# Evaluate the complex integral $\int_{C_R}\frac{z^3}{(z-1)(z-4)^2}$

Let $C_R$ be the positively oriented circle with centre $3i$ and radius $R > 0$. Use the Cauchy Residue Theorem to evaluate the integral $$\int_{C_R}\frac{z^3}{(z-1)(z-4)^2}$$ Your answer should state any values of R for which the integral cannot be evaluated.

Now I can find the residues of the function easily enough. They are $\frac{1}{9}$ at $z=1$ and $\frac{80}{9}$ at $z=4$. However I have no idea at what radius the points $z=1$ and $z=4$ are included in the region being integrated. Is there a way to calculate the radius at which these points need to be taken into account?