Computing the integral using cauchy's theorem I need to integrate
$$\oint _{|z|=1} \frac {\sin z}{z}\, dz$$
I write $\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-...$, then I get by dividing by $x$, the series $1-\frac{z^2}{3!}+\frac{z^4}{5!}-...$. I am confused as to how to use this to integrate. What happens to $\frac{z^2}{3!}+\frac{z^4}{5!}-...$ now?
Thanks in advance!
 A: We manage write the function $\frac{\sin z}{z}$ as a power series
$$
F(z) = 1-\frac{z^2}{3!}+\frac{z^4}{5!}-...,
$$
which converges in $\mathbb C$. That means $F(z)$ is analytic and by Cauchy's theorem
$$
\oint_{|z|=1}F(z)\, dz = 0.
$$

To be more formal note that $\frac{\sin z}{z} = F(z)$ when $z\in\mathbb C\setminus 0$, and, in particular when $|z|=1$. So 
$$
\oint_{|z|=1}\frac{\sin z}{z}\, dz = \oint_{|z|=1}F(z)\, dz = 0.
$$
A: The function $f(z)=sin(z)$ is entire. In particular, it is analytic on the unit disk. By Cauchy's Formula, we have 
$$
f(z_0)=\frac{1}{2\pi i}\int_{\|z\|=1}\frac{f(z)}{(z-z_0)}dz
$$
for any $z_0$ in the interior of the unit disk. Now apply this for a suitable choice of $z_0$. 
A: Note that $\lim_{z \to 0} \frac{\sin z}{z} = 1$, and that the function
$$
f(z) = \begin{cases}
\sin z & z \neq 0 \\
1 & z = 0
\end{cases}
$$
is entire (that is, $\sin z/z$ has a removable discontinuity at $z = 0$).  Because $f$ is entire, we have
$$
\oint \frac{\sin z}{z}\,dz = \oint f(z)\,dz = 0
$$
