Homotopy invariance of winding number in complex analysis.

Question

In topology, the winding number is homotopy invariant under the definition $n(\gamma,a)=\frac{\tilde{\theta}(\beta)-\tilde{\theta}(\alpha)}{2\pi}.$

I assume the must be true in the framework of complex analysis. Suppose you take as definition for the winding number $n(C,a)$ of a curve $C$ through $a$ to be $$ n(C,a)=\frac{1}{2\pi i}\int_C\frac{dz}{z-a}. $$

Is is still true that $n(C,a)$ is hopotopy invariant under smooth curves $C$ not going through $a$? Thank you.

It seems intuitively that it would be the case that $n(C,a)$ is homotopy invariant under $C$, but I don't know how to prove it. — Samuel Reid, Feb 25 '12 at 07:12

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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Both definitions agree, so the winding number is homotopy invariant. See Stewart and Tall, "Complex Analysis", section 7.5. You might be also interested in this question and, perhaps, my answer therein.

edited Apr 13 '17 at 12:20

Community

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answered Feb 25 '12 at 07:14

Agustí Roig

17,959

Thanks, is it possible to show it without referring back to the topological definition? – Dedede Feb 27 '12 at 11:00
Indeed: look at op.cit., theorems 9.3 and 9.4. – Agustí Roig Feb 27 '12 at 13:09

Homotopy invariance of winding number in complex analysis.

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