# Finding value of contour integral, if given values of function

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?

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