Compute the integrals using the residue theorem Compute the following integrals:
$I:=\int_{|z|=2}\frac{1}{(z-3)(z^{13}-1)}dz$
$J:=\int_{|z|=10}\frac{z^3}{z^4-1}dz$
I do not know where to begin. I know I am supposed to use the substitution $z\rightarrow \frac{1}{t}$. I can see that $I$ has 13 poles inside the circle of radius 2.
Using a hint provided by Winther I was able to get that
$I=2\pi i\cdot\sum\limits_{n=1}^{13}\frac{z_n}{13(z_n-3)}$ where $z_n^{13}=1$
 A: For

$$J:=\int_{|z|=10}\frac{z^3}{z^4-1}dz$$

$$J:=\int_{|z|=10}\frac{z^3}{(z-1)(z+1)(z-i)(z+i)}dz$$
So the roots are: $z=1,z=-1,z=i,z=-i$
Now use the Residue formula for the roots:
$$\text{Res}(f,1)=\lim\limits_{z\to 1}\frac{z^3}{(z+1)(z-i)(z+i)}=\frac{1}{(1+1)(1-i)(1+i)}=\frac{1}{4}$$
$$\text{Res}(f,i)=\lim\limits_{z\to i}\frac{i^3}{(i+1)(i-1)(i+i)}=\frac{-i}{-4 i}=-4$$
$$\text{Res}(f,-i)=\lim\limits_{z\to -i}\frac{(-i)^3}{(-i-1)(-i+1)(-i-i)}=\frac{i}{4i}$$
$$\text{Res}(f,-1)=\lim\limits_{z\to -1}\frac{(-1)^3}{(-1-1)(-i-i)(-1+i)}=\frac{1}{4}$$
So $$J=2\pi i\sum_{j=1}^{4}\text{Res}(f,a_j)$$
A: For 

$$I:=\int_{|z|=2}\frac{1}{(z-3)(z^{13}-1)}dz$$

First you need to find all the roots that in $|z|<2$  then apply the Residue formula, I can't find an easy way to do this (maybe there is). 
A: For $$J=\int_{|z|=10}\frac{z^3}{z^4-1}dz,$$
we see there are poles at $r^4e^{i4\theta}-e^{i2\pi k}=0,$ so that $r=1$ and $$4\theta=2\pi k\implies\theta=\frac{\pi k}{2}\implies\theta\in\left\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\right\}.$$
Since $\left|e^{i\theta}\right|<10$ for all $\theta$ then the residue theorem gives us
$$J=2\pi i\sum_{i=0}^3\text{Res}(f,a_i)=2\pi i.$$
