integrating $\displaystyle\int_{\gamma}\frac{1+(-iz-2)^9}{16-(-iz-2)^2}\,dz$

Question

$$\int_{\gamma}\frac{1+(-iz-2)^9}{16-(-iz-2)^2}\,dz$$

Where $\gamma$ is a half circle entered at $2i$ of radius $2$ connection $0$ to $4i$

$$\int_{\gamma}\frac{1+(-iz-2)^9}{16-(-iz-2)^2}\,dz=\int_{\gamma}\frac{1+(-iz-2)^9}{16-(iz+2)^2}\,dz$$

To find points where the function is not analytic:

$$16-(iz+2)^2=0\iff 16-(-z^2+4iz+4)=0\iff z^2-4iz+12=0$$

So $z=\dfrac{4i\pm\sqrt{-16-48}}{2}=\dfrac{4i\pm 8i}{2}$

$z_1=6i$ and $z_2=-2i$

So we get

$$f'(2i)=\frac{-9i(-iz-2)^8(z-6i)+1+(-iz-2)^9}{(z-6i)^2}=-\frac{1}{16}$$

Is it correct?

Answer 1

You might want to check again your calculation when you plug in $2i$, but this is basically how you solve. Don't forget to multiply by $2\pi i$If you had the pole $2i$ inside the domain you would get that $$I=2\pi i \frac{-72\cdot4^8+1+4^9}{-64}$$ But since there are no poles the integral is 0 due to Cauchy theorem since no poles are inside the domain of integration.