Can the residue of a singularity be zero?

Question

If a complex-valued function $f$ has a singularity at some point $z_0$, it has a Laurent expansion $$ f(z) = \sum_{n = -\infty}^\infty a_nz^n, $$ where the coefficient $a_{-1}$ is known as the residue. Because each term of the Laurent expansion has a primitive except $a_{-1}z^{-1}$, we can compute the integral along any closed curve containing $z_0$ with $$ \tag{$\star$} \int_\gamma f(z)dz = 2\pi ia_{-1}. $$ My question is, can the residue of a singularity be zero? It doesn't seem so; otherwise, by $(\star)$, the integral along any closed curve in the domain of $f$ would be $0$, which means $f$ at least behaves like a holomorphic function. If this is correct, how might I prove it?

Answer 1

Yes, it can be $0$. For instance, since$$\frac1{z^2}=\cdots+0\times\frac1{z^4}+0\times\frac1{z^3}+\frac1{z^2}+\color{red}0\times\frac1z+0+\cdots,$$the residue of $\frac1{z^2}$ at $0$ is $0$.

And, yes, the integral of $\frac1{z^2}$ along a closed path is always $0$. That's not surprising, since $\frac1{z^2}$ has an antiderivative ($-\frac1z$).