remove.htm Removable and Nonremoveable Singularities

Removable and Nonremovable Singularities
Let Ω be an open set. Let f(z) be analytic in Ω. A complex number z0 in the boundary of Ω is a removable singularity for f [with respect to Ω] if there is a neighborhood1 Bz0, ϵ of z0, and a function g(z), analytic in Bz0, ϵ, such that g(z) = f(z) in Bz0, ϵ ∩Ω.
In this case the function f(z) can be continued analytically to a larger open set, Ω∪Bz0, ϵ. The process is called analytic continuation.


Examples
•
For example, if Ω = B0,r\{0}, and f(z) is analytic and bounded in Ω, then the point z=0 is a removable singularity for f(z).
•
if Ω = B0,r, and z0 ∈ ∂Ω and f(z) is analytic in Ω, but f(z) is unbounded for z near z0 then the point z=z0 is a nonremovable singularity for f(z) with respect to Ω.
•
The function f(z) = ln(z) may be defined in the set
Ω = {z = x + iy | x > 0}
as
ln(z) = ln|z| + i arg(z), − π
2
 < arg(z) < π
2
 .
The only point on the imaginary axis, ∂Ω, which is not a removable singularity for f(z) with respect to Ω is z = 0.
•
The function f1(z) = ln(z) may be defined in the set
Ω1 = C\{z = x + i 0 | x ≤ 0}
as
ln(z) = ln|z| + i arg(z), −π < arg(z) < π.
Then all points on the nonnegative real axis, ∂Ω1, are nonremovable singularities for f1(z) with respect to Ω1. This example shows that whether the singularity is removable depends on the original domain of definition.


Radius of Convergence for Power Series
Suppose that f(z) is analytic in |z| < R and there is a point z0, |z0| = R, such that z0 is a nonremovable singularity for f(z) with respect to B0,R. Then the radius convergence for the series
f(z) = ∞
∑
n=0 
 an zn,
is exactly R.
Proof: We know that the radius of convergence is at least R. If the series converges for |z| < R1, R1 > R, let
g(z) = ∞
∑
n=0 
 an zn, |z| < R1.
Then g(z) = f(z), |z| < R1, but g(z) is analytic at z=z0.

Footnotes:

1We use the notation
Bz0, ϵ = {z ||z − z0| < ϵ}.
for the open disk centered at z0 of radius ϵ.

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On 01 Oct 2012, 19:46.