Laurent series expansion of complex functions

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suppose is an isolated singularity of f, then
is removable the Laurent expansion is a Taylor series - there are no negative powers of in the expansion.
is a pole of order n there are exactly n terms in the Laurent expansion with negative powers of
is an essential singularity there is infinite number of terms in the Laurent expansion with negative powers of

1.
(1.1)
There are two singularities for this function, namely and .
In order to expand it into a Laurent series around we need to exclude the singularity, hence we expand it for
Note that is already a Laurent expansion, so all that is left is to expand around . Since the function is analytic in the domain it is a simple Taylor expansion:
(1.2)
 
Therefore:
(1.3)
The point is a pole of order 1 point since there is only one term of with a negative power.
The residue of the function at this point is the coefficient of the term which in this case it is simply
(1.4)

2.
The function is:
(2.1)
Obviously the singular point is
We can use the Taylor's expansion for the exponent:
(2.2)
Therefore:
(2.3)
The domain is
There is infinite number of terms with negative power, hence this is an essential singularity.
The residue is the coefficient of the term which in this case:
(2.4)

3.
The function is
(3.1)
the function is analytic at , so we can just use its Taylor series expansion:
(3.2)
Thus:

(3.3)
The domain is
If we explicitly write the terms with negative powers we get:

(3.4)
We see that there is one term with a negative power, so the singular point is a pole of order 1 and the residue is:
(3.5)
4.
The function is:
(4.1)
At the numerator is analytic, so we can expand it into a Taylor series:
(4.2)
Thus:
(4.3)
there are no other singularities for this function, hence the domain is
Again, we explicitly write the terms with the negative powers:
(4.4)
there are two terms with negative powers of , hence this singular point is a pole of order 2.
The residue is:
(4.5)

5.
Riemann theorem of removable singularities:
If is analytic in the punctured disk and
(5.1)
Then is a removable singularity.
Further, we can write in terms of its Laurent series:
(5.2)
 
Since is a removable singularity then:
(5.3)
We can bring the limits into the sums:
(5.4)
Hence all the coefficients for must be identically zero. For the sum converges to zero for any finite
So we are left with:
(5.5)
But this coefficient is the residue of at so we get for a removable singularity:
(5.6)

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