Friday, November 29, 2019

complex numbers - Did Euler discover the Euler's identity




We know that in $1748$ Euler published the "Introductio in analysin infinitorum", in which, he released the discovery of the Euler's formula:



$$e^{ix} = \cos x+i \sin x$$



But who was the first mathematician to convert this to the form we all know and love, the Euler's identity:



$$e^{i\pi}+1=0$$
When was this formula first explicitly written in this way?




Was it Bernoulli, Euler's teacher and mentor, or another more modern mathematician?


Answer



For a reference pointing to Euler, see :





discussing Bernoulli's thesis that $l (-1)=l(+1)=0$ :




le rayon [du cercle] est à la quatrieme partie de la circonference, comme $\sqrt -1$ à $l \sqrt -1$. Donc posant le rapport du diametre à la circonference $= 1 : \pi$, il sera $\frac 1 2 \pi = \dfrac {l \sqrt -1}{\sqrt -1}$ et pertant $l \sqrt -1 = \frac 1 2 \pi \sqrt -1$.





In a nutshell, he derive for the area of the first quadrant of the unit circle the formula :




$\dfrac {\pi} 4 = \dfrac 1 {4 \sqrt -1} l (-1)$,




from which : $l (-1) = \pi \sqrt -1$.







See also page 165-on where, starting from his formula "dont la vérité est suffissament prouvée ailleurs" :




$$x= \text {cos} \phi + \sqrt -1 \ \text {sin} \phi \ ,$$




posing $C$ as the real logarithm of the positive quantity $\sqrt {(aa+bb)}=c$, he derives the general formula for the logarithm of :





$$a+b \sqrt -1 = C + (\phi + p \pi) \sqrt -1.$$




With $c=1$ and $C=0$ he get :




$$l (\text {cos} \phi + \sqrt -1 \ \text {sin} \phi) = (\phi + p \pi) \sqrt -1.$$





Finally, with $\phi = 0$ [and thus : $\text {cos} \phi = 1$ and $\text {sin} \phi = 0$] :




$l(+1) = p \pi \sqrt (-1)$ and thus [for $p=0$] : $l(+1) = 0$




and, with $\phi = \pi$ :




$l(-1) = \frac + - \pi \sqrt -1$.








Euler in : Introductio in analysin infinitorum, Tomus Secundus (1748), Ch.XXI, page 290, uses $i$ for an imaginary quantity :




Cum enim numerorum negativorum Logarithmi sint imaginarii (...) erit $l(-n)$, quantitas imaginaria, quae sit $= i$.





But he does not say that the symbol $i$ is such that $i^2 = -1$.



In the same Introductio, Tomus Primus, §138, the formula is written as :




$e^{v \sqrt -1}= \text {cos} v + \sqrt -1 \text {sin} v$.








In conclusion, Euler "knows" the identity and he is the "iventor" of $i$ to name an imaginary quantity, but it seems that he never writed it in the "modern form", at least because he constantly writes $\sqrt -1$.









Note



See also Cuchy's Cours (1821) for Euler's identity ; again, $\sqrt -1$ is used.




I've not made an extensive research but, due to the fact that Cauchy uses systematically $i$ for denoting an increment [see : Résumé des leçons sur le calcul infinitésimal (1823) ] :




$\Delta x = i$,




my conjecture is that we hardly find any use of $i$ as imaginary.


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