How Do I Understand $e^i$, the Euler Form of Complex Number

Solution 1:

Essentially, the complex exponential characterizes rotations. I will not go into the mathematical "reasons" for why the equations are "true" but rather expand on how to get an intuitive feel for what they mean. This discussion will be quite informal.

The imaginary exponential is characterized by $$e^{iy} = \cos(y) + i\sin(y)$$ where $y\in\mathbb{R}$ and the full exponential can then be defined as $$e^z = e^{x+iy} = e^x\cos(y) + ie^x\sin(y)$$ where I have written $z\in\mathbb{C}$ in terms of its real and imaginary parts $x,\ y\in\mathbb{R}$.

Now complex numbers are really just elements of $\mathbb{R}^2$ imposed with a special structure. So if you regard $e^{iy}$ as a point on the complex plane, you see that the point lies on the unit circle, with coordinates $\left(\cos(y),\ \sin(y)\right)$. More precisely, with our traditional convention of measuring angles counter clock-wise from the positive $x$ (in this case Real) axis, we see that $e^{iy}$ is a point on the unit circle which forms angle $y$ (in Radians) with the positive Real axis.

In this respect, $e^i$ is just a point on the unit circle which is precisely $1$ radian counter clock-wise from the positive real axis. This also helps to explain why $e^{i\pi} = - 1$. A rotation of $\pi$ from the positive real axis is just the opposite side, i.e. the negative real axis.

Now with the full complex exponential, the imaginary part of $z$ signifies the rotational aspect of the number while the real part of the number signifies the magnitude. Each complex number is assigned a magnitude and an angle (called the argument). This is done precisely with the complex exponential.

You may recall that multiplying two complex numbers is equivalent to rotating one number by the angle of the second (and then applying the proper stretches and compressions). But notice that when we rotate something, all we're really doing is adding the appropriate angles together. So complex multiplication is connected to real angle addition. In hindsight, it probably seems natural that the exponential should be somehow related to complex rotations: The map which takes angle to point must satisfy $$f(\theta_1 + \theta_2) = f(\theta_1)f(\theta_2)$$ which is a property which characterizes the exponential function.

Solution 2:

The other answers are very nice. I'd just like to add how this works, because it's very nifty and somewhat surprising if you see it the first time. Look at the series definition of $\exp(x)$:

$$ \exp x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \ldots $$

When $x$ is real, this has a fairly simple behavior, it grows monotonously. However, if you allow complex numbers, you can get minus signs in there because of $i^2 = -1$. Let $x = i\alpha$, and then:

$$ \exp i\alpha = \sum_{k=0}^\infty \frac{(i\alpha)^k}{k!} = 1 + i\alpha - \frac{\alpha^2}{2} - i\frac{\alpha^3}{3!} + \frac{\alpha^4}{4!} + i\frac{\alpha^5}{5!} + \ldots $$

Two interesting things happen:

• All odd powers of $\alpha$ get an $i$, while all even powers are real (because $i^2$ is real). This lets you single out the even part (symmetric to the Y-axis), and the odd part (point symmetric wrt. the origin) of the function.
• Every two terms the sign switches, so you no longer have unbounded growth, but can have periodicity.

If you collect all the odd terms, which are imaginary, you get:

$$ \mathrm{Im}\,(\exp i\alpha) = \alpha - \frac{\alpha^3}{3!} + \frac{\alpha^5}{5!}- \frac{\alpha^7}{7!}\pm \ldots = \sin \alpha $$ and this happens to be exactly the Taylor expansion of $\sin\alpha$! You start with the value of $\sin(0)=1$ as a first approximation, and keep adding Taylor terms. Since $\sin'0 = \cos 0 = 0 $ there is no $\alpha^1$ term, but $\sin''0 = -1$, so the $\alpha^2$ term is negative, and so on. Every term flips the sign and adds two bends, and you approximate $\sin \alpha$ better and better. A nice picture from the wikipedia article Taylor series:

All the even power terms are real, and they give $\cos \alpha$:

$$ \mathrm{Re}\,(\exp i\alpha) = 1 - \frac{\alpha^2}{2!} + \frac{\alpha^4}{4!} -\frac{\alpha^6}{6!} \pm \ldots = \cos \alpha $$

Which allows you to write the nice formula of Euler:

$$ \exp i\alpha = \cos \alpha + i\sin \alpha $$

As already said in the other answers, you can think of $\alpha$ as the angle from the x-axis (counterclockwise). Since for complex numbers, conventionally $z= x + iy$, it follows that $x = \cos \alpha$ and $y = \sin \alpha$. If I'm doing e.g. some computer graphics calculations, often all I need to remember is "$x = \cos$" and all else falls into place. Another Wikipedia illustration of this: