<s>[INST] Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation.
What was  shown in class:  $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz=\large \int_{|z-2|=2} \frac {5z+7}{(z+3)(z-1)}=\cdots= \int_{|z-2|=2} \frac {2}{z+3}dz+\int_{|z-2|=2} \frac {3}{z-1}dz $
And then it is said that $\large \int_{|z-2|=2} \frac {2}{z+3}dz=0$ while $\large \int_{|z-2|=2} \frac {3}{z-1}dz=3 \cdot 2 \cdot \pi \cdot i$.
Can you please help me understand the last step? What's the difference bewteen the integrals?     
 [/INST] Let $D=\{z: |z-2|\leq2\}$, thus $\partial D=\{|z-2|=2\}$. Since $f(z)=2/(z+3)$ is clearly analytic in $D$, then by Cauchy's Theorem $$ \int_{\partial D} f(z)dz=0.$$
However, if $g(z)=3/(z-1)$, since $1 \in D$, by Cauchy's Integral Formula, letting $g(z)=h(z)/(z-1)$ with $h\equiv 3$ we get
$$
\int_{\partial D} g(z)dz=\int_{\partial D} \frac{h(z)}{z-1} dz= 2\pi  i \cdot h(1) = 2\pi i \cdot 3 = 6\pi i
$$
</s>