Simple contour intergal along a circular path

Question

Am I correct that the the following integral evaluates to 0 since the domain of integration is a closed loop and the integrand is continuous over the loop? $$\int_{C(0,7)}\frac 1{(z-1)(z-3)} dz$$

Answer 1

It's just a coincidence to have the correct result, but your reason is wrong, indeed the poles $1$ and $3$ are in $C(0,7)$ so by Residue theorem we have $$\int_{C(0,7)}\frac 1{(z-1)(z-3)} dz=2i\pi(Res(f,1)+Res(f,3))=2i\pi(-\frac{1}{2}+\frac{1}{2})=0$$

Answer 2

Let $C(0,\,7)$ be a circle with the center at $z=0$ and radius $r=7.$ Function $$f(z)=\frac {1}{(z-1)(z-3)}$$ has simple poles at $z_1=1,\;\;z_2=3,$ so $f(z)$ is not analytic in the $\operatorname{int} C(0,\,7).$ However, $$ \int\limits_{C(0,7)}\frac 1{(z-1)(z-3)} dz=0 $$ since $\operatorname{Res}\limits_{z=\infty}{\frac {1}{(z-1)(z-3)}}=0.$