Cauchy's integral formula on unit circle $|z|$=2

Question

I want to calculate an integral $$\int_{|z|=2}\frac{\sin(z)}{z^3-z^2}dz.$$ I am pretty sure that this is a pretty straight forward Cauchy integral, but what I want to know is this: what do I do different when I calculate over $|z|=2$ compared to if I were to calculate the integral over $|z|=1$?

Answer 1

The denominator vanishes when $z=0$ 0r $1$. Both of them are in side $|z|=1$. Both are poles for the integrand. The value of the integral is $2\pi i(A+B)$ where $A$ and $B$ are the residues at these poles. Do you know how to find the residues? The answer is $A=-1$ and $B =\sin 1$ so the integral is $2\pi i(-1+\sin 1)$.

For a pole $z_0$ of order $1$ of $f(z)$ the residue is just $\lim_{z \to z_0} (z-z_0)f(z)$,

Note: the integrand has a pole on $|z|=1$. The integral over the cirlce $|z|=1$ does not even exist.

Answer 2

Let $f(z):= \frac{\sin z}{z^3-z^2}$, then $f(z)=\frac{\sin z}{z^2(z-1)}$.

We see that $f$ has poles in $z=0$ and $z=1$ of order $1$ (oberve that $ \frac{ \sin z}{z} $ has a removable singularity in $z=0.$)

Now invoke the residue theorem to compute $\int_{|z|=2}\frac{\sin(z)}{z^3-z^2}dz.$