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anonymous

  • 5 years ago

Lokisan, I have another complex analysis question for you....what is the integral of 1/z dz if that z is imaginary again....

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  1. anonymous
    • 5 years ago
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    try robapps.com

  2. anonymous
    • 5 years ago
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    thanks!

  3. anonymous
    • 5 years ago
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    Laurie, if you mean z is complex, then the integral of 1/z dz is log(z) for a suitable branch cut. Usually we would take Log(z) which is the principle logarithm. Log(z) is the integral to 1/z because it can be shown that the derivative of log(z) is 1/z (sounds circular, but we have to use the Fundamental Theorem of the Calculus). You would have then\[\int\limits_{}^{}\frac{dz}{z}=Log (z) + c\]where z is in the domain\[D=\left\{ z:|z|>0,-\pi<\arg(z)<\pi \right\}\] and c is a complex constant. The logarithm of z is\[\log z = \ln |z| + i \arg (z)\]and here, specifically, \[-\pi<\arg(z)<\pi\](which is the entire complex plane except for the negative real number line and 0)

  4. anonymous
    • 5 years ago
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    Technically, you have to show that\[\frac{d}{dz}\log(z)=\frac{1}{z}\]which you can do using derivative theorems (I won't go too much more into it - may lose the point here).

  5. anonymous
    • 5 years ago
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    *principal* logarithm, not principle... ><

  6. anonymous
    • 5 years ago
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    ok, sort of see.....the prof told us to use integral of z dz (is complex)

  7. anonymous
    • 5 years ago
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    Hi Laurie, log(z) is complex. You can see it above when I write,\[\log(z)=\ln |z| + i \arg (z)\]This number is complex. You have it in the form,\[x+iy\]where\[x=\ln|z|\]and\[y=\arg(z)\]

  8. anonymous
    • 5 years ago
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    ok, thanks!

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