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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

try robapps.com

2. anonymous

thanks!

3. anonymous

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

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

*principal* logarithm, not principle... ><

6. anonymous

ok, sort of see.....the prof told us to use integral of z dz (is complex)

7. anonymous

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

ok, thanks!

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