Cauchy Integral Formula to evaluate a Taylor Coefficient

Question

I need to compute the Cauchy Integral Formula $${1\over{2{\pi}i}}\int {f(z)\over(z-a)^{k+1}}$$

$$Where\, k=3\,and\,z=e^{sin(z)}\,at\, z=i$$

I need help figuring out how to do this exactly, I get to this point: $${1\over{2{\pi}i}}\int {e^{sin(z)}\over(z-i)^{4}}$$ From here I am unsure of what I am suppose to do to evaluate this integral. I have done the differentiation side and gotten this ugly thing: $$e^{sin(i)}(cos^3(i)-3sin(i)cos(i)-cos(i))\over 6$$ Basically I want to show that my integral equals the $3^{rd}$ Taylor coefficient but am unsure of how.

Answer 1

To see this write out the Taylor series

$$f(z) = \sum_{n=0}^\infty {f^{(n)}(a)(z-a)^n\over n!}$$

Now divide and integrate term-by-term (you should have already proven this is a legal maneuver in class)

$${1\over 2\pi i}\oint {f(z)\over (z-a)^{k+1}}\,dz = \sum_{n=0}^\infty {1\over 2\pi i}\oint {f^{(n)}(a)(z-a)^{n-k-1}\over n!}$$

Since all terms except the $k=n$ term have analytic primitives on the entire curve, we see their integrals are zero, and we are left only with

$${1\over 2\pi i}\oint {f^{(k)}(a)\over k!(z-a)}={f^{(k)}(a)\over k!}$$

Now in your case $k=3$ and the result follows.