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Hi I was asked by a friends child who is in middle school why $e^{i\pi}+1=0$. Now I couldn't think of a way to explain it so he would understand. Albert Einstein once said “If you can't explain it simply, you don't understand it well enough” so how would I explain it.

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What about trying to explain it using polar coordinates, a good picture of the unit circle and a good geometric explanation of vector addition in the plane by using the parallelogram law? – Adrián Barquero Mar 10 '13 at 5:03
@AdriánBarquero I support that one. – Pedro Tamaroff Mar 10 '13 at 5:04
@amr Define better – Ethan Mar 10 '13 at 5:12
Absolute first point is "exponential doesn't mean what you think it means. You don't multiply something $i$ times. Or $\pi$ times, for that matter". Once you find a way to intuit the basic idea behind generalizations of the exponential (i.e. is its own rate of change), the rest flows out effortlessly by comparison. – Robert Mastragostino Mar 10 '13 at 5:13
Are you allowed to spank the child? Or, at least, the friend? – Will Jagy Mar 10 '13 at 5:23
up vote 10 down vote accepted

If the middle school student doesn't know calculus, he's just going to have to take your word that $e^{i\theta}=\cos{\theta}+i\sin{\theta}$. At best, you could try showing him the proof with limits (which I think is a lot more satisfying than the power series proof, especially with pictures), but it may be best to ask him to take this part on faith.

The idea of the complex plane isn't too hard to explain, though. Every complex number has a real part and an "$i$ part," so you can put a dot representing a complex number $z$ between two axes showing "how much real" and "how much $i$" $z$ has. Then, assuming he knows what $\sin$ and $\cos$ are, you can show him how you can write every complex number in polar coordinates by substituting and expanding: $$|z|e^{i\theta}=|z|\cos{\theta}+i|z|\sin{\theta}$$ Drawing a triangle to $z$ from the origin and emphasizing which sides correspond to which will solidify this.

If he understands that, it should be completely clear why the $-1=e^{i\pi}$ identity works. (You may have to explain why $\cos\pi=-1$, but I don't think that's hard. $\pi$ is just half of $2\pi$, the circumference of a circle.) Showing him $e^{i\pi/2}=i$ and $e^{i3\pi/2}=-i$ would support this and give him a few related identities which aren't as well known.

If he doesn't know what $\sin$ and $\cos$ are, though, you'll probably have to let it go. That's a lot to learn in one sitting, and I really don't think there is a simpler, honest way of teaching someone the meaning of that identity. This is okay. Though it may be hard to accept, there are some things which simply take too much background to be explained to a layman. (Albert Einstein, by the way, also said, "If I could explain it to the average person, I wouldn't have been worth the Nobel Prize.") I think in mathematics things have a tendency to look very easy once you've already learned them, but it's important to keep in mind how difficult and impossible they seemed beforehand when explaining things to people.

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Don't mystify it.

Define complex exponentiation $\exp : \mathbb{C} \rightarrow \mathbb{C}$ via power series, then show that $\exp(i\theta) = \cos(\theta)+i\sin(\theta)$, also by power series.

Point out that complex numbers can be viewed as points in the plane, so in some sense it holds that $\exp(i\theta) = (\cos \theta, \sin\theta).$

Then draw an Argand diagram so that its clear that as $\theta$ varies between $0$ and $2\pi$, the expression $\exp(i\theta)$ sweeps out one full revolution of the unit circle.

Then look at values for $\theta$ that are especially interesting. $2\pi,\pi,\pi/2$ etc. Work out what $\exp(i\theta)$ equals (for that $\theta)$ just from the diagram.

Edit: Even better, instead of using $\pi$, use $\tau=2\pi$, which is in some sense more useful, since it corresponds to one full revolution. So perhaps consider the cases $\theta = \tau$, $\theta = \tau/2$ and $\theta=\tau/4$.

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This picture helped him understand

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@Gruber: you can justify $e^{it} = \cos t + i \sin t$ as follows.

Let $i$ be the entity that satisfies $i^2 = -1$. Then $f(t) := \cos t + i \sin t$ has the property $f'(t) = i f(t)$. Integrate $\frac{f'(t)}{f(t)} = i$ from $0$ to $t$ to get $\ln f(t) - \ln f(0) = i t$. Since $f(0) = 1$, we have $f(t) = e^{i t}$. From there, $e^{i \pi} = -1$!

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