Cauchy Integral Formula for $\oint_{\gamma_i} \frac{z^2+1}{z(z-8)}dz~~~\gamma_i = \mathcal C(3,i), ~~i=1,4,6$

I have a question that i'd like to check my working on. Calculate the integral of $$\oint_{\gamma_i} \frac{z^2+1}{z(z-8)}dz~~~\gamma_i = \mathcal C(3,i), ~~i=1,4,6$$

(a) $\gamma_1$ is the circular contour, positively oriented, with centre 3 and radius 1.

(b) $\gamma_4$ is the circular contour, positively oriented, with centre 3 and radius 4.

(c) $\gamma_6$ is the circular contour, positively oriented, with centre 3 and radius 6.

My attempt:

For a) I wrote $a=0$ and $a=8$ are critical points but are outside the sketch so the answer is $0$

For b) I wrote $a=0$ is inside the circle so I wrote $f(z)=\frac{z^2+1}{z}$ and so = $\int \frac{f(z)}{z-8}$ using the formular we get $2\pi if(8)$ = $\frac{69}{4}\pi i$

and for c) both critical points $a=0$ and $a=8$ lie in the circle but i'm not sure how to do this bit

• similar method to (b), only this time you have to apply it twice, that is, you have to define $f(z)$ twice and add up the answers you obtain. Commented Nov 18, 2017 at 13:47
• I made some progress I think would it be 69/4πi -1/8? Commented Nov 18, 2017 at 14:06
• @SarahAngel It that not $\pi i\frac{65}4$ instaed of $$\pi i\frac{69}4$$ Commented Nov 18, 2017 at 14:44

For c) you can decompose the fraction as follows $$\frac{z^2+1}{z(z-8)} =1+- \frac{1}{8z}+\frac{65}{8(z-8)}$$
Thus, $$\oint_{\gamma_6} \frac{z^2+1}{z(z-8)}dz=\oint_{\gamma_6} dz-\frac{1}{8}\oint_{\gamma_6} \frac{dz}{z}+\frac{65}{8}\oint_{\gamma_6} \frac{dz}{(z-8)}\\-\frac{2\pi i}{8}+\frac{2\pi i*65}{8} =\color{red}{16\pi i}$$
Given that $0, 8\in D(3,6)$ and $\oint_{\gamma_6} dz=0$.