Euler's formula

[hobz]

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?

[Dave]

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

[hobz]

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?

[timo]

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

[Kyrisch]

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]

[hobz]

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]?

[Kyrisch]

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?

[hobz]

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.