removable singularity

Question

Let $C$ be the positively oriented boundary of the square with vertices $(1,0)$, $(1,-i)$, $(-1,-i)$ and $(-1,0)$. If $$ f(z)=\frac{\sin(z)}{z}, $$ then clearly $f$ has a removable singularity on $z=0$. This means that $f$ is analytic on $z=0$? My real question is, since $f$ is analytic inside $C$, can I apply cauchy's thm to say that $$ \int_C f(z)dz =0 ? $$ The fact that on $C$ there exist a singularity makes me think I can´t, however, since it is a removable singularity I think that I can…. What is the correct thing to do here?

Answer 1

In fact,

$$\frac{\sin z}z=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots+(-1)^n\frac{z^{2n}}{(2n+1)!}+\cdots$$

Since the radius of convergence of this power series is $\infty$, this function is entire.

(Of course, the above equality has no sense for $z=0$, but RHS is defined at $z=0$ and is equal to $\sin z/z$ elsewhere).