$\begin{align}
(x,y) &= \text{ distance } r \text{ from origin thru the } rt \text{ unit rotated by angle } \theta \\\\
&= \text{ distance } r \text{ from origin thru } \mathrm{rot}(\theta)
\end{align}
$$
Of course, \\(r\\) and \\(\\theta\\) can be calculated from the \\(x, y\\) coordinates
exactly using basic high school trigonometry:
$$\begin{align}
r &= \sqrt{x^2 + y^2} \\\\
\theta &= \arccos{ \(\frac{x}{r}\) }
\end{align}
$$
By the way "arccos" is just a fancy way to mean the inverse cos,
popularly learned in school as \\(\\cos^{-1}\\). I use "arccos"
because it's easier to write on computer (and it has a
[fancy meaning](https://math.stackexchange.com/questions/33175/etymology-of-arccos-arcsin-arctan)).
We don't have a symbol to mean "distance from origin thru"
so we'll again abuse notation, and say it's the *dot operator*.
So the dot operator (kind of like multiplication) we'll use to
combine a number and the result of the **rot** function, just to abbreviate
the above, like so:
$$\begin{align}
(x,y) &= \text{ distance } r \text{ from origin thru } \mathrm{rot}(\theta) \\\\
&= r \cdot \mathrm{rot}(\theta)
\end{align}
$$
Why the "dot" multiplication operator? Because "the
distance from the origin thru the point at the end
of the arrow rotated by the **rot** operator" is just like
multiplying the **rt** arrow after rotation --- but a lot
more abbreviated with similar meaning!
## Multiplying rotations
Let's stick with points on the unit circle.
Notice that a point on the unit circle
at angle \\(\theta\\) when rotated
by an angle \\(\phi\\) ends up being the point at angle
\\((\theta + \phi)\\). Of course, the angle \\(\phi\\)
itself also defines a point on the unit circle.
So this means:
$$\mathrm{rot}(\theta + \phi) = \mathrm{rot}(\theta) \text{ rotate by } \mathrm{rot}(\phi)$$
We don't have a symbol to mean "rotate by"
so we'll abuse notation and say it's the dot operator
combining two **rot** function values. Thus:
$$\mathrm{rot}(\theta + \phi) = \mathrm{rot}(\theta) \cdot \mathrm{rot}(\phi)$$
Very interestingly, if we take the abbreviated notation at face value,
this means **rot** respects the **Sum of Exponents** (SOE) rule ---
remember that from high school math, e.g., \\(2^{3 + 4} = 2^3 \cdot 2^4\\),
well imagine \\(\\mathrm{rot}\\) is kind of like "2 to the power of",
then notice it also respects the SOE rule.
Why do we use the "dot" multiplication symbol to mean
a rotation "rotate by" another rotation? No reason, other
than the fact that it means we get to respect the SOE rule.
Whenever we can respect existing arithmetic rules, we do so
to make the symbols easier to understand, and to leverage
existing intuitions.
## Rectangular View = Circular View
So far we've defined rectangular and circular views
by way of addressing points on the 2D plane using
the grid view coordinates.
But this must mean we can also equate rectangular
with circular views (using grid coordinates as a bridge).
$$(x,y) = x \cdot\mathrm{rt} + y \cdot\mathrm{up}$$
and,
$$\begin{align}
(x,y) &= r \cdot \mathrm{rot}(\theta) \\\\
\end{align}
$$
with the usual calculations to get *r* and \\(\\theta\\):
$$\begin{align}
r &= \sqrt{x^2 + y^2} \\\\
\theta &= \arccos{ \(\frac{x}{r}\) }
\end{align}
$$
meaning:
$$
x \cdot\mathrm{rt} + y \cdot\mathrm{up} = r \cdot \mathrm{rot}(\theta)
$$
and both the left-side and right-side of the equality referring to
the same point at coordinate *(x,y)*.
## Rectangular View of Unit Circle
If we consider **only points on the unit circle again** in
considering the link between the rectangular and circular
views from above, then because the unit circle has radius
of 1, we get by substitution and simplification:
$$\begin{align}
r = 1&= x^2 + y^2 \\\\
\theta &= \arccos{ \(\frac{x}{r}\) } = \arccos(x)
\end{align}
$$
Therefore we get the following two equalities:
$$\begin{align}
\cos(\theta) &= x \\\\
1 - \cos^2(\theta) &= y^2
\end{align}
$$
Now if we consult some
[trig identities](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Pythagorean_identity)
or from working out geometrically from a diagram, we'd derive this:
$$\begin{align}
\sin(\theta) &= y
\end{align}
$$
So by substitution into *x* and *y* we **finally get**:
$$\begin{align}
\cos(\theta) \cdot\mathrm{rt} + \sin(\theta) \cdot\mathrm{up} &= \mathrm{rot}(\theta)
\end{align}
$$
### Rather silly but seriously *deep* questions
Assuming the age old **zero times anything is zero** rule, now we can have
some fun by asking silly questions like:
What is (1,0)?
$$\begin{align}
1 \cdot\mathrm{rt}
&= 1 \cdot\mathrm{rt} + 0 \cdot\mathrm{up}\\\\
&= (1,0) \\\\
&= \mathrm{rot}(0)
\end{align}
$$
What is (0,1)?
$$\begin{align}
1 \cdot\mathrm{up}
&= 0 \cdot\mathrm{rt} + 1 \cdot\mathrm{up}\\\\
&= (0,1) \\\\
&= \mathrm{rot}(\frac{\pi}{2}) \text{ --- that's a rotation by 90}^{\circ}
\end{align}
$$
What is (-1,0)?
$$\begin{align}
-1 \cdot\mathrm{rt}
&= -1 \cdot\mathrm{rt} + 0 \cdot\mathrm{up}\\\\
&= (-1,0) \\\\
&= \mathrm{rot}(\frac{\pi}{2}+\frac{\pi}{2}) \text{ --- that's a rotation by 180}^{\circ}\\\\
&= \mathrm{rot}(\frac{\pi}{2}) \cdot \mathrm{rot}(\frac{\pi}{2})
\end{align}
$$
What is (0,-1)?
$$\begin{align}
-1 \cdot\mathrm{up}
&= 0 \cdot\mathrm{rt} + -1 \cdot\mathrm{up}\\\\
&= (0,-1) \\\\
&= \mathrm{rot}(\frac{3\pi}{2}) \text{ --- that's a rotation by 270}^{\circ}\\\\
&= \mathrm{rot}(\frac{\pi}{2}) \cdot \mathrm{rot}(\frac{\pi}{2}) \cdot \mathrm{rot}(\frac{\pi}{2})
\end{align}
$$
These four sets of equalities are easy to prove. Just draw it out
on paper yourself if you're not convinced that, rotating the *rt*
unit arrow by 270° in standard position (i.e. 3 quarters of a
full turn counterclockwise, i.e. rotating counterclockwise by 90°
three times) gives you an arrow
pointing at the coordinate *(0,-1)* for example.
## Multiplying right with up: surprising properties of Rectangular View of Rotations
We haven't defined a multiplication symbol for mixing
the two units **rt** and **up** together yet,
because what could it possibly mean to multiple a right
with an up direction?
But from the above "silly questions", we see these four equalities:
$$\begin{align}
1\cdot\mathrm{rt} &= \mathrm{rot}(0) \\\\
1\cdot\mathrm{up} &= \mathrm{rot}(\frac{\pi}{2}) \\\\
-1\cdot\mathrm{rt} &= \mathrm{rot}(\frac{\pi}{2}) \cdot \mathrm{rot}(\frac{\pi}{2}) \\\\
-1\cdot\mathrm{up} &= \mathrm{rot}(\frac{\pi}{2}) \cdot \mathrm{rot}(\frac{\pi}{2}) \cdot \mathrm{rot}(\frac{\pi}{2})
\end{align}
$$
The second equation above, thru substitution with the third and
fourth equations above, implies these two equations:
$$\begin{align}
-1\cdot\mathrm{rt} &= (1\cdot\mathrm{up}) \cdot (1\cdot\mathrm{up}) \\\\
-1\cdot\mathrm{up} &= (1\cdot\mathrm{up}) \cdot (1\cdot\mathrm{up}) \cdot (1\cdot\mathrm{up}) \\\\
\end{align}
$$
Then substituting the above equations together, we get:
$$\begin{align}
-1\cdot\mathrm{up} &= (-1\cdot\mathrm{rt}) \cdot (1\cdot\mathrm{up}) \\\\
\end{align}
$$
Notice the above substitutions rids the right-sides of any rot functions!
The only equation left to rid the rot function from by substitution requires a trick:
$$\begin{align}
1\cdot\mathrm{rt}
&= \mathrm{rot}(0) \\\\
&= \mathrm{rot}(0 + 0) \\\\
&= \mathrm{rot}(0) \cdot \mathrm{rot}(0) \text{ --- haha! By SOE rule from way, way above!} \\\\
&= (1\cdot\mathrm{rt}) \cdot (1\cdot\mathrm{rt})
\end{align}
$$
So two rights don't make a wrong, but two ups make a negative right. *har har*
## Where's the real number line in the 2D plane?
Noticing that:
$$
1\cdot\mathrm{rt} = (1\cdot\mathrm{rt}) \cdot (1\cdot\mathrm{rt})
$$
And of course by the definition of "+" we had started with
for adding the units **rt** and **up** together to begin with.
We can see that the **rt** unit along the x-axis is just the
real number line with its normal elementary school arithmetic
properties. It's as if the real number line is *embedded*
within the rectangular view of the 2D plane as its x-axis.
And so we can drop the use of the **rt** unit whenever we want
if we agree to identify real numbers with points on the
x-axis in the rectangular view of the 2D plane.
Thus:
$$\begin{align}
x + y \cdot\mathrm{up} &= r \cdot \mathrm{rot}(\theta) \\\\
\text{where } r &= \sqrt{x^2 + y^2} \\\\
\theta &= \arccos{ (\frac{x}{r}) }
\end{align}
$$
## Those "up" units take imagination
Two **up**s multiply to make a **negative right** makes
multiplication very strange.
After all, what could possibly be multiplied with itself
to result in -1? Well, square root of -1 works, but we
all know square root of -1 is undefined from grade-school
math.
So the **up** unit gives us an *imaginative*, different kind of math.
Let's abbreviate the **up** unit with the letter **i** then.
Assuming the age old **one times anything is itself** rule, we thus have:
\begin{align}
-1
&= -1\cdot\mathrm{rt} \\\\
&= (1\cdot\mathrm{up}) \cdot (1\cdot\mathrm{up}) \\\\
&= (1\cdot i) \cdot (1\cdot i) \\\\
&= i \cdot i
\end{align}
## The "Rot" function, because "rot" is not invertible
You've probably learned in high school what it means for a
function to not be invertible.
Similarly here, not invertible means if we knew \\(\theta\\),
we can know definitely what **x** and **y** *really* is in
\\((x,y) = \mathrm{rot}(\theta)\\).
But the reverse is not true, i.e. if we knew what **x** and **y** is of a
point on the unit circle, there's many, many, *many* angles of rotations
\\(\theta\\) that satisfies \\((x,y) = \mathrm{rot}(\theta)\\) --- just
keep spinning around the circle another full turn!
Suppose we define a *restricted* **Rot** function (note the capital "R"
to emphasize it's restricted) that is just like the old **rot** function,
but only ever deals with the angles between 0 and \\(2\pi\\) (i.e. 360°).
Then obviously **Rot** is invertible because we
can't keep spinning around the circle anymore!
(As in there's only ever one angle possible to
rotate to to shoot at any given point.)
## Attempt 1 at making the "rot" function work with "up" units (i.e. "i")
With the latest abbreviations in place, we have:
$$
\begin{align}
x + y \cdot i &= r \cdot \mathrm{rot}(\theta) \\\\
r &= \sqrt{x^2 + y^2} \\\\
\theta &= \arccos{ (\frac{x}{r}) }
\end{align}
$$
The "rot" function seems unnatural in our increasingly symbolic and
algebraic math since it encodes our *intuitive
physical* understanding of rotating the **rt** unit starting at the
origin (i.e. rotating counterclockwise the rightward pointing arrow
of length 1).
A bigger problem is that the **rot** function only makes sense when we
use it with regular real numbers. After all, it's applying a rotation
of a certain angular distance measured in radians.
So this makes sense:
$$\mathrm{rot}(1)$$
and this currently doesn't really make sense, although it's okay if we squint
hard enough and ignore the *rt* unit given our understanding of embedding the
real number line into the 2D plane:
$$\mathrm{rot}(1\cdot\mathrm{rt})$$
but certainly this is completely meaningless (right?), because we defined *rot* to *mean*
rotating counterclockwise the *rightward* pointing arrow of length 1:
$$\mathrm{rot}(1\cdot\mathrm{up})$$
Can we can do better than that though?
Suppose we agree that **rot** only makes
sense in terms of a real number of radian measured angles, then if we
agree to strip the **rt** and **up** units away when using the **rot**
function, suddenly the **rot** function will work with **rt** and
**up** (nee, **i**) units too!
Thus can we explicitly just *say* the following equality is okay?
$$
\mathrm{rot}(\theta) = \mathrm{rot}(\theta\cdot\mathrm{rt}) = \mathrm{rot}(\theta\cdot\mathrm{up}) = \mathrm{rot}(\theta\cdot\mathrm{i}) \text{ ???}
$$
But if we allow the above equality as "okay", we'd have to
also allow the following equality using the previously defined
invertible **Rot** function:
$$
\mathrm{Rot}(\theta) = \mathrm{Rot}(\theta\cdot\mathrm{rt}) = \mathrm{Rot}(\theta\cdot\mathrm{up}) = \mathrm{Rot}(\theta\cdot\mathrm{i}) \text{ ???}
$$
and that clearly makes Rot not invertible again
despite our requiring Rot to be invertible!
See, we can't abuse meaning when we abuse notation...
So trying to extend the **rot** function with the above equalities is
a **terrible strategy. We need a different, better strategy**, although this first
attempt reminds us a valuable fact about of definition of the **rot** function:
it means the counterclockwise rotation of a rightward pointing arrow of length 1.
This first attempt also teaches us that we can abuse notation,
as we have done many times before, but we cannot abuse *meaning*!
## Attempt 2 at generalizing the "rot" function: make it work with any point on the 2D plane
We wish to make the **rot** function work not just with regular real
numbers (i.e. the \\(x\cdot\mathrm{rt}\\) numbers), but also
generalized to work with any point on the 2D plane.
Currently the following makes perfect sense for any **x** that is a regular real number:
$$\mathrm{rot}(x) = \text{ the rt unit rotated by angle of x radians}$$
Currently, using **rot** on numbers involving the **rt** unit makes little sense, e.g.:
$$\mathrm{rot}(x\cdot\mathrm{rt}) = \text{ ???}$$
but if we recall how the real number line is embedded into the 2D plane,
i.e. \\(x = x\cdot\mathrm{rt}\\), then it's easy to *extend* the
**rot** function like this:
$$\mathrm{rot}(x\cdot\mathrm{rt}) = \mathrm{rot}(x)$$
Another reason why we can extend the **rot** function like that is
that the **rot** function only makes sense in terms of a real number
of radian measured angles, so if we agree to strip the **rt** unit
away when using the **rot** function, suddenly the **rot** function
will work with **rt** units too! We're just making it formal with the
above definition.
But still this is meaningless:
$$\mathrm{rot}(1\cdot\mathrm{up}) = \text{ ???}$$
So we certainly cannot make sense of using **rot** on any
point on the 2D plane, i.e. at:
$$x + y\cdot i$$
Our first attempt at giving the above meaning was sunk because it
made what should've been an invertible function no longer invertible.
Our second attempt uses a different strategy: define a **different
function** (named **Rote** --- the "e" means "expanded"),
and hope the original **Rot** function is just
a special case of **Rote**.
### "Rot Expanded": defining the "Rote" function
If **Rote** must be different than **Rot**, then obviously this is
true of real numbers *x*:
$$
\mathrm{Rote}(x) \neq \mathrm{Rot}(x)
$$
That rules out what we did with attempt 1.
Instead, let's stipulate and define **Rote** so that this is true:
$$
\mathrm{Rote}(i\cdot x) = \mathrm{Rot}(x)
$$
We *purposely* leave out what \\(\mathrm{Rote}(x)\\) means without the "i" in
between the parentheses for now (by the way, the stuff in between
the parentheses is called that function's *argument*).
You may ask why we make such an initial definition, but it's for the
same reason cited in attempt 1 before, that **Rot** makes sense only
in terms of real number of radian measured angles. So we *make* sense of
**Rote**, literally create meaning for it, by just plugging its argument
into the **Rot** function after stripping off the **up** (i.e. the **i**)
unit --- and hope this makes **Rote** play nice with the rest of the
system we've been setting up... (spoiler alert: it'll work fine).
### "Rote" properties
Since \\(\mathrm{Rote}(i\cdot x) = \mathrm{Rot}(x)\\), and recall that **Rot** respects
**Sum of Exponents** Rule (SOE), thus \\(\mathrm{Rote}(i\cdot x)\\) respects SOE so
long as the unit "i" is in the argument.
Let's just impose that **Rote** respects SOE more generally, for any
argument (again to leverage our existing intuitions about the SOE rule).
Then we can see by SOE rule that:
$$\begin{align}
\mathrm{Rote}(a + i\cdot x) &= \mathrm{Rote}(a) \cdot \mathrm{Rote}(i\cdot x) \\\\
&= \mathrm{Rote}(a) \cdot \mathrm{Rot}(x)
\end{align}
$$
Recall we wish that **Rote** be invertible. That means we must be able to define
an inverse function **Lote** that respects:
$$\mathrm{Lote}(\mathrm{Rote}(x)) = x$$
By substitution into the previous equality, we see that:
$$\begin{align}
\mathrm{Lote}(\mathrm{Rote}(a)) + \mathrm{Lote}(\mathrm{Rote}(i\cdot x)) &= a + i\cdot x \\\\
&= \mathrm{Lote}(\mathrm{Rote}(a+i\cdot x)) \\\\
&= \mathrm{Lote}(\mathrm{Rote}(a)\cdot \mathrm{Rote}(i\cdot x)) \\\\
&= \mathrm{Lote}(\mathrm{Rote}(a)\cdot \mathrm{Rot}(x))
\end{align}
$$
Thus **Lote** respects the **Sum of Logs** (SOL) Rule! Don't see it?
Let's pull out the equality from above that demonstrates it:
$$
\mathrm{Lote}(\mathrm{Rote}(a)) + \mathrm{Lote}(\mathrm{Rote}(i\cdot x)) = \mathrm{Lote}(\mathrm{Rote}(a)\cdot \mathrm{Rote}(i\cdot x))
$$
Remember from high school math that, e.g., \\(\log(3) + \log(4) = \log(3 \times 4)\\),
well imagine \\(\\mathrm{Lote}\\) is kind of like \\(\\log\\),
then notice it also respects the SOL rule.
Even though our definition didn't include information about what
**Rote(a)** means without the "i" in the argument, but from the above
we can at least see that \\(\mathrm{Lote}(\mathrm{Rote}(a)) = a\\)
for any real number \\(a\\).
We're getting close to knowing what **Rote** must be.
### So who is "Rote"?
So it's a mystery novel ending: what pair of functions do we know of
that has the above qualities of:
0. works with real numbers,
1. **Rote** respects SOE,
2. **Lote** respects SOL, and
3. **Rote** and **Lote** are inverses of each other???
Obviously **exponential and logorithmic functions**! This means that for *real numbers a*:
$$
\mathrm{Lote}(a) = \log(a)
$$
and
$$
\mathrm{Rote}(a) = e^a
$$
So now **Rote** works with both real numbers and numbers of the form
\\(i\cdot x\\), and even \\(a + i\cdot x\\) numbers!!!
$$\begin{align}
\mathrm{Rote}(a) &= e^a\text{ --- we just discovered, for real numbers }a \\\\
\mathrm{Rote}(i\cdot x) &= \mathrm{Rot}(x)\text{ --- by definition above, for real numbers }x \\\\
\mathrm{Rote}(a+i\cdot x) &= \mathrm{Rote}(a) \cdot \mathrm{Rote}(i\cdot x) = e^a \cdot \mathrm{Rot}(x)\text{ --- by SOE and substitution}
\end{align}
$$
Notice that grade-school exponential function work ONLY for real
numbers...
We had **Rote** defined by stipulating initially its meaning when
applied to numbers of the form \\(i\cdot x\\) but not for regular real
numbers *x*, but then we discovered that \\(\mathrm{Rote}(x)\\) is \\(e^x\\).
Now we can synergistically generalize the grade-school expoential
function to work with numbers of the form \\(i\cdot x\\) as well!
Since grade-school math teaches us the meaning of \\(e^x\\) but not
the meaning for \\(e^{i\cdot x}\\), we'll just do the reverse
stipulation definition that is natural since we identified **Rote** with
**e**: Namely,
$$
e^{i\cdot x} = \mathrm{Rote}(i\cdot x) = \mathrm{Rot}(x)
$$
So now the **e** function works with both real numbers and numbers of
the form \\(i\cdot \theta\\), and even numbers of the form
\\(a + i\cdot \theta\\) via application of the SOE Rule!!!
$$\begin{align}
e^{a}&\text{ --- just grade-school math function}\\\\
e^{i\cdot \theta} &= \mathrm{Rote}(i\cdot \theta) = \mathrm{Rot}(\theta)\text{ --- by definition above}\\\\
e^{a+i\cdot \theta} &= e^a \cdot e^{i\cdot \theta} = e^a \cdot \mathrm{Rot}(\theta)\text{ --- by SOE and substitution}\\\\
\end{align}
$$
But we still have the pesky **Rot** function there!
## Observe: Euler's Formula
Recall that:
$$
\mathrm{rot}(\theta) = \cos(\theta) \cdot\mathrm{rt} + \sin(\theta) \cdot\mathrm{up}
$$
Or using our newest abbreviations:
$$
\mathrm{rot}(\theta) = \cos(\theta) + \sin(\theta) \cdot\mathrm{i}
$$
So this means:
$$\begin{align}
e^{a + i\cdot \theta}
&= e^a \cdot e^{i\cdot \theta}\\\\
&= e^a \cdot \mathrm{Rot}(\theta)\\\\
&= e^a \cdot (\cos(\theta) + \sin(\theta) \cdot\mathrm{i})
\end{align}
$$
Now we can ask a silly question like, what if *a = 0*? Well, then we'd get
$$\begin{align}
e^{i\cdot \theta}
&= e^{0 + i\cdot \theta}\\\\
&= e^0 \cdot e^{i\cdot \theta}\\\\
&= \cos(\theta) + \sin(\theta) \cdot\mathrm{i}\text{ --- since }e^0 = 1
\end{align}
$$
From the above equality, we thus observe **Euler's Formula**:
$$
e^{i\cdot \theta} = \cos(\theta) + \sin(\theta) \cdot\mathrm{i}
$$
I say observe, not discover, because we got here by *stipulating* the
*e* function to work with numbers of the form \\(i\cdot x\\) via the definition:
$$
e^{i\cdot x} = \mathrm{Rote}(i\cdot x) = \mathrm{Rot}(x)
$$
and **Rot** was defined quite physically to work a certain way
resulting in
$$
\mathrm{rot}(\theta) = \cos(\theta) + \sin(\theta) \cdot\mathrm{i}
$$
So all that work we did initially to equate the rectangular view and the
circular view of the 2D plane comes back to award us
an easy way to understand what \\(e^{i\cdot \theta}\\) is doing!
### An easy way to understand what \\(e^{i\cdot \theta}\\) is doing
\\(e^{i\cdot \theta}\\) applies a rotation of \\(\theta\\) to the
**rt** unit. And multiple rotations by angles of \\(\theta\\),
\\(\phi\\), etc. can thus be written as the product of
\\(e^{i\cdot \theta} e^{i\cdot \phi}\\) etc.
Whereas multiplying by a real number
multiplies its distance from the origin --- all thanks to our
equating the rectangular view and the circular view of the 2D plane
using physical geometric intuitions!
That makes **Euler's Identity** obvious:
$$
e^{i\pi} = -1
$$
Why obvious? Well that's just equal to \\(\mathrm{Rot}(\pi)\\),
i.e. rotating the **rt** unit by an angle of \\(\pi\\) radians,
which rotates the **rt** unit to point leftward to -1.
## Unity of the Grid, Rectangular, and Circular Views of the 2D Plane
Now we can dispense with the **Rot** function entirely, by noticing that
$$
\mathrm{Rot}(x) = e^{i\cdot x}
$$
Thus we can unify the rectangular view with the circular view, in
terms of grid view coordinates as follows. In fact, it's as if
**every point on the 2D plane can be represented using four different
addresses**:
$$\begin{align}
(x,y)&\text{ --- grid view coordinate}\\\\
&= x + y \cdot\mathrm{i} \text{ --- rectangular view "complex numbers"}\\\\
&= r \cdot e^{i \cdot\theta} \text{ --- circular view "complex numbers"}\\\\
&= (r, \theta) \text{ --- circular view "polar coordinate"}\\\\
\text{with these defintions:}&\\\\
r &= \sqrt{x^2 + y^2} \\\\
\theta &= \arccos{ (\frac{x}{r}) }
\end{align}
$$
Or we can unify rectangular and circular views, in terms of circular
view polar coordinates of a point as follows:
For a point at distance *r* from the origin, and with an angle in
standard position of \\(\theta\\):
$$\begin{align}
(r, \theta)&\text{ in polar coordinates}\\\\
&= r \cdot e^{i \cdot\theta} \\\\
&= r \cdot \cos(\theta) + r \cdot \sin(\theta) \cdot\mathrm{i}\\\\
&= (r \cdot \cos(\theta), r \cdot \sin(\theta))\text{ in grid coordinates}
\end{align}
$$
## Afterword
This was a long and windy road to those few formulas at the end,
Euler's Formula, and Euler's Identity, that are
probably usually just given to students to memorize, maybe with a few
short proofs to help understand "why" they're true.
I wanted to understand *why* these algebraic formulas are true in terms
of physical and geometric intuitions: arrows and rotations.
I was pleasantly surprised that all that was needed to understand them
was nothing more than high school math. It's a bonus that it pulled in just enough
polar coordinates ideas, which are usually taught in high school but
never given any reason for existing, as in students are taught about them,
how they work a little bit, but never for *why* we should care about
polar coordinates --- turns out they were crucial for us today.
Then we could build up the whole thing without ever needing to have learned
vector math or calculus, and build up an intuition of complex numbers and
even complex exponentials.
That's in opposition to, e.g.,
[Euler's Formula in Wikipedia](https://en.wikipedia.org/wiki/Euler%27s_formula).
After giving you the formula, the first section (History) immediately assumes
you know about imaginary numbers, complex numbers, and calculus (and I mean
integrals, which isn't usually covered, or covered well, in high school math);
the next sections also assume knowledge of complex numbers, and even the
complex exponential which is explained as a
[unit complex number](https://en.wikipedia.org/wiki/Unit_complex_number)
which actually links to [Circle group](https://en.wikipedia.org/wiki/Circle_group),
a Lie group in Group Theory...or something.
Of course, this physical, geometric intuition built up here won't
work for hard core pure math understanding (and yes, I know there were parts
where the derivations weren't just hand wavey, but needed some serious squinting
to ignore certain technical issues), but for working practically
with things like 3D graphics, or even just some applications that require
complex exponentials (they show up a lot), I think a good intuition is better
than mathematical rigour.
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