$\sin z/z$ has no poles

For \(z\ne0\), the quotient \(\sin z/z\) is holomorphic. At \(z=0\), use the Taylor series \[ \sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots. \] Then \[ \frac{\sin z}{z} =1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots, \] which converges near \(0\) and extends the quotient holomorphically there with value \(1\). Hence the apparent singularity at \(0\) is removable, and \(\sin z/z\) has no poles anywhere.