# Where does complex exponential come from?

The complex exponential function is defined as : $$e^{ix} = \cos x + i\sin x$$ It shares most of its properties with real exponential and it allows a lot of trigonometric calculations such as de Moivre's formula : $$(\cos x+i\sin x)^n = \cos{nx}+i\sin{nx}$$

But where does this definition come from and why does it work ?

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That identity is known as Euler's formula, you can read more about its history and what not on Wikipedia. –  Willie Wong Jun 7 '12 at 10:51

Another way to look at it is to view the exponential and trigonometric functions as defined by a power series:

$$\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$

This has the advantage that the $x$ can be anything, as long as we know how to multiply two of them, add two of them together, and divide them by a real number. In particular, it makes sense for both real and complex numbers.

Now you can put $ix$ into the definitions in place of $x$, and compute:

\begin{align} \exp(ix) & = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \cdots \\ & = \left( 1 - \frac{x^2}{2!} + \cdots\right) + i \left( x - \frac{x^3}{3!} + \cdots\right) \\ & = \cos x + i \sin x \end{align}

so the formula you quoted is seen to be a theorem rather than a definition. If we now assume that the familiar law

$$\exp(a+b) = \exp(a) \exp(b)$$

holds for arbitrary $a$ and $b$ (it does, and you can prove it from the power series definition) then we now have a way to compute the exponential of any complex number:

$$\exp(x+iy) = \exp(x) (\cos y + i\sin y)$$

where $x$ and $y$ are real.

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I disagree that the quoted formula cannot be a definition. –  Phira Jun 7 '12 at 10:52
I don't think I say that it can't be a definition - only that if you take the power series to be the definition, then the quoted formula becomes a consequence of that definition. –  Chris Taylor Jun 7 '12 at 10:53
I personally agree that the definitive definition of $e^{\mathrm{i}x}$ should be thru a series. Euler's identity is a bit ambiguous, if you want to define $e^{z}$ for any $z \in \mathbb{C}$. –  Siminore Jun 7 '12 at 11:10
I would say the definition of the exponential should be the solution of a differential equation: $f\prime(x)=f(x)$ and $f(0) = 1$. A lot of the nice properties (including the power series) come out of that equation. –  Ben Millwood Jun 7 '12 at 11:54
Hmm. I'm not sure I agree (inasmuch as I see the complex derivative as not a problem because it's easy enough, and the arbitrary Banach algebra as not a problem because it's out of scope), but I don't feel particularly strongly about it, since probably one of the first things you would do with either definition is prove the other as a theorem. It's certainly at least interesting to know that it can be done that way. Unfortunately my first-year notes are buried under a pile of paper somewhere so I can't cite the neat proofs as evidence :) –  Ben Millwood Jun 7 '12 at 17:05

People have given you some good algebraic answers that show that the equations work out; here's a hand-wavy justification that may make you more comfortable with the idea in general.

The derivative of $f(x)$ is how $f(x)$ changes as you change $x$ a little.

What's the derivative of $f(x) = e^{ix}$? It's $ie^{ix}$, which is $i$ times $f(x)$ itself. So the way in which $f$ changes as you wiggle $x$ by $dx$ is $i \cdot f \cdot dx$.

So what does it mean to multiply something by $i$? It means rotating it 90 degrees counterclockwise in the complex plane. (Try this yourself with some simple complex numbers if you didn't notice this already.)

So when $x=0$, and $f(x)=1$, $f'(0) = i \cdot f(0) = i \cdot 1 = i$; $f(0)$ is changing to the north when its value is to the east. The same argument works for other values of $x$; $f(x)$ will be changing at a 90 degree clockwise angle from the current value of $f(x)$.

The equation of motion that satisfies this rule (velocity is always perpendicular to the direction from the origin to the current point) is a circle. $f(x) = e^{ix}$ moves in a counterclockwise circle around the complex plane, and that's exactly what $\cos x + i \sin x$ does too.

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I knew this sounded familiar, but I couldn't place it for a minute, and then I realized you got it from Needham. –  MJD May 21 '13 at 2:01

One approach is to start with $$e^z = \lim_{n\to\infty}(1+\frac{z}{n})^n$$

Now, when $z=ix$ and $n$ is large, you can show geometrically that $1+\frac{ix}{n}$ is "very close" to $\cos \frac{x}{n} + i\sin\frac{x}{n}$. You just need that "close enough" to be enough to show:

$$e^{ix} = \lim_{n\to\infty} (\cos \frac{x}{n} + i\sin\frac{x}{n})^n = \cos x + i\sin x$$

So this can ultimately be seen as a result of the fact that if $\theta$ is small, $\cos \theta \approx 1$ and $\sin \theta \approx \theta$. In particular, the error in both cases is $O(\theta^2)$, which turns out to be all you need.

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A simple way to prove this requires the use of Maclaurin's Series, which state that for any n-times differentiable function, $$f(x)\approx\sum_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}$$ Where $f^{(k)}(0)$ is $\frac{d^{k}f}{dx^{k}}$ evaluated at zero.
It is easy to show that: $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+....$$and also that$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$ and hence that $$e^{ix}=1+ix-\frac{x^2}{2}-\frac{ix^3}{6}+...=\cos(x)+i\sin(x)$$

I would note that this probably isn't entirely rigorous (what with us throwing infinite series and complex numbers about with reckless abandon) but it's definitely accurate.

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Well, the series on both sides are absolutely convergent. So re-arranging and re-grouping is perfectly okay. –  Willie Wong Jun 7 '12 at 10:55
I know that in this case everything is justified, I was just pointing out that we hadn't explicitly shown it. –  Daniel Littlewood Jun 7 '12 at 11:03
The formula at the top does not hold. In general a function need not be equal to its power series: the property of being so is called being analytic. Even in that case, your formula suggests that, since $e^x$ is twice differentiable, $e^1 = 1 + 1 + 1/2$. You certainly can't truncate the power series after $n$ terms and expect to get $f$ back. –  Ben Millwood Jun 7 '12 at 11:57