- #1
thelema418
- 131
- 4
I'm trying to understand how to use Taylor series expansion as a method to solve complex integrals. I would appreciate someone looking over my thoughts on this. I don't know if they are right or wrong or how they could be improved. I suppose that my issue is that I don't feel confident in my solution because it is so abstract and I don't have anything to justify my answer with in the real world...
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.
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.