removable singularity

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?

• Yes $f$ is analytic and you can use the Cauchy Integral formula. – Tim Raczkowski Jun 5 '15 at 19:07
• @TimRaczkowski thanks a lot! – Leo Sera Jun 5 '15 at 19:36

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