- #1

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Say I want to solve [itex]\oint_C \frac{\sin z}{z^2} dz[/itex] where C is the unit circle. I do a Taylor series expansion and get:

[itex]\frac{1}{z^2} [ \frac{1}{1!}z-\frac{1}{3!}z^3+ \frac{1}{5!}z^5-\ldots] = \frac{1}{z}-\frac{1}{3!}z+ \frac{1}{5!}z^3-\ldots[/itex]

My next step in reasoning is that I can use the Cauchy Integral Theorem on all the terms except [itex]\frac{1}{z}[/itex]; in other words, these terms are all equivalent to 0.

I only need to use the Cauchy Integral Formula on the first term. This gives the solution of [itex] 2 \pi i [/itex].

Is this a proper / appropriate rationale? Thanks.