Thread: non-existence of an antiderivative of a complex function

1. non-existence of an antiderivative of a complex function

Explain in detail why the fact that the integral S(1/z)dz=1/(2(pi)i) taken on the unit circle with center zero i.e lzl=1 ensures that f(z)=1/z does not have an antiderivative on the region {z: z does not equal 0}.

my thoughts
f(z) is continuous on the region as the region does not include z=0 and the contour lzl=1 lies entirely in the region so the integral should be path independent and so the integral should have an antiderivative (i.e logz) and the integral around closed surves lying entirely in the region should all have values of 0 but 1/(2(pi)i) does not equal 0 so the antiderivative does not exist (also at (1,0) on the circle 1/(2(pi)i) is undefined).

2. Originally Posted by chocaholic
Explain in detail why the fact that the integral S(1/z)dz=1/(2(pi)i) taken on the unit circle with center zero i.e lzl=1 ensures that f(z)=1/z does not have an antiderivative on the region {z: z does not equal 0}.

my thoughts
f(z) is continuous on the region as the region does not include z=0 and the contour lzl=1 lies entirely in the region so the integral should be path independent and so the integral should have an antiderivative (i.e logz) and the integral around closed surves lying entirely in the region should all have values of 0 but 1/(2(pi)i) does not equal 0 so the antiderivative does not exist (also at (1,0) on the circle 1/(2(pi)i) is undefined).

3. Only as a complement: the differential form $\displaystyle w=dz/z$ is closed in $\displaystyle \mathbb{C}-\{0\}$ i.e. for every $\displaystyle z_0\neq 0$ there exists a determination of $\displaystyle \log z$ in a neighbourhood of $\displaystyle z_0$ and that determination is a a primitive of $\displaystyle w$ , but that primitive does not exist on $\displaystyle \mathbb{C}-\{0\}$ as already has been pointed out.