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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.

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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 7:12
up vote 1 down vote accepted

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.

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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. – a.r. Feb 27 '12 at 13:09

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