integrals with cauchy theorem

Solution 1:

You have the wrong formula in the post. The formula should be $$f(z)=\frac1{2\pi{i}}\oint_{\gamma}\frac{f(w)}{w-z}\,\mathrm{d}w,$$ where $\gamma$ is a closed contour in $U,$ the domain of $f,$ such that the region $D$ enclosed by $\gamma$ contains $z.$ It is very important that $f$ be holomorphic in the region enclosed, for this formula to work, and it is necessary that there be no singularities in the region enclosed either. If the region does enclose singularities, then instead, $$\frac1{2\pi{i}}\oint_{\gamma}\frac{f(w)}{w-z}\,\mathrm{d}w=\sum_{m=0}^n\mathrm{I}(\gamma,c_m)\mathrm{Res}\left(\frac{f(w)}{w-z},c_m\right),$$ where the $c_m$ are the singularities enclosed, and $\mathrm{I}(\gamma,c_m)$ is the corresponding winding number.

I assume, from the context of your post, is that the contour given to you is a closed curve $\gamma$ as a function from $[0,2\pi]$ or $[-\pi,\pi]$ to $\mathbb{C},$ whose graph is $\gamma(t)=2+3e^{it},$ since you said the curve was a circumference with center $2.$ The domain of $f$ is then some open set $U$ that contains this circumference, and the region enclosed by this curve, $D,$ is the region where you need to check for singularities. The specifics of $U$ do not actually matter for the purpose of your exercise, only the region $D,$ which is the region enclosed by the given $\gamma,$ matters.