Simple Closed Contour Integral with a simple pole equalling 0?

Question

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?

Answer 1

To answer your very first question, sure. Take $\frac{\sin z}{z}$ $-$ it has a residue of $0$ at its pole.

Answer 2

Let $z_0 \in \mathbb C$, r>0, $D=\{z \in \mathbb C: 0<|z-z_0|<r\}$ 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 $g: \{z \in \mathbb C: |z-z_0|<r\} \to \mathbb C$ holomorphic.

Now let $0< \rho <r$. 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

$$ \int_{|z-z_0|= \rho} f(z) dz= a 2 \pi i \ne 0.$$

Answer 3

I prefer to look at this from the point of view of Cauchy's integral.

Cauchy's integral: Let $f$ be analytic around the origin, then $$ f(v) = \oint\frac{f(z)}{z-v}dz $$ if everything is taken in the domain of analyticity and $v$ is enclosed by the contour of integration. We can differentiate this relation and we get $$ f^{(n)}(v) = \oint\frac{n!f(z)}{(z-v)^{n+1}}dz. $$ This implies that $$ f^{(n)}(0) = \oint\frac{n!f(z)}{z^{n+1}}dz $$ which can be rewritten as $$ \oint\frac{f(z)}{z^{n+1}}dz = \frac{f^{(n)}(0)}{n!} $$ Now let $f$ be constant, $f(v)=c$ then we have $$ \oint\frac{f(z)}{z^{n+1}}dz = \oint\frac{c}{z^{n+1}}dz = \frac{f^{(n)}(0)}{n!}. $$ This means that unless $n=0$, the integral vanishes.