Is there a geometric proof of the de Moivre's formula?

Question

I am trying to understand (intuitively) this formula here: $$e^{ic} = cos (c) + i.sin (c)$$

I understand the infinite sum (traditional) approach 1, but I am looking for something more geometric, maybe because of the involvement of the trig functions. I found another approach 2 which uses the fact that (assuming $f(x)=e^{ix}$): $$f'(x)=i.e^{ix}=i(g(x)+i.h(x))=i.g(x)-h(x) $$

Since the pair of functions $g$ and $h$ for which $h' = g$ and $g' = -h$ happen to be sine and cosine, we get the proof.

I know that exponentiation involving imaginary numbers are more easily dealt with using the power series, but is there a more visual approach to this?

To better understand your question: Are you happy with the fact that $e^{it}$ traces the unit circle or is that the main point you are struggling with? — Klaus, Nov 16 '20 at 13:57
The question 'Euler's Formula, from Needham's Visual Complex Analysis" shows a geometric representation of the identity. A GeoGebra project linked from this answer gives a dynamic form of the illustration. I believe there are other instances scattered about Math.SE. — Blue, Nov 16 '20 at 14:03
@Klaus Yes, I understand that. And I know this question is pretty trivial- sorry for that... — Bipasha, Nov 16 '20 at 14:22

score 1 · Answer 1 · answered Nov 16 '20 at 15:13

Complex numbers admit the matrix representation $x+yi=\left(\begin{array}{cc} x & -y\\ y & x \end{array}\right)$. They can then be seen as a generalization of $2\times2$ rotation matrices that also including scaling. Writing rotation matrices as something exponentiated then just means anticlockwise rotations by $\theta,\,\phi$ compose to an anticlockwise rotation by $\theta+\phi$. Why the base is $e^{i}$ is the hard part, which I think needs Taylor series, albeit seen again in terms of matrices.

Is there a geometric proof of the de Moivre's formula?

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