Euler's first proof of $e^{ix}=\cos(x)+i\sin(x)$

What was Eulers first proof of his famous formula?

In Euler's book on complex functions he used the following proof. But was this his first proof?

Euler starts with writing down De Moivre's Formula (can be proven by simple induction using some basic trig identities).

$$\cos(nx)+i\sin(nx)=\left( \cos(x)+i\sin(x)\right)^n$$

He says that $n$ is very large ($n \to \infty$) and $x$ is very small ($x\to 0$). The product of both will be a finite number called $\omega =nx$. Then he applies this as substitution for De Moivres Formula: $$\cos(\omega)+i\sin(\omega)=\left( \cos(\frac{\omega}{n})+i\sin(\frac{\omega}{n})\right)^n$$

Euler now applies the limit $n\to \infty$: $$\cos(\omega)+i\sin(\omega)=\lim_{n\to \infty}\left( \cos(\frac{\omega}{n})+i\sin(\frac{\omega}{n})\right)^n$$

using small angle aproximations $\cos(x)\approx 1$ and $\sin(x)\approx x$: $$\cos(\omega)+i\sin(\omega)=\lim_{n\to \infty}\left( 1+\frac{i\omega}{n}\right)^n=e^{i\omega}.$$

In the last line he applied the limit representation $e^x=\lim_{n\to \infty}(1+\frac{x}{n})^n$.

• I thought limits did not exist until at least a few decades after Eulers death. Is this how he wrote the proof? – Avatrin Jun 16 '16 at 20:55
• He didn't use the limit notation, but he actually used a limit by saying that $n$ is very large or $x$ is very small. – MrYouMath Jun 20 '16 at 7:25

According to Boyer's A History of Mathematics, the identity first appeared in Euler's Introductio in analysin infinitorum of 1748. Euler started from the infinite series for $e^x$, $\sin x$, and $\cos x$, from which the identity quickly follows. Thus from:
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$ $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$
substituting $ix$ into the series for $e^x$ and rearranging terms quickly leads to the result.
Boyer also notes that Roger Cotes had derived this identity, given by him in a Philosophical Transactions article of 1714 in the equivalent form $\ln(\cos \theta + i \sin \theta) = i\theta$.