Example for complex integral

Question

In my question I use the following notations:

$r,R\in\mathbb{R};$ $\alpha,\beta\in\mathbb{C};$

$A'\left(\alpha,R\right)\dot{=}\left\{ z:z\in\mathbb{C},\left|z-\alpha\right|\leq R\right\};$

$A\left(\alpha,R\right)\dot{=}\left\{ z:z\in\mathbb{C},\left|z-\alpha\right|<R\right\};$

$C\left(\alpha,R\right)\dot{=}\left\{ z:z\in\mathbb{C},\left|z-\alpha\right|=R\right\}$.

I found the following theorem: Let $f$ be an analytic function on a simply connected domain $D$ and $A'\left(\alpha,R\right)\setminus\left\{ \beta\right\} \subset D$, where $\beta\in A\left(\alpha,R\right)$. If $0<r<R-\left|\beta-\alpha\right|$, then

$$\int_{C\left(\alpha,R\right)}f\left(z\right)dz=\int_{C\left(\beta,r\right)}f\left(z\right)dz.$$

Can you give me an example, where the two sides of the equation above is not zero?

Answer 1

As indicated by @JoséCarlosSantos this is not possible since the function is assumed to be analytic throughout the domain $D$ and by the Cauchy Goursat theorem integration along any closed path will be zero. The situation is more interesting when the function has a point on non-analyticity within the domain such as a pole. Then, if you assume such a point is contained within the interior of both curves (the point has winding number equal to say 1 with respect to both curves if you like) then the statement still holds and both integrals may not be zero. This is an example of a deformation of contour which is a wonderfully useful technique used throughout complex analysis.

Answer 2

Since $D$ is simply connected, $f$ has an antiderivative, and therefore its integral along any loop is equal to $0$. So, no, nobody can do that.