Suppose $P$ is a covex pentagon enclosing the point $0$ in the complex plane. If $f$ is analytic in $\mathbb C \setminus \{0\}$ and $\int_{P}f(z)dz \neq 0,$ Then
(a) $0$ is a pole or an essential singularity.
(b) $0$ is a removable singularity.
(c)$f$ has an analytic extension to $\mathbb C$
(d)$f$ has a continuous extension to $\mathbb C$
My Solution: If I take $f(z)=\frac{1}{z}$ condition satisfied. It has simple pole. There is no analytic extention nor continuous extension to $\mathbb C$. So, $(a)$ is the correct answer. Am I correct? But How do I prove it?
The singularity is not removable: if $0$ is removable, after removing it, f has an analytic extension to $\mathbb C$. Then by Cauchy integration theorem, the integral should be zero. So (b) and (c) are both wrong.
As for (d), if $f$ is made continuous at $0$, then as $\epsilon \to 0$, $$\int_{\partial B(0,\epsilon)}f(z)dz\to0, \int_{P-\partial B(0,\epsilon)}f(z) dz=0.$$ (the 2nd equality is Cauchy integration theorem) So, $$\int_Pf(z)dz=\int_{\partial B(0,\epsilon)}f(z)dz+\int_{P-\partial B(0,\epsilon)}f(z)\to 0$$
So (d) is also wrong.
So we can only conclude that $0$ is a pole or an essential singularity.