Assume that: $f(z)=\frac{1}{(z-2)^2(z-4)}$ , with has singularities at $2$ and $4$ and suppose we have to compute: $$\oint_{C}^{}f(z)dz$$ with $C$ being the positively oriented circle $\mid{z}\mid=3$, or $\mid{z}\mid=5$ (contour containing one or both singular points).
I know that if the contour $C$ does not enclose any singular points, then $f(z)$ is holomorphic inside $C$ and continuous on its boundary, so the Cauchy-Goursat Theorem can be applied and the integral equals zero.
Now, examining the case of the contour enclosing singular points, I can only come up with a principle (how's that called?) which states that, if:
$C_1, C_2$ are two positively oriented curves with $C_2$ completely inside $C_1$
$f$ is holomorphic in the domain between $C_1, C_2$
$f$ is continuous on both their boundaries
then: $$\int_{C_1}f(z)dz=\int_{C_2}f(z)dz$$
Tried it, but the parameterization turns 'complex'. Is there an easier way?
Why not try residue theorem? These are exactly the cases where residue theorem comes into play!