Show that \(\sin(z)/z\) has no poles.
Solution
The function \(\sin(z)/z\) is holomorphic on \(\CC\setminus\theset{0}\), and \(\sin(z)\) has a simple zero at \(0\). The Laurent expansion about zero is \[ \inverseof{z} \sin(z) = \inverseof{z}\qty{ z - {z^3 \over 3!} + {z^5\over 5!} - \cdots} = 1 - {z^2\over 3!} + {z^4 \over 5!} - \cdots ,\] which has no terms \(z^{-k}\) with \(k\geq 1\). So \(z=0\) is a removable singularity, and \(\sin(z)/z\) has no poles.