Here is the problem. How do I choose the correct denominator?
Compute $$ \int_{\Gamma} \frac{\cos(z)} {z^2(z-3)}\, dz $$ along the contour indicated in Fig below.
Here is my answer. $$ \begin{align} \int_{\Gamma} \frac{\cos(z)}{z^2(z-3)}\,dz &= \int_{\Gamma}\frac{\frac{\cos(z)}{(z-3)}} {z^2} \\ &= \frac{2\pi}{1!} f'(0) \end{align} $$
$$ \begin{align} f(z) &=\frac{\cos(z)} {z-3}\\ f'(z) &=\frac{-\sin(z) (z-3) - \cos(z)}{(z-3)^2} \end{align} $$
$$ \frac{2\pi}{1!} f'(0) = \frac{-2 \pi i}{9} $$
However, the answer given by Chegg wrote like this below. $$ \begin{align} \int_{\Gamma} \frac{\cos(z)}{z^2(z-3)}\,dz &= \int_{\Gamma}\frac{\frac{\cos(z)}{z^2}} {z-3}\\ &= \frac{2\pi i}{0!} f(3) \end{align} $$
$$ \begin{align} f(z) &=\frac{\cos(z)} {z^2}\\ f(3) &=\frac{\cos(3)}{9} \end{align} $$
Since 3 lies outside $\Gamma$, $\frac{\cos(z)}{z^2(z-3)}$ is not analytical on $\Gamma$. $\int_{\Gamma} \frac{\cos(z)}{z^2(z-3)}\,dz = 0$
Which one is correct?