- (5 points) Give a counterexample to the assertion: If f is holomorphic on the annulus \(\{ z : 1 < | z | < 3 \}\) , then f extends holomorphically to the disc \(\{ z : | z | < 3 \}\)
- (15 points) Determine whether the following is true: If f is holomorphic on the annulus \(\{ z : 1 < | z | < 3 \}\) , then f extends meromorphically to the disc \(\{ z : | z | < 3 \}\)
Solution
Both assertions are false.
1. Part (a) fails for \(f(z)=1/z\).
The function \[ f(z)=\frac1z \] is holomorphic on the annulus \(1<|z|<3\). Suppose it had a holomorphic extension \(F\) to the disk \(|z|<3\). On the annulus one would have \(zF(z)=1\). Since both sides are holomorphic on the whole disk, the identity theorem [@SS03] would imply \[ zF(z)=1 \] for every \(|z|<3\). Evaluating at \(z=0\) gives \(0=1\), a contradiction. Thus no holomorphic extension exists.
2. Part (b) fails for \(f(z)=e^{1/z}\).
The function \[ f(z)=e^{1/z} \] is holomorphic on \(1<|z|<3\). Suppose there were a meromorphic function \(F\) on \(|z|<3\) agreeing with \(f\) on that annulus. On the punctured disk \(0<|z|<3\), both \(F\) and \(e^{1/z}\) are meromorphic, and they agree on the nonempty open subset \(1<|z|<3\). The identity theorem for meromorphic functions [@SS03] therefore gives \[ F(z)=e^{1/z}\qquad(0<|z|<3). \] But \[ e^{1/z}=\sum_{n=0}^{\infty}\frac{1}{n!}z^{-n} \] has infinitely many nonzero negative Laurent coefficients at zero, so zero is an essential singularity. A meromorphic function at zero can have only a removable singularity or a pole. This contradiction shows that no such meromorphic extension exists.