Find the orders of zeros of the following functions:
- \((e^z-1)^3\)
Solution
- The zeros are \(z=2\pi i k\), \(k\in\ZZ\), each of order 3: if \(z_0\) is a zero of order \(n\) for \(f\), then it is a zero of order \(kn\) for \(f^k\). The zeros of \(e^z-1\) are exactly the points \(2\pi ik\), and \(\dd{}{z}(e^z-1)\mid_{z=2\pi ik} = e^{2\pi ik}=1\neq 0\), so each is a zero of order 1.