To whom do we owe this construction of angles and trigonometry?

Solution 1:

Yes. In fact, $\operatorname{SO}(2)$ is isomorphic to $S^1$ (the circle) as a subset of $\mathbb{C}$. An amusing way of seeing this is to consider all the $2\times 2$ matrices of the form $$Z:=\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$ for any choice of $a,b\in\mathbb{R}$. Exercise: check that the set of all such matrices is closed under addition and matrix multiplication and thus forms a so-called algebra, whose only singular element is $(0,0)$. In fact this algebra is isomorphic to $\mathbb{C}$. Now you will recall from basic complex arithmetic that any complex number $z=a+ib$ can be written as $z=\rho e^{i\theta}$ where $\rho^2=a^2+b^2$, $\rho>0$ and $\cos\theta=a/\rho$, $\sin\theta=b/\rho$, $\theta\in[0,2\pi)$ are uniquely determined, but there is always something upsetting about this approach. Think instead of the complex number $z$ as a matrix $Z$ of the above form, then $\rho^2=\det Z$ and $\exp(\theta J)=Z/\rho$ where the $\exp$ function of matrices is defined as a power series (of matrices) $$\exp(M):=I+M+\frac{M^2}2+\frac{M^3}{3!}+\dotsb$$ and $$J=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\text{ hence } Z=\sqrt{\det Z}\exp(\theta J).$$ If you don't like that exponential, just think of it as $$Z=\sqrt{\det Z} E\text{ for some matrix }E.$$

Now thinking geometrically, a matrix $Z$ of the above form represents a linear transformation in the plane and $\det Z$ is the corresponding transformation of area, so $\sqrt{\det Z}$ is the change in length, provided length changes uniformly in all directions (isotropic). But isotropic change occurs if and only if angles (or their cosines) are preserved by the linear transformation... and they are, take two vectors $u,v\in\mathbb R^2$ and check that $u^*Z^*Zv=\det Z\,u^*v$. So in fact, the matrices $Z$ are nothing but a stretching by $\sqrt{\det Z}$ combined with a rotation $E:=Z/\sqrt{\det{Z}}$. (At this point you see that this rotation $E$ corresponds to one of angle $\theta$, but you don't need to think of $\theta$ yet.) If you impose the extra constraint that $\det Z=1$, the transformations $Z$ describe all possible rotations in the plane. The set of matrices of the form of $Z$ with $\det Z=1$ is known as $\operatorname{SO}(2)$ and this shows that it is isomorphic to the circle $S^1$.

If you prefer complex analysis, you may introduce the so called unitary group of order $1$, $\operatorname{U}(1)$ by taking all complex numbers $z$ of unit length $|z|=1$. But as we've seen above this is nothing but $\operatorname{SO}(2)$ viewed from a different angle (pun unintended).

I'm not sure this clarifies things, but I find it fascinating to introduce complex numbers in this geometric way, as orientation-preserving linear and conformal (OPLAC) mappings, i.e., as plane transformations (mappings) that preserve orientation, straight lines (linear) and angles (conformal), rather than the mechanical $i^2=-1$ algebraic way.

To finish, let's revisit the matrix exponential function. If $L$ is an OPLAC map let's look at $\exp L$. First we may decompose \begin{equation} L=\begin{bmatrix}m&-n\\n&m\end{bmatrix} =\begin{bmatrix}m&0\\0&m\end{bmatrix}+\begin{bmatrix}0&-n\\n&0\end{bmatrix} =mI+nJ,\text{ with $J$ as above}. \end{equation} Since $I$ and $J$ commute (check) it follows that \begin{equation} \exp L=\exp(mI)\exp(nJ)=e^m\exp(nJ). \end{equation} Now given another OPLAC map $Z$ one may ask the question of whether it is possible to find $L$ such that $Z=\exp L$. If $Z=0$, no chance. But if $Z\neq0$ it is sufficient to take $m=\log\sqrt{\det Z}$ and $n$ such that $\exp(nJ)=Z/\sqrt{\det Z}$. Working out the exponential we see Euler's formula in matrix form: \begin{equation} \exp(nJ) = I+nJ+\frac{n^2J^2}2+\frac{n^3J^3}{3!}+\frac{n^4J^4}{4!}+\dotsb \\ = I+nJ-\frac{n^2I}2-\frac{n^3J}{3!}+\frac{n^4I}{4!}+\dotsb \\ = (\cos n)I+(\sin n)J . \end{equation} Again this connects to the rotation and the trigonometry, but by using the cosine and sine as power series.