## e, i, pi

Here’s a mysterious fact.

$\large \dpi{300} \bg_white e^{i \pi} = -1$

What’s up with that?

The answer is forthcoming, but first I’ll remind you what all the letters mean.

e e e e e e

e = 1 + 1/1 + 1/12 + 1/123 + 1/1234 + 1/12345 + 1/123456 + 1/1234567 + 1/12345678 + …

I’m writing 1/234 to mean “One over 2×3×4.” With the (2×3×4) in parentheses. Get me?

$\dpi{200} \bg_white e &= 1 + {1 \over 1} + {1 \over 1 \cdot 2} + {1 \over 2 \cdot 3} + {1 \over 2 \cdot 3 \cdot 4} + {1 \over 2 \cdot 3 \cdot 4 \cdot 5} + {1 \over 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} + \ldots \\ &= \sum_{n=0}^\infty {1 \over n!}$

Why is e useful, though?  Because the derivative of e^x is itself.  D[e^x] = e^x.  In other words, the function e^x looks like 1 to the operator D (derivative operation).  Because of that fact, answers to the differential equations that describe the interesting and complicated world we live in, are stated easiest in terms of e^x.  (More specifically, functions like e^x, e^2x, e^ix form a linear basis for the solution space of ODE’s.)

π π π π π

2π is the length around a circle.  (Assuming the circle has a radius of 1 in the chosen units.)

i i i i i i i i i

i = √-1.  What does that mean—half-negative?  Answering this question provoked a flourishing of mathematics in the 19th and 20th centuries.  √-1 is used in electrical engineering, quantum mechanics, solving equations, understanding fluid flows, non-Euclidean geometry, and according to some (Roger Penrose & friends), more and more areas of physics will ultimately be explained with complex numbers.

For now, think about this.  When the power company sends electricity to you, it is traveling in the direction “+1”.  (With AC I guess it switches between +1 and −1.)  If it traveled in the direction “i”, the wire would heat up but no useful energy would be transmitted.

In fact, e^{ i × any number of ° degrees} behaves like a unit circle — a circle with radius one — in the complex plane.

So e^{ i × 90° } = i, e^{ i × 270° } = −i, and e^{ i × 360° } = 1.  In terms of distance around the circle , that’s a quarter-turn ↶, a three-quarters turn ⟲, and a full rotation ⥀.  Or, in π terms:

$\LARGE \dpi{200} \bg_white \begin{matrix} e^{ \pi \over 2} &= i \\ e^{3 \pi \over 2} &= -i \\ e^{2 \pi} &= 1 \end{matrix}$

Similarly, going , , 8π, or 17362 times around the circle leaves you just where you started — whether that was from a quarter-turn anticlockwise ⤿ or at the three o’clock starting point. −1 is 180 degrees ↶ or π arc-length away from the starting point of +1 so voilà,

$\LARGE \dpi{300} \bg_white e^{i \pi} = -1$.

The Punchline

That’s all fine, for parlor tricks.  But there’s more, much more.  Seeing the gamut of numbers as concentric circles in the complex plane allows you to solve every equation, ever*.  I’ll build this up for a paragraph or two.

First, consider that there are two solutions to x²=1, which are ±1.  Meaning, x=1 satisfies the condition and x=−1 satisfies the condition, and those are the only two values of x that satisfy the condition.

Second, there are four solutions to x⁴=1:  1, −1, i, and −i.  Also known as e^{ i × 90°, 180°, 270°, and 360° ⟲}.  And what would 7, −7, 7i, and −7i solve?  x⁴ = 2401.  (Two four oh one is 7⁴.)

Following the same pattern, there are five solutions to x⁵=1 (intervals of 72°), thirteen solutions to x¹³=1 (intervals of 2π/13), and thirty-six solutions to x³⁶=1 (intervals of 10°).  And again, if you doubled the circle’s radius, the solutions would be multiplied by 2⁵, 2¹³, or 2³⁶.

So the answer to any problem of powers is shaped like a ship’s steering wheel:  evenly spaced points on a circle, possibly a slightly wider circle if the question is x^p = 999.

The answer to polynomial problems is then built up from these sorts of objects.  (But just imagine laying two different-radius wheels on a rectangular grid, and being asked to compute the (x,y) coordinates of sums of pairs of handles.  ouch! gimme a computer)  Anyway, it’s nice to know that the answer exists, and it’s good to form your imagination around this kind of shape, in case you ever want to explore something practical that involves this lay of the land.

* That’s an exaggeration, but at least you can solve every equation you’ve ever seen.

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