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Where'd I go wrong?

  1. Feb 1, 2005 #1
    I was messing around and decided to prove Euler's 'formula' using a method that doesn't involve power series. Here's how I did it:

    [tex]z=\cos\theta + i \sin\theta[/tex]

    [tex]\frac{dz}{d\theta} = -\sin\theta + i\cos\theta[/tex]

    [tex]i\frac{dz}{d\theta} = -i\sin\theta + i^{2}\cos\theta = -i\sin\theta - \cos\theta = -(\cos\theta+i\sin\theta) = -z[/tex]

    [tex]-i\int \frac{1}{z} \; dz = \int \; d\theta[/tex]

    [tex]-i \log|z| = \theta + C[/tex]

    When [itex]\theta=0[/tex], [itex]z=1[/itex], so [itex]C=0[/itex]. Now:

    [tex]\log|z| = \frac{-\theta}{i}[/tex]


    Last edited by a moderator: Feb 1, 2005
  2. jcsd
  3. Feb 1, 2005 #2
    Nice work. Calculate 1/i in terms of a+ib, where a and b are real. :smile:
  4. Feb 1, 2005 #3


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    Nothing as far as I can tell; you just didn't finish.

    [tex] \frac{-\theta}i = \frac{-\theta i}{i^2} = \frac{-i\theta}{-1} = i\theta [/tex]

    ETA: Oops, too slow on the draw.
  5. Feb 1, 2005 #4
    Nice! I was too busy checking my first steps to notice that I had to simplify my answer!

  6. Feb 2, 2005 #5

    matt grime

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    There are, though, several things you have sidestepped when doing that integral of 1/z, such as the fact that log is an many valued function, and why when you exp log of |z| you don't get |z|, for instance. I mean, in

    log|z| = \theta/i

    the lhs is strictly real and the rhs is strictly imaginary.
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