\begin{equation*}
\int _{|z|\ =\ 1}\frac{sin( z)}{z}\\
\\
=\int _{|z|\ =\ 1} \ \frac{sin( z)}{z\ -\ 0}\\
\\
Set\ \zeta ( z) \ =\ sin( z)\\
\\
So,\ by\ Cauchy's\ formula,\\
\\
f( z) \ =\ \frac{1}{2\pi i}\int _{\gamma } \ \frac{\zeta ( z)}{\zeta \ -\ z}\\
\\
\int _{|z|\ =\ 1} \ \frac{sin( z)}{z\ -\ 0} \ =\ ( 2\pi i) \zeta ( 0) \ \\
\\
=\ ( 2\pi i)( sin\ 0) \ =\ 0\ \ \\
\end{equation*}
