Simple Closed Contour Integral with a simple pole equalling 0? Out of curiosity, can a function with simple poles have their contour integral equalling to 0?
I ask because, say for the function $\ f(z) = \frac{1}{z^2}$, if we integrate it around the closed contour $|z| = 1$, it will be equal to 0 by the fundamental theorem of calculus. However, for $g(z) = \frac{1}{z}$ we cannot say the same because the antiderivative would not be holomorphic along the contour - it has a discontinuity along the negative real axis. Can this be generalised to all functions with simple poles within the contour?
Let $z_0 \in \mathbb C$, r>0, $D=\{z \in \mathbb C: 0&lt;|z-z_0|&lt;r\}$ and suppose that $f:D \to \mathbb C$ is holomorphic and has a simple pole at $z_0$. Then , by Laurent, we have

$$f(z)= \frac{a}{z-z_0}+g(z)$$

with $a \ne 0$ and $g: \{z \in \mathbb C: |z-z_0|&lt;r\} \to \mathbb C$ holomorphic.

Now let $0&lt; \rho &lt;r$. Then we have

$$ \int_{|z-z_0|= \rho} f(z) dz= a\int_{|z-z_0|= \rho} \frac{dz}{z-z_0}+\int_{|z-z_0|= \rho} g(z) dz.$$

Since $\int_{|z-z_0|= \rho} g(z) dz=0$ and $\int_{|z-z_0|= \rho}\frac{dz}{z-z_0}=2 \pi i$, we derive 

$$ \int_{|z-z_0|= \rho} f(z) dz= a 2 \pi i \ne 0.$$
To answer your very first question, sure. Take $\frac{\sin z}{z}$ $-$ it has a residue of $0$ at its pole.