Note
to readers: I reiterate once more that this set of blog posts isn’t intended to
be comprehensive nor should they be seen as replacing the hard work of
diligently working problems on your own using available text books. They are
merely intended to spur on curiosity and perhaps offer a few different
perspectives from what one might have seen already.

Included
here are the coefficients in typical Laurent series which are generally
obtained by other means (i.e. than appealing directly to integral
representations). Some examples I have included below might suffice to expose
this.

Ex.
(1): You
have the Maclaurin series:

^{¥}

_{n = 0}(z)

^{n}

_{ }

**/**n! = 1 + z + z

^{2}/ 2! + z

^{3}/ 3! +…..

Now,
replace z by 1/z in the expansion:

exp(
1/z) = å

+ 1/ 3! z

^{¥}_{n = 0}1_{ }**/**n! z^{n}**= 1 + 1/ 1! z + 1/ 2! z**^{2}+ 1/ 3! z

^{3}+…..
for: 0 <

**÷**z**÷****< ¥**
Which then becomes a

__Laurent series__expansion.
To
establish this on inspection note that no positive powers of z appear, only
negative, i.e. then 1/z is

z

So, in effect we can say that the

^{- 1 }^{}So, in effect we can say that the

*coefficients of the positive powers are zero*. It’s also important to note here that the coefficient of :
1/ 1! z = 1/ z

is unity, so according to Laurent’s theorem we designate that the coefficient:

c

**= 1/2 pi**_{n}**ò**_{C}^{ }exp( 1/z) dz
where
C is any positively oriented simple closed contour around the origin. Now, since from our prior Dec. 12, Introducing Laurent Series) examination c

**= 1 then:**_{n}
1 =1/2 pi

**ò**_{C}^{ }exp( 1/z) dz and:**ò**_{C}^{ }exp( 1/z) dz = 2 pi
Ex.
(2): Consider
the function:

f(z) = 1/ (z – i)

^{2}

This
is already in the form of a Laurent series , where z

**= i. That is,**_{0}
å

^{¥}_{n = -}**c**_{¥}**(z – i)**_{n}^{n}(For: 0 <**÷**z - i**÷****< ¥ )**Where in this case, c

**= 1 and all other coefficients**

_{- 2}*are zero*. Then from the previous theorems, relations we’ve explored (blog post of December 12, ‘Introduction to Laurent Series’, sub-header: ‘

*More intricacies and singularities’*):

c

**= 1/2 pi**_{n}**ò**_{C}^{ }dz/ ( z – i)^{n +3 }n = 0,__+__1,__+__2,__+__3 …..
where
C denotes any positively oriented circle

**÷**z - i**÷****= R about the point: z****= i**_{0}
Then
it follows from this:

a)

**ò**_{C}^{ }dz/ ( z – i)^{n +3 }= 0 (when n ¹ 2)
b)

**ò**_{C}^{ }dz/ ( z – i)^{n +3 }= 1 (when n= 2)__Problems for the Math Maven__:

1)
The
function: f(z) = -1/ (z – 1) (z – 2)

a) Rewrite
it using the partial fraction form

__the two singular points__.

c) If the function is analytic in the domains:

**÷**z

**÷**< 1; 1 <

**÷**z

**÷**< 2 and: 2 <

**÷**z

**÷**< ¥,

Draw
a sketch showing the different domains.

2) Find
the Laurent series that represents the function:

f(z)
= z

^{2 }sin (1 / z^{2})
In
the domain: : 0 <

**÷**z**÷****< ¥**
Hint: Recall the series for sin(z) = z - z

^{3}/ 3! + z^{5}/ 5! …..-
(-1)

^{n- 1}z^{2n -1}/ (2n -1)! + …… (For**÷**z**÷****< ¥ )**
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