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

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? Nov 16, 2020 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, 2020 at 14:03
• @Klaus Yes, I understand that. And I know this question is pretty trivial- sorry for that... Nov 16, 2020 at 14:22
• @Blue Thank you, I found my answer! Nov 16, 2020 at 14:23
• Here is the best visual explanation I have seen.
– Joe
Nov 16, 2020 at 15:08

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.