Proof of Euler's theorem in complex numbers

Question

I am an engineer and not a mathematician, but I have used the formula stated below many number of times.$$e^{i\theta} = \cos{\theta} + i\sin{\theta}$$ Just out of curiosity, what is the proof to this theorem? I know that adding the Taylor series of the sines and the cosines and comparing it to the Taylor series of $e^{i\theta}$ is a proof, but is there some other way to go about this?

Answer 1

The complex function $\exp(z) = e^z$ is a simple generalization of the real valued function $\exp(t) = e^t$. It is an incredibly useful property that $\exp'(t) = \exp(t)$, so we might hope to find a complex-valued generalization such that $\exp'(z) = \exp(z)$, for complex values of $z$. Now by the chain rule, this would tells us that the derivative of the function $f(t) = \exp(it)$ (mapping the real numbers to the complex plane) satisfies $f'(t) = i f(t)$. Since $i$ is a rotation 90 degrees anticlockwise, what this tells us is that the path $f(t)$ always moves anticlockwise perpendicular to its position in space relative to the origin. If we now consider $g(t) = |f(t)|^2$, we find $g'(t) = 2 \langle f, f' \rangle = 2 \langle f, if \rangle = 0$, so that $f$ always remains at a constant distance to the origin.

Now we make our final assumption about $\exp$, that $\exp(0) = 1$ (which is true for the real valued exponential). Then we see that $|f(t)| = 1$ for all $t$, because $|f(0)| = 1$, and therefore that $f$ moves perpendicularly around the unit circle at uniform velocity. Now which functions do we know that have the same property?

Answer 2

The top-voted answer to this question is very intuitive and a fine example of what makes mathematics so great as the link between seemingly unrelated concepts.

The proof of Euler's law that I have seen is algebraic and a little simpler, admittedly not quite as elegant but still fascinating in how complex numbers, exponents and trigonometry come together in a cyclic calculus-based manner.

Here goes:

Define $f(\theta)=\cos \theta + i \sin \theta$

So $f'(\theta)=-\sin \theta + i \cos \theta$

and since $i^2=-1$, we have

$f'(\theta)=i(\cos \theta + i \sin \theta)$

i.e. $f'(\theta)=if(\theta)$

or $\displaystyle\frac{f'(\theta)}{f(\theta)}=i$.

Note that the RHS follows a certain pattern in calculus.

Integrating both sides,

$\ln{f(\theta)}=i\theta + C$

and since $f(0)=\cos 0 + i \sin 0 = 1$ (and so $\ln{f(0)}=0$), $C=0$ and we are left with

$\ln{f(\theta)}=i\theta$

or

$e^{i\theta}=f(\theta)$

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

Answer 3

It can also be seen with series. Robert mentioned this but here are the details. It needs to be noted that the series expansions for $e^z$, $\sin x$, and $\cos x$, with $z \in \Bbb C$ and $x \in \Bbb R$, each have an infinite radius of convergence. Also, these series all converge absolutely. Absolute convergence makes it safe to add the sine and cosine series like we do in the end.

Recall: \begin{align*} e^x &= \sum_{n=0}^{+\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}2 + \frac{x^3}{3!} + \cdots\\[0.3cm] \cos x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}2 + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\\[0.3cm] \sin x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \end{align*} Therefore: \begin{align*} e^{ix} &= 1 + ix + \frac{(ix)^2}2 + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots\\[0.3cm] &= 1 + ix - \frac{x^2}2 - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \cdots\\[0.3cm] i\sin x &= ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - i\frac{x^7}{7!} + \cdots \end{align*} And so: \begin{align*} \cos x + i\sin x &= \left(1 - \frac{x^2}2 + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) + \left(ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - i\frac{x^7}{7!} + \cdots\right)\\[0.3cm] &= 1 + ix - \frac{x^2}2 - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \cdots\\[0.3cm] &= e^{ix} \end{align*}

Answer 4

The proof will depend on what you are using as definition of the exponential function on complex numbers. Tayor series is one approach. Another way is to define $e^{z}$ as the solution of the differential equation $$\dfrac{dy}{dz} = y$$ with $y(0)=1$. A third is to define $$ e^z = \lim_{n \to \infty} (1 + z/n)^n$$

Answer 5

By DeMovre's theorem $$\cos \theta +i \sin \theta= \left(\cos \frac{\theta}{n} +i \sin \frac{\theta}{n}\right)^n$$

and

$$ \left(\cos \frac{\theta}{n} +i \sin \frac{\theta}{n}\right)^n = \left(1+ \frac{z_n}{n}\right)^n $$ where $$z_n=n(\cos \frac{\theta}{n}-1)+i n \sin \frac{\theta}{n} \to i\theta$$