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?
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 $z = i$.