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Complex Analysis: Singularities and Casorati-Weierstrass Theorem (University of Leicester)

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Antri Stylianou

on 18 December 2012

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Transcript of Complex Analysis: Singularities and Casorati-Weierstrass Theorem (University of Leicester)


Charalampidi Marianna
Christou Athanasia
Dimitriou Georgia
Loizidou Marina
Papadaki Ifigenia
Polykarpou Thekla
Stylianou Antri Singular Point
Singular point of a function f(z) is a point at which f(z) is not analytic.

A function f(z) is said to be analytic at some point z0 if it is differentiable at every point of a certain neighborhood of z0.

f(z) is analytic if and only if there exists a neighborhood N(z0;ε), ε>0 such that f’(z) exists for all zєN(z0;ε). Non-isolated singularity
A singular point z=z0 of a function f(z) is non-isolated if every neighborhood
of z0 contains at least one singularity of f(z) other than z0. Example:

1. f(z)= is nowhere analytic so that every point in is a
non-isolated singularity.

2. , the point z=0 is a non-isolated singularity
of Lnz since every neighborhood of z=0 contains points
on the negative real axis.
Principal part
Laurent Series Taylor Series Removable singularity
The principal part vanishes and the Laurent series is essentially a Taylor series. The series represents an analytic function in |z-z0|<R.

Example: = has an essential singularity at the origin. Essential singularity
The principal part has infinitely many non-zero terms.

Example: Examples:

1. f(z)= has 0 as an isolated singularity.

2. f(z)= has 2 as an isolated singularity. Isolated Singularity
Existence of a neighborhood of z0 in which z0 is the only singular point of f(z). with bk ≠ 0. Note: k=0 removable singularity
eg. f(z)=z

k=1 simple pole
eg. f(z)= 1/z Pole of order k
Pole is a type of singularity in which the Laurent expansion in the deleted neighborhood of z0 is
In general, if we multiply f(z) by (z-z0)m and take the limit z→z0, then Method of finding the order of a pole

If z0 is a pole of order k, then Example 2: Suppose that z0 is an essential singularity of a
function f, and let ω0 be any complex number.
Then, for any positive number ε,
|f(z)-ω0|< ε
is satisfied at some point z in each deleted
neighborhood of z0. Proof (by contradiction):

f is analytic in any deleted neighborhood of z0 and z0 is an essential singularity such that
0 <|z- z0|< δ, for some δ.

Claim: |f(z)-ω0|≥ ε, 0 <|z- z0|< δ for any z. Case 2:
If g(z0)=0, then g must have a zero of some finite order m at z0.
Define , ψ(z0) ≠ 0.
=>f has a pole of order m. Case 1:
If g(z0) ≠ 0, then , 0<|z-z0|<δ,
becomes analytic at z0 if .
=>z0 is a removable singularity of f. Define , 0 <|z-z0|< δ, which is bounded and analytic.
Hence, z0 is a removable singularity of g.
Let g be defined at z0 such that it is analytic there. Both cases lead to a contradiction. Hence our Theorem holds. Example 1:
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