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The Euler formula how was it developed?

  1. Apr 17, 2009 #1
    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.
     
  2. jcsd
  3. Apr 17, 2009 #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.
     
  4. Apr 17, 2009 #3
    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.
     
    Last edited: Apr 17, 2009
  5. Apr 17, 2009 #4
    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]
     
    Last edited: Apr 17, 2009
  6. Apr 18, 2009 #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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