We are asked to evaluate the following integral:
$$\int_{\gamma}^{} \frac{z}{(z+9)^{2}}$$
Where $\gamma$ is the region $\gamma(t) = 2i + 4e^{it}$
Not quite sure where to take it from here, I know there is a pole of order 2 at $z=-9$ but when trying to find the residue at this pole I end up with a 1. But as this residual point corresponds to a pole which is not in $\gamma$ then it can't be used. Implying by the residue theorem that the integral is 0.
Using the C.I.F $$\int_{\gamma}\frac{f(z)}{z-a} = 2\pi if(a)$$ but I can't apply this here as $z=-9$ is not an element of $\gamma$
Is the entire integral just equal to zero?
Any help is much appreciated, thank you!
Hint : Actually $\gamma:|z-2i|=4$. So $-9 \not \in \gamma$. That is $\displaystyle f(z)=\frac{z}{(z+9)^2}$ is analytic in $\gamma$.
So what about Cauchys theorem ?