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In the reals, $e^{nx}$ explodes to infinity very fast. But, $e^{inx}$ is bounded and periodic.

I am familiar with Euler's formula $e^{ix} = \cos x +i\sin x $. Yet, could you give me some intuition what's happening in the transition from the reals to the complex when multiplying by $e^i$?

Thanks.

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  • $\begingroup$ tauday.com/tau-manifesto $\endgroup$ – user301988 Dec 18 '16 at 22:45
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    $\begingroup$ "multiplying by $e^i$" is not what happens. $\endgroup$ – Yves Daoust Aug 26 '19 at 21:06
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You already see periodicity when you think about $n \mapsto i^n$.

\begin{align} i^0 & = 1 \\ i^1 & = i \\ i^2 & = -1 \\ i^3 & = -i \\ i^4 & = 1 \\ & {}\,\vdots \end{align} If one can imagine such a thing as $\log_e i$ then this would say $n\mapsto e^{n\log_e i}$ is periodic with period $4$, and it is clearly an exponential function of $n$. If one were to define it for non-integer $n$ then its values for non-integer real $n$ would have to have absolute value $1$. What, for example, is its value when $n=1/10$? Clearly it would have to be a $10$th root of $i$, so if multiplication by $i$ is a $90^\circ$ counterclockwise rotation, then that would be a $9^\circ$ counterclockwise rotation --- thus a multiplication by $\cos9^\circ+i\sin9^\circ$. Now we start to see circular motion.

But how fast should it move around the circle? Since $\left.\dfrac d {dz} e^z\right|_{z=0}=1$, we have that when $z=0$ then $e^{0+dz}= 1 + 1\cdot dz$. If $dz$ is a pure imaginary infinitesimal, this says $e^{0+dz}$ is directly above or below $1$ in the plane. But how far above or below? Clearly $|dz|$, i.e. the rate at which it moves is $1$ times the rate at which $z$ moves along the imaginary axis.

That tells use that $e^{i\theta}$ must treat $\theta$ as being in radians, so that the arc length equals $\theta$. ${{}}$

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  • $\begingroup$ Thank you for the enlightening answer! $\endgroup$ – AlonAlon Dec 23 '14 at 20:32
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Here's a physical interpretation: Imagine a particle lying inside $\Bbb C$, where its position is described parametrically by $\mathrm V(t)$, where $t$ stands for time. Precisely define

$$\mathrm V(t)=e^{it}.\tag{1}$$

Now recall from physics that the velocity vector $\mathrm Z(x)$ of this particle has a direction and length given by the instantaneous speed, and the instantaneous direction of motion (tangent to the trajectory) of the moving particle. Recall also that

$$\begin{align*} \dfrac{\mathrm d}{\mathrm dx}\mathrm V(x) &=\dfrac{\mathrm d}{\mathrm dx}e^{it}\\ &=ie^{it}\\ &=i\mathrm V(x)=\mathrm Z(x).\tag2 \end{align*}$$

This means that the velocity vector is just the position vector rotated through a right angle (remember that multiplication by $i$ in the complex plane is a rotation by $\tfrac{\pi}{2}$). Since the initial position is $e^{i0}=e^0=1$, and thus its initial velocity is $i$. So it is moving vertically upward A moment later the particle will have moved very slightly in this direction, and its new velocity will be at right angles to its new position vector. Continuing to construct this motion you'll rapidly see that this is a cicle. And since $|\mathrm V(t)|=1$, it follows by $(2)$ that $|\mathrm Z(t)|=1$. Therefore after some time $t=\theta$ the particle will have traveled a distance $\theta$ around the unit circle, hence

$$e^{i\theta}=\cos\theta+i\,\sin\theta.\tag3$$

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Source: Tristan Needham, Visual Complex Analysis, pp.10-2.


You may ask yourself why this holds precisely for $e$, why not for $2$ or some other constant? The answer boils down to the fundamental property of the exponential function which is that it satisfies

$$\dfrac{\mathrm d}{\mathrm dx}f(kx)=kf(kx),$$

hence it satisfies $(2)$ and the argument follows.

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First: You made a mistake

I think your lack of intuition could be caused by the following error:

$$ e^{ix} \neq e^i\cdot e^x$$

Of course, if this were the case, $e^{ix}$ would also explode.

What stops it from blowing up?

$e^{x}$ for real or complex $x$ is defined by

$$ e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots $$

This explains what happens on the transition from real to imaginary $x$. For real, positive $x$ this is a series of positive terms that get larger with $x$ and blow as up $x\rightarrow\infty$. If $x$ is imaginary, e.g. $x=iy$ with $y\in\mathbb R$, then the very definition of $i$ ($i^2=-1$ , which is impossible for real numbers) will make half of the terms negative. If you actually write this out you end up with

$$ e^{iy} = \cos y + i\sin y $$

This is proof that the negative terms stop the function from blowing up.

Periodicity

Sine and cosine are defined as the first and second components of a point on the unit circle. Thus the function is periodic and if you interpret a complex number as a point on the 2D plane you get exactly that unit circle.

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The exponential enjoys the "sum-to-product" property,

$$e^{z+w}=e^ze^w,$$ from which you draw by induction

$$e^{nw}=(e^w)^n.$$

This explains why the function "blows" for the reals (for $w>0$, $e^w>1$, and you get a growing geometric sequence).

But it turns out that in the complex, there are numbers such that

$$e^w=1$$ (namely $w=2ik\pi$). With such $w$, you have

$$e^{nw}=(e^w)^n=1$$ for integer $n$ and this explains the periodic behavior.


In the complex plane, multiplication by $e^{x+iy}$ is the combination of a scaling by the factor $e^x$, and a rotation by the angle $y$.

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