For each integer \(n>0\), give an example of an analytic function \[ f:\Delta^*\to\mathbb C \] such that:
- \(f\) is nowhere zero;
- \(f\) has an essential singularity at \(z=0\);
- \(\operatorname{Res}(f'/f,0)=n\).
Solution
For the prescribed integer \(n>0\), take \[ \boxed{f(z)=z^n e^{1/z}}. \]
1. The function \(f\) is analytic and nowhere zero on \(\Delta^*\).
Both factors \(z^n\) and \(e^{1/z}\) are analytic on \(\Delta^*\). They are also both nonzero there: \(z^n\neq0\) for \(z\neq0\), and the exponential function never vanishes. Hence their product is analytic and nowhere zero on \(\Delta^*\).
2. The singularity of \(f\) at \(0\) is essential.
The Laurent expansion is \[ f(z)=z^n e^{1/z} =z^n\sum_{k=0}^{\infty}\frac{z^{-k}}{k!} =\sum_{k=0}^{\infty}\frac{z^{n-k}}{k!}. \] There are infinitely many negative powers of \(z\), namely all terms with \(k>n\). Therefore the isolated singularity at \(0\) is essential.
3. The logarithmic derivative of \(f\) has residue \(n\) at \(0\).
Direct differentiation gives \[ f'(z) =nz^{n-1}e^{1/z}-z^{n-2}e^{1/z}. \] Dividing by the nowhere-zero function \(f(z)=z^ne^{1/z}\) yields \[ \frac{f'(z)}{f(z)}=\frac nz-\frac1{z^2}. \] The coefficient of \(z^{-1}\) is \(n\), so \[ \operatorname{Res}\!\left(\frac{f'}f,0\right)=n. \]