# Chaitanya's Random Pages

## August 4, 2010

### My Six Favourite Formulas – #1

Filed under: mathematics — ckrao @ 1:59 pm

Going back to my earlier post listing my favourite formulas, it’s time to explore the beauty in them. First of all let’s look at Euler’s identity.

$e^{i\pi} + 1 = 0$

Why is this a favourite of mine and so many others? For me it is because it is highly unexpected on first look, and it brings together no fewer than nine of the most important mathematical concepts in a manner more succinct than one could imagine:

1. the number $e \approx 2.7183$, which is the limit of the sequence $a_n = (1 + 1/n)^n$ as $n$ tends to infinity
2. the number $\pi \approx 3.1416$, which is the ratio of a circle’s circumference to its diameter
3. the imaginary unit $i$ (defined by $i^2 = -1$)
4. the multiplicative identity $1$
5. the additive identity $0$
6. the exponentiation operation $a^b$
8. multiplication ($\times$)
9. equality (=)

This equality is a special case of Euler’s formula, but it can also be seen with physical ideas without the use of trigonometric functions. The exponential function $y = e^z$ has the property that its rate of change is equal to itself (i.e. $\frac{dy}{dz} = e^z$), so for $z = it$,

$\frac{d}{dt} e^{it} = i\frac{d}{dz}e^{z} = ie^{it}.$

We visualise $e^{it}$ as the position of a point particle in the two-dimensional complex plane, with position $e^0 = 1$ at time $t = 0$. The above equation describes the motion of the particle and says that its velocity is always $i$ times its position.

Since multiplication by $i$ corresponds to an anti-clockwise rotation by 90 degrees in the complex plane, this is saying that the instantaneous motion is in a direction perpendicular to its position relative to 0, which will neither increase nor decrease its distance from 0. (Here I visualise turning a handle that you might use when opening a car window, in the days before power windows! The incremental motion is always perpendicular to the length of the handle which is fixed in size.)

Hence as $t$ increases from 0, $e^{it}$ will trace out a circular arc, centred at 0, in an anticlockwise direction, at speed equal to the magnitude of its velocity, or

$|dz/dt| = |i e^{it}| = |e^{it}| = |e^0| = 1.$

Here the second last equality comes from the earlier statement that $e^{it}$ does not change in magnitude with t (see footnote below for a more rigorous argument). Therefore $z = e^{it}$ traces out a unit circle at unit speed. At $t = \pi$, the particle will have covered length $\pi$ along the circular arc. By the definition/property of $\pi$, it will have covered half the circumference of the circle, and be diametrically opposite 1, at -1. Hence $e^{i\pi} = -1$ from which our magical formula arises.

Footnote: If $e^{it} = x(t) + iy(t)$ where x and y are real functions of t, then taking derivatives or multiplying both sides by i gives $ie^{it} = dx/dt + i dy/dt = ix(t) -y(t)$. Matching real and imaginary parts leads to $dx/dt = -y, dy/dt = x$, and so

$\frac{d}{dt} |e^{it}|^2 = \frac{d}{dt} (x^2(t) + y^2(t)) = 2(x \frac{dx}{dt} + y \frac{dy}{dt}) = 2(-xy + yx) = 0.$

Therefore when t is real, the magnitude of $e^{it}$ is constant and equal to its value at t=0.