hobz Posted October 22, 2008 Share Posted October 22, 2008 I am having problems with understanding how the formula can have a geometrical interpretation by having an imaginary axis perpendicular to the real axis. I mean how can an exponent (of e) multiplied by the imaginary unit be related to an angle? Link to comment Share on other sites More sharing options...
Dave Posted October 24, 2008 Share Posted October 24, 2008 It might be a bit hard to visualise, but I suggest an analytic approach. Take the power series of [imath]e^x[/imath], and substitute [imath]x=i\theta[/imath]. You'll soon see power series for both sine and cosine appear as if by magic Link to comment Share on other sites More sharing options...
hobz Posted October 25, 2008 Author Share Posted October 25, 2008 Truely magic However this holds for ANY value of [math]i[/math] doesn't it? I mean, the derivation of the fact doesn't involve any characteristic if [math]\sqrt{-1}[/math] does it? Link to comment Share on other sites More sharing options...
timo Posted October 25, 2008 Share Posted October 25, 2008 (edited) I would consider the 4-periodicity of exp(ix) under derivatives (i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5=i^1=i) quite an important characteristic for reproducing the sine and the cosine which also have this 4-periodicity under derivatives. Edited October 25, 2008 by timo i¹=i not 1, of course Link to comment Share on other sites More sharing options...
Kyrisch Posted October 25, 2008 Share Posted October 25, 2008 Truely magic However this holds for ANY value of [math]i[/math] doesn't it? I mean, the derivation of the fact doesn't involve any characteristic if [math]\sqrt{-1}[/math] does it? Of course it does, because you get multiple instances of [math]i^2[/math] for which you substitute -1. [math]e^x=[/math] [math]e^{i\theta}=1 + i\theta + \frac{i^2\theta^2}{2!} + \frac{i^3\theta^3}{3!}...[/math] [math]= 1 + i\theta + \frac{-\theta^2}{2!} + \frac{-i\theta^3}{3!}...[/math] [math]= (1 - \frac{\theta^2}{2!} ...) + (i\theta - \frac{i\theta^3}{3!}...)[/math] [math]= \cos\theta + i\sin\theta[/math] 1 Link to comment Share on other sites More sharing options...
hobz Posted October 25, 2008 Author Share Posted October 25, 2008 Ah yes of course. Thanks for pointing that out. So then, the geometrical property of the imaginary axis being perpendicular to the real axis, can be viewed as a consequence of the angular relation between [math]e[/math] and [math]i[/math]? Link to comment Share on other sites More sharing options...
Kyrisch Posted October 26, 2008 Share Posted October 26, 2008 Is the perpendicularity of the two really a geometric property? I thought it was just a convenient way to graph complex numbers, to adopt the cartesian coordinate system, and axes by definition are orthogonal. I mean, what do you mean by the "angular relation" between e and i? Link to comment Share on other sites More sharing options...
hobz Posted October 26, 2008 Author Share Posted October 26, 2008 Well, I suppose you could argue that the term perpendicular is just as valid in an algebraic sense. For example orthogonal functions, and vectors spaces, need only to satisfy a certain criterium to be called orthogonal, but the criterium has roots in geometry analysis. I mean [math]e^{i\phi}[/math] relates [math]e[/math] to an angle (in the complex plane) [math]\phi[/math] via [math]i[/math], such that e.g. [math]e^{i\pi}=-1[/math]. In essence you've already explained this relation in your previous answer, where [math]e^{i}[/math] was related to trigonometric functions. Link to comment Share on other sites More sharing options...
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