- Suppose \(f ( z )\) is holomorphic in a punctured disc \(D _ { r } ( z _ { 0 } ) - \{ z _ { 0 } \}\) . Suppose also that
\[ | f (z) | \leq A | z - z _ {0} | ^ {- 1 + \epsilon} \]
for some \(\epsilon > 0\) , and all \(z \ \mathrm { n e a r } \ z _ { 0 }\) . Show that the singularity of \(f\) at \(z _ { 0 }\) is removable.
Solution
Translate so that \(z_0=0\). Write the Laurent expansion \[ f(z)=\sum_{n=-\infty}^{\infty}a_n z^n \] in a punctured disc. For \(m\ge1\), Cauchy’s coefficient formula on \(|z|=\rho\) gives \[ a_{-m}=\frac1{2\pi i}\int_{|z|=\rho} f(z)z^{m-1}\,dz. \] Using the assumed bound, \[ |a_{-m}|\le \rho\cdot A\rho^{-1+\epsilon}\rho^{m-1} =A\rho^{m-1+\epsilon}. \] Since \(m-1+\epsilon>0\), letting \(\rho\downarrow0\) gives \(a_{-m}=0\) for every \(m\ge1\). Thus the principal part vanishes, so the singularity is removable.