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 $e^{ix}$
- Height controls height -- this is a charmingly simple explanation used, e.g. here. In exponential growth, "height controls height" -- the higher you are, the higher you grow in height. In circular motion, "height controls width" -- the higher you are, the further you go leftwards. The rigorous formulation of this is that exponential motion is modeled as $dx/dt=x$ while circular motion is modeled as $dx/dt=ix$.
- 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, which I'll cover below, 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.
(Above edited on 28 April 2018 -- older text with further identities follows.)
From complex multiplication
Suppose we haven't yet heard of the identity $e^{i\theta} = \cos\theta + i\sin\theta$, but only know that the right-hand-side is the polar-co-ordinate representation of a unit complex number in the complex plane (i.e. the form taken by $\frac{z}{|z|}$ for any complex number z), and are playing around with it for a geometric understanding of multiplication on the complex plane.
For simplicity, we'll stick to unit complex numbers $z_1 = \cos\theta_1 +i\sin\theta_1$ and $z_2=\cos\theta_2+i\sin\theta_2$. Multiplying these two gives us:
$$\begin{array}{c}{{\hat z}_1}{{\hat z}_2} = (\cos \theta + i\sin \theta )(\cos \phi + i\sin \phi )\\ = \left( {\cos \theta \cos \phi - \sin \theta \sin \phi } \right) + i\left( {\cos \theta \sin \phi + \sin \theta \cos \phi } \right)\end{array}$$
Well, they're all wearing real nice hats, but if you notice, the real part is equal to $\cos (\theta + \phi )$, and the imaginary part is equal to $\sin (\theta + \phi )$, by the angle addition identities from trigonometry.
$${\hat z_1}{\hat z_2} = \cos (\theta + \phi ) + i\sin (\theta + \phi )$$
This just gives us another unit complex number, of direction (argument) $\theta+\phi$. This gives us our nice geometric interpretation of complex number multiplication -- dilate by the length of the other complex number, and rotate by its angle. More interestingly, though, this means that ${\hat z}(\theta) = \cos\theta+i\sin\theta$ as a function of its argument satisfies the relation ${\hat z}(\theta){\hat z}(\phi)={\hat z}(\theta+\phi)$, and considering non-unit complex numbers, $z(r,\theta)z(s,\phi)=z(rs,\theta+\phi)$.
If you recall, this identity is satisfied by functions of the form $re^{a\theta}$. Since this yields real values for all real values of a, it makes sense to expect the value of a to be complex (it happens to be i).
It is also satisfied by zero, but obviously, all complex numbers can't be zero.
Note that this is not a proof of Euler's formula -- which is something you should already know from learning calculus and Taylor series -- but an attempt to explain why we shouldn't be surprised that it holds.
Derivative on the unit circle
You can find more proofs of the identity by considering other properties of $e^{i\theta}$. For instance, its derivative is i times itself, which also holds for $\operatorname{cis}\theta$. Since $f(x)=e^{ax}$ is the only solution to $f'(x)=af(x)$ among the reals, if we want this (defining) property of $e^{ax}$ to remain true for complex $a$, we would set $e^{i\theta}=\operatorname{cis}\theta$.
A geometric interpretation of this is found in Needham, pg. 13 -- we let $\theta$ be the time of a particle moving around the unit circle in the complex plane. The instantaneous velocity of the particle at any point is $\frac{\pi}2$ counterclockwise to the complex number itself, and of the same magnitude.
(Don't just listen to me -- prove it!)
Exercise 1
This is another way to prove that the angles add up when you multiply two complex numbers -- in the standard Cartesian system, we write complex multiplication as (a + ib)(c + id) = (ac - bd) + i(ad + bc). To show that the arguments add up, you must show that $\arctan\frac{b}{a}+\arctan\frac{d}{c}=\arctan\frac{ad+bc}{ac-bd}$, which is reduced to a plain-old trigonometry problem.
Since we're talking about the sum of two angles and their tangent, we're inclined to think about something like two triangles combined in one or something like that. We could pile one triangle on the hypotenuse of another, but then we'd have to write the dimensions of the piled-up triangle things in a surd-y way (try it), so we'd rather pile the triangles with the bases matching. For this, we'd need the common denominator ac for the angles on the left-hand side.
In other words, we need to prove that the angle on the left, subtended by bc + ad, has a tangent of $\frac{bc+ad}{ac-bd}$. To do so, we could construct another point such that the angle at the point subtended by the same line segment is the same, but the resulting triangle would be a right-angled triangle. For the angles to be the same, the points must lie on the same segment of a circle for which bc + ad is a chord, i.e.
(a reader was so outraged at this diagram that he sent me an alternate version made in PowerPoint)
Thus our problem is reduced to proving that the horizontal line segment at the top has length "ac - bd".
Well, prove it!
Exercise 2
Many trigonometric identities can be shown through Euler's formula. For example,
$$\begin{array}{c}\cos \left( {2\theta } \right) + i\sin \left( {2\theta } \right) = {e^{2i\theta }}\\ = {\left( {{e^{i\theta }}} \right)^2}\\ = {\left( {\cos \theta + i\sin \theta } \right)^2}\\ = \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + \left( {2\cos \theta \sin \theta } \right)i\end{array}$$
Which carries both double-angle identities for the cosine and the sine in the real and imaginary parts respectively.
Try to derive all the trigonometric identities you know this way.
This is another way to prove that the angles add up when you multiply two complex numbers -- in the standard Cartesian system, we write complex multiplication as (a + ib)(c + id) = (ac - bd) + i(ad + bc). To show that the arguments add up, you must show that $\arctan\frac{b}{a}+\arctan\frac{d}{c}=\arctan\frac{ad+bc}{ac-bd}$, which is reduced to a plain-old trigonometry problem.
Since we're talking about the sum of two angles and their tangent, we're inclined to think about something like two triangles combined in one or something like that. We could pile one triangle on the hypotenuse of another, but then we'd have to write the dimensions of the piled-up triangle things in a surd-y way (try it), so we'd rather pile the triangles with the bases matching. For this, we'd need the common denominator ac for the angles on the left-hand side.
(a reader was so outraged at this diagram that he sent me an alternate version made in PowerPoint)
Thus our problem is reduced to proving that the horizontal line segment at the top has length "ac - bd".
Well, prove it!
Exercise 2
Many trigonometric identities can be shown through Euler's formula. For example,
$$\begin{array}{c}\cos \left( {2\theta } \right) + i\sin \left( {2\theta } \right) = {e^{2i\theta }}\\ = {\left( {{e^{i\theta }}} \right)^2}\\ = {\left( {\cos \theta + i\sin \theta } \right)^2}\\ = \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + \left( {2\cos \theta \sin \theta } \right)i\end{array}$$
Which carries both double-angle identities for the cosine and the sine in the real and imaginary parts respectively.
Try to derive all the trigonometric identities you know this way.
If you define the trigonometric functions in the traditional way, some of these proofs would be circular arguments. Which ones?
Related, interesting:
At 5:58, you will notice that the usage of $a-bi$ (as opposed to $a+bi$) is linked to the minus sign in the expansion of $\cos(\theta +\phi)$ and the plus sign in the expansion of $\cos(\theta - \phi)$ -- this gives us another explanation for the sign reversal in the cosine-of-a-sum formulae: because $i\cdot -i=1$.
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