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I was reading this primer not the Fourier transform and it discusses an intuitive way to think about way to think about the $e^{-i\omega t}$ function. By picking a series of inputs for $\omega t$ and using Euler's formula to calculate the $\sin x$ function on the complex plane. Why is this?

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  • $\begingroup$ Because $e^{it}=\cos(t)+i\sin(t)$??? $\endgroup$ – David C. Ullrich Dec 26 '17 at 1:17
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It comes from power series:

$$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$

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Why: you use the complex representation because it is more natural by an objective measure: the number of written operations in involved expressions becomes significantly less than if you were to do the dual component $\Bbb{R}^2$ rep.

So it's as if in elementary school they should be teaching us arithmetic in $\Bbb{C}$ at times because of its many applications.

Another reason: it's easier to work with exponentials than $\sin/\cos$. Nearly all trig identities are 10x easier to prove knowing some basic properties of $e^{it}$. Therefore the shortest path of proof to a lot of problems is by using the complex numbers.

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The supposed mysteriousness of Euler's formula is overrated, and actually harmful if you want to get through the light and smoke and actually learn the basic math.

Euler's formula relates exponentials to periodic functions. Although the two kinds of functions look superficially very different (exponentials diverge really quickly, periodic functions keep oscillating back and forth), any serious math student would have noted a curious relation between the two -- periodic functions arise whenever you do some negative number-ish stuff with exponentials.

For instance --

  • Simple harmonic motion -- the differential equation $F=kx$ represents exponential motion when $k>0$, periodic motion when $k<0$. This is just a special case of the idea that the derivatives of the trigonometric functions match up with what you'd expect from $e^{ix}$
  • Negative exponential bases -- although exponential functions like $e^x$ and $a^x$ for any positive $a$ would seem to diverge nuttily at some infinity, it turns out that $(-1)^x$ is actually a periodic function, at least for integer $x$ (other negative integers give you a periodic function times a crazy diverging function).
  • Conic sections -- Trigonometric functions are defined on the unit circle, if you defined similar functions on the unit rectangular hyperbola, you'll get linear combinations of exponentials.

There are others, based on trigonometric identities (I wrote a blog post covering some examples recently, see "Making sense of Euler's formula"), but the point is that this relationship is really natural, something you should expect, not some bizarre coincidence that arises from manipulating Taylor series around.

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