I want to find the value of the contour integral $$ \int_C \frac{[g(z)]^4}{(z-i)^3} \,\mathrm{d} z $$ where $ C $ is the circle centred at origin with radius 2. If I have some values of the function $ g $, how can I find this contour integral?

Specifically, I know $ g $ is an entire function, and $ g(i) = 2, g(4i) = 5, g'(i) = 3, g'(4i) = 6, g''(i) = 4, g''(4i) = 7 $. Is there also a way to use Cauchy's integral formula for this?

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    $\begingroup$ Are you sure of the $|\cdots|$? In general, |g| isn't holomorphic. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 31 at 16:30
  • $\begingroup$ This is a rather odd integral. $\endgroup$ – copper.hat Mar 31 at 17:03
  • $\begingroup$ Sorry meant to type square brackets instead of absolute. Edited the integral accordingly! $\endgroup$ – Breton Thomas Apr 1 at 0:57
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    $\begingroup$ You can use that for analytic $f(z)$ and a counterclockwise contour $C$ encircling the point $w$: $$f^{(n)}(w) = \frac{n!}{2\pi i}\oint_C\frac{f(z)dz}{(z-w)^{n+1}}$$ $\endgroup$ – Count Iblis Apr 1 at 1:09
  • $\begingroup$ @CountIblis thanks! if you can add this as an answer I'll mark it as accepted $\endgroup$ – Breton Thomas Apr 1 at 1:47

For a function $f(z)$ that is complex differentiable, Cauchy's integral formula

$$f(w)=\frac{1}{2\pi i}\oint_C\frac{f(z)dz}{z-w}$$

where $C$ a counterclockwise contour that encircles the point $w$, is valid. From this formula, one can deduce that a complex differentiable function is in fact infinitely differentiable, with the $n$th derivative given by:

$$f^{(n)}(w)=\frac{n!}{2\pi i}\oint_C\frac{f(z)dz}{(z-w)^{n+1}}$$

The integral in the question can thus be expressed in terms of the second derivative of $g(z)^4$ evaluated at $z = i$.


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