I want to integrate $f(z)=\dfrac{cosh(4z)}{(z-4)^3}$ with respect to $z$ over $C$, where $C$ consists of $|z|=6$ counterclockwise and $|z-3|=2$ clockwise.
$z=4$ is not in $C$ (it is inside $|z-3|=2$). So, $f(z)$ is analytic in $C$.
From what I understand, my book says that since $f(z)$ is analytic, we could write $\int_{|z|=6}f(z)=\int_{|z-3|=2}f(z)$ if they would both be integrated the same direction. But since we have one clockwise and one counterclockwise, i guess we have $\int_{|z|=6}f(z)=-\int_{|z-3|=2}f(z)$.
So, $\int_{|z|=6}f(z) +\int_{|z-3|=2}f(z) = 0$
If I can write $\int_{C}f(z) = \int_{|z|=6}f(z) +\int_{|z-3|=2}f(z)$, my integral would be zero (which is the right answer according to the book), but I'm not sure if and why I would be able to do that?
When you use the Cauchy integral formula, the singularity is inside both disks, and the positively-oriented integrals give the same value. It isn't even necessary to work out exactly what it is. Then when we subtract the integrals, we get $0$.