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(Sorry if the question was already asked. I don't seem to find the right keywords, so after the first few pages of search I gave up)

For complex numbers, we know that $$(\cos \phi + i \sin \phi) (\cos \psi + i \sin \psi) = \cos (\phi + \psi) + i \sin (\phi + \psi)$$ I noticed that, on some level, I don't really understand why it's the case.

  • It's easy to prove using trigonometry. Not very insightful though.
  • We know that $\cos \phi + i \sin \phi = e^{i \phi}$, and then we can use the properties of the exponent. The problem here is that it's not intuitive why in fact $\cos \phi + i \sin \phi = e^{i \phi}$. While we can prove this via the Taylor series, for me it has the same problem as the previous proof: the math works out, but it's not clear why.
  • Now, the explanation which makes intuitive sense. Let's think about the complex numbers as matrices, i.e. $1 = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$ and $i = \begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix}$. It's easy to check that $i^2 = -1$, $1i = i1 = i$ and $11 = 1$ (so the operations match that of the complex numbers), and also $\cos \phi + i \sin \phi = \begin{pmatrix}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}$, which is the rotation matrix by angle $\phi$.

While the last proof kind of works (although it relies on the knowledge that the matrix is the rotation matrix, this is easy to understand geometrically), I believe that there must be an elementary explanation. Any ideas?

Bill Dubuque
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Dmitry
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  • It depends upon the definition - how do you define $\cos$, $\sin$ and $e^z$? How do you intuitively think about what $\cos\phi$, $\sin\phi$ and $e^{i\phi}$ even mean? – ultralegend5385 Jun 10 '23 at 14:34
  • I define $\cos \phi$ as an $x$-coordinate of a point on the unit circle with angle $\phi$. There is no $e^{i \phi}$ in the statement that I want to prove; you can say that it's by definition $\cos \phi + i \sin \phi$, but then you need to prove that the properties of the exponent hold. – Dmitry Jun 10 '23 at 21:46

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