Is it possible to intuitively explain, how the three irrational numbers $e$, $i$ and $\pi$ are related?
I read a bit about this equation: $e^{i\pi}=-1$ For someone knowing high school maths this perplexes me. How are these three irrational numbers so seemingly smoothly related to one another? Can this be explained in a somewhat intuitive manner? From my perspective it is hard to comprehend why these almost arbitrary looking irrational numbers have such a relationship to one another. I know the meanings and origins behind these constants.
Just think about it this way: $\pi$ is related to the circle, whose equation is $x^2+y^2=r^2$. Euler's constant e is related to the hyperbola, whose equation is $x^2-y^2=r^2$. In order to turn $y^2$ into $-y^2$ we need a substitution of the form $y\mapsto iy$.
The most intuitive explanation I know involves a combination of three facts:
If a particle has position $p(t)$ proportional to its velocity $p'(t)$, say $p'(t) = kp(t)$, then $p(t) = A e^{kt}$ for some constant $A$. We can take this as a definition of the exponential.
In the complex plane, multiplication by $i$ is the same as a counter-clockwise rotation by $90$ degrees.
A particle whose velocity is perpendicular to its position travels at constant speed, tracing a circle centered at the origin.
Combining these three, we find that the particle with position $p(t) = e^{it}$ travels uniformly in a circle. Since $p(0) = 1$ and $p'(0) = i$, we must have $p(t) = \cos{t} + i\sin{t}$, so that $p(\pi) = -1$.
The natural definition of $e$ is not a definition of that single number, but rather of a special function $x\mapsto \exp(x)$, which has the property that $\exp(0)=1$ and that it is its own derivative: $\exp'(x)=\exp(x)$. It turns out, that $\exp$ is uniquely determined by this.
It follows that for any constant $k$, $\exp'(kx)=k\exp(kx)$ and $\exp''(kx)=k^2\exp(kx)$ and so on. Now both sine and cosine are functions with $f''(x)=-f(x)$, which matches the above if one picks $k$ so that $k^2=-1$, i.e., let $k=i$. A closer look show sthat $f(x)=\exp(ix)$ has $f(0)=1$ and $f'(0)=i$, so it looks exactly like $\cos x+i\sin x$ (and by another uniqueness argument, "looks like" means "equals" here). Therefore $\exp(i\pi)=\cos \pi+i\sin\pi=-1$ is a simple value again - in the end this is by the very definition of $2\pi$ as the period of sine and cosine.
Writing $\exp(x)$ instead of $e^x$, this may look less like magic. One can motivate by the fact that the properties and uniqueness of $\exp$ imply the functional equation $\exp(x+y)=\exp(x)\exp(y)$ (which looks like the rules for powers) that one customary uses the suggestive notation $e^x$ for $\exp(x)$ where $e$ is defined as $\exp(1)$.