Cauchy Integral Formula Question $\int_{C(2,1)} (z^4-\frac 1 z)dz$

Question

I'm looking to compute the following integral

$$\int_{C(2,1)} (z^4-\frac 1 z)dz$$.

My question here is does Cauchy’s Integral Formula hold or even apply here?

Is $f(z)=(z^4-\frac 1 z)$ entire and therefore differentiable on $D(2,1 + \epsilon)$ for any $\epsilon > 0$.

If so my answer would just simply be $I = 2\pi i f(i) = 2\pi i(1-\frac 1 i) = 2\pi i - 2\pi$

Any help would be great thanks.

Answer 1

Since the disk centered at $2$ with radius $1$ is simply connected and it is contained in the domain of $z^4-\dfrac1z$, Cauchy's integral theorem tells us that that integral is equal to $0$.

Answer 2

The integrand is analytic on the disk of radius $1$ centered at $2$. Thus by Cauchy's theorem, the integral is zero.

If instead you take a closed contour enclosing zero, you get $-2\pi i$, since $z^4$ is entire and $1/z$ has a simple pole.