The Euler formula how was it developed?

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  • #1
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Main Question or Discussion Point

I'm reading a book called The Road to Reality by Roger Penrose, and I'm on the chapter for complex logarithms. What I don't understand is how the identity e[tex]\theta i[/tex] = cos [tex]\theta[/tex] + i sin [tex]\theta[/tex] is found through the use of complex logarithms. I also don't understand how if w = ez, z = log r + i[tex]\theta[/tex] if w is in [r, [tex]\theta[/tex]] form.

Basically, he explains in this book that e was chosen as a base in the general form w = bz because it reduces the ambiguity of bz.

I guess I'm just not seeing how the complex logarithm comes into play for the Euler formula being developed.

EDIT: Sorry about the format of the post, I was just trying latex. I hope someone can clear this up for me, because I feel it's something obvious I'm not understanding.
 

Answers and Replies

  • #2
mathman
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Off hand I don't know how Euler first discovered his identity, but I don't think natural logs were involved. However, the easiest way to prove the identity is to expand both sides into power series and see that these series are identical.
 
  • #3
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First you have to think about what it means to write [itex]e^{z}[/itex] for complex z. One way is via power series as already mentioned. Another is in terms if the differential equation [itex]\frac{d}{dz}e^z = e^z[/itex], e0=1 (ez is its own derivative), which I imagine is how the concept arose historically.

If you express ei z = u(z) + iv(z) for real u and v and use the differential equation definition, then simple computation will show that ei z has length 1, its derivative has length 1, and [itex](\frac{d}{dx}e^x)(0)=i[/itex], so the curve z->ei z must trace out the unit circle in the u-v plane counterclockwise at unit-speed. (sin,cosine) also traces out the unit circle at unit speed, so Euler's formula follows.
 
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  • #4
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Maybe this is the proof you mean (I saw this somewhere on mathworld I think)?
let [tex]z=\cos{\theta}+i\sin{\theta}[/tex]
then [tex]\text{d}z = -\sin{\theta} + i\cos{\theta} \, \text{d}\theta= i(\cos{\theta}+i\sin{\theta}) \, \text{d}\theta[/tex]

The integral
[tex]\int{\frac{\text{d}z}{z}}[/tex]
can be evaluated in two ways. First you can say
[tex]\int{\frac{\text{d}z}{z}} = \log{z} = \log{(\cos{\theta}+i\sin{\theta})}[/tex]
On the other hand you can substitute in dz and z to get
[tex]\int{\frac{i(\cos{\theta}+i\sin{\theta})}{\cos{\theta}+i\sin{\theta}}} \, \text{d}\theta =\int{i \, \text{d}\theta[/tex]
so
[tex]\log{(\cos{\theta}+i\sin{\theta})} = i\theta[/tex]

[tex]e^{i\theta} = \cos{\theta} + i\sin{\theta}[/tex]
 
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  • #5
I believe I have seen the page of Euler's introductio in analysin infinitorum where he gives the proof in terms of power series.
 

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