Let V=C−[0,1]V = \mathbb{C} – [0,1]. Show there exists a function ff that is holomorphic in VV such that f′(z)=1z−1z−1f'(z) = \frac{1}{z} – \frac{1}{z-1}.

Attempt: Let γ⊂V\gamma \subset V be a piecewise smooth curve. Then consider ∫γf′(z)dz=∫γ1z−1z−1dz=∫γ1zdz−∫γ1z−1dz=2πi∗Win(γ,0)−2πi∗Win(γ,1) \int^{}_{\gamma}f'(z)dz = \int^{}_{\gamma} \frac{1}{z} – \frac{1}{z-1}dz = \int^{}_{\gamma} \frac{1}{z}dz – \int^{}_{\gamma} \frac{1}{z-1}dz = 2 \pi i *Win(\gamma , 0) – 2 \pi i *Win(\gamma, 1) . Now I am not sure what to do.

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Show that the integral depends only on the endpoints of the curve, and not the specific curve chosen.

– Daniel Fischer♦

2 days ago

The endpoints of the curve γ\gamma?

– James.Welch

2 days ago

Yes. That is equivalent to showing that the integral over every (piecewise smooth) closed curve in VV vanishes.

– Daniel Fischer♦

2 days ago

Does this have anything to do with homotopy?

– James.Welch

2 days ago

It has something to do with homotopy, but it’s more basic. Winding numbers are all you need.

– Daniel Fischer♦

2 days ago

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