I was pointed to this excellent series on complex numbers from Welch labs, following my last post on complex numbers. It in the 3Blue1Brown mold, with just the right dose of insight and animation. The complex number series starts with basic ideas, and ends with a discussion of Riemann surfaces.
I also came across an interesting way of proving exp(ix) = cos x + i sin x (@fermatslibrary), which I feel compelled to share, since we are already talking about complex numbers.
Let \(f(x) = e^{-ix} (\cos x + i \sin x)\).
The derivative of this function is \[f'(x) = e^{-ix} (i\cos x - i \sin x) - i e^{-ix} (\cos x + i \sin x) = 0.\] Since \(f'(x) = 0\), the function is a constant.
Also f(0) = 1, which implies f(x) = 1.
Thus, \(e^{ix} = \cos x + i \sin x\).
PS: One of my students told me last week about the new podcast (Ben, Ben, and Blue) that Grant Sanderson (of 3Blue1Brown) hosts on math, computer science and education. It is delightful.
I also came across an interesting way of proving exp(ix) = cos x + i sin x (@fermatslibrary), which I feel compelled to share, since we are already talking about complex numbers.
Let \(f(x) = e^{-ix} (\cos x + i \sin x)\).
The derivative of this function is \[f'(x) = e^{-ix} (i\cos x - i \sin x) - i e^{-ix} (\cos x + i \sin x) = 0.\] Since \(f'(x) = 0\), the function is a constant.
Also f(0) = 1, which implies f(x) = 1.
Thus, \(e^{ix} = \cos x + i \sin x\).
PS: One of my students told me last week about the new podcast (Ben, Ben, and Blue) that Grant Sanderson (of 3Blue1Brown) hosts on math, computer science and education. It is delightful.
No comments:
Post a Comment