Why does e to pi i equal -1? (2015) [video]
This explanation strikes me as a little too aggressive in throwing out the notation with the bathwater, only reaching its result by redefining the terms we already have intuition for into space-stretching operations that don&#x27;t work like arithmetic does in my head.<p>In that sense I think I have the same problem with this proof that I do with the standard one, where you add the Maclaurin series of cos(θ) and i * sin(θ) and match term-by-term with the series for e^(i * θ). The problem is, at the point you can actually show equality, the things on one side aren&#x27;t obviously a rotation and the things on the other side aren&#x27;t obviously an exponential.<p>I&#x27;m not just hear to yell at clouds. I was given a proof that I truly love by a professor I adore, which I think really does give insight into what all these operators are doing. The best video I can find with it is here:<p><a href="https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=-dhHrg-KbJ0" rel="nofollow">https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=-dhHrg-KbJ0</a> (Skip to 7:30 if you&#x27;re already comfortable with the limit definition of e^x)<p>The basic summary is:<p>1) e^iθ  is equal to  (1 + iθ&#x2F;n) ^ n  for large n<p>2) That base, (1 + iθ&#x2F;n), plotted as a complex number, has length approaching 1, angle approaching θ&#x2F;n<p>3) The base squared, (1 + iθ&#x2F;n)^2, by de moivre&#x27;s theorem, forms another point as if the transformation from (0, 1) were repeated twice — that is, the length stays one, and another tiny angle is added for a total of 2θ&#x2F;n<p>4) The full result is therefore n transformations, taking the path along the unit circle, traveling θ and arriving at cos(θ) + i * sin(θ)