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A comment on Why does e to pi i equal -1? (2015) [video]. Parent comment
For f(x)=e^(kx), double derivative d^2f/dx^2 = k^2 f(x)
Meanwhile for g(x)=sin(kx), d^2g/dx^2 = -k^2 g(x), and similarly for cosine.
So if k is imaginary, from a differential equations point of view, the exponential behaves exactly like a sine or cosine.
That shows the general idea, and further consideration of boundary conditions gives e^ix = cos(x) + isin(x).
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For f(x)=e^(kx), double derivative d^2f/dx^2 = k^2 f(x)
Meanwhile for g(x)=sin(kx), d^2g/dx^2 = -k^2 g(x), and similarly for cosine.
So if k is imaginary, from a differential equations point of view, the exponential behaves exactly like a sine or cosine.
That shows the general idea, and further consideration of boundary conditions gives e^ix = cos(x) + isin(x).