Question

Suppose f(z) is holomorphic in a punctured disc Dᵣ (z₀)−{z₀}. Suppose also that ∣f(z)∣≤A∣z−z₀∣⁻¹⁺^ϵ for some ϵ>0, and all z near z 0. Show that the singularity of f at z₀ is removable.

Answer 1

To show that the singularity of f at z₀ is removable, we need to prove that f can be extended to be holomorphic at z₀. This can be done by finding a value for f(z₀) such that the principal part of the Laurent series is zero.

Step-by-step explanation:

To show that the singularity of f at z₀ is removable, we need to prove that f can be extended to be holomorphic at z₀. In other words, we need to find a value for f(z₀) such that f is holomorphic in a disc containing z₀.

Since ∣f(z)∣≤A∣z−z₀∣⁻¹⁺^ϵ for all z near z₀, we can conclude that the singularity at z₀ is a pole of order at most 1. This means that the Laurent series expansion of f around z₀ has a principal part with a single term of the form c₁⁄(z−z₀), where c₁ is a constant.

To show that the singularity is removable, we can set f(z₀) = c₁. This value will ensure that the principal part of the Laurent series is zero, and thus f(z) will be holomorphic in a punctured disc Dᵣ (z₀)−{z₀}.