MTH 5103 COMPLEX VARIABLES
LECTURE NOTES, WEEK 6
6
Taylor Series and Laurent Series
In this Lecture, we shall investigate complex Taylor series, as well as a generalization of power series to include negative powers, called Laurent series. These series’
relationships with holomorphic functions are also presented.
6.1
Motivation
To motivate use of Taylor and Laurent series, we shall first present a few examples
related to the previous Lecture on power series.
Example 1. Determine the convergence or divergence of the following series
n
∞ X
4+i
n=0
6
n
∞ X
4+i
,
n=2
56
,
n
∞ X
4+i
n=0
3
.
We recognize each of the above series as being a geometric series, so convergence
will be determined by whether the modulus of the general term is less than 1. In
the first example, we have
√
4 + i r 42 + 12
17
|z| = =
< 1,
=
2
6
6
6
1
6
6
1
=
=
= (2 + i).
1−z
1 − (4 + i)/6
2−i
5
In the next series, we may argue in the same manner to conclude that the
series converges absolutely, but note that the sum begins at n = 2. We use the
following observation to determine the value of the sum: for |z| < 1,
therefore the series converges absolutely to
∞
X
n=k
n
z =
∞
X
n=0
n
z −
k−1
X
n=0
zn =
1
− (1 + z + z 2 + · · · + z k−1 ).
1−z
1
1
4+i
−1−
.
1 − (4 + i)/56
56
Finally, in the last sequence we have
√
4 + i r 42 + 12
17
|z| = > 1,
=
=
3
32
3
Thus, the middle series converges to
whereby the series diverges.
Let us adjust this example slightly to include a variable z ∈ C, making the
geometric geries above into power series.
n
∞ X
4+i
(z − 3)n
Example 2. For what values of z does the power series
6
n=0
converge? In this case, we are studying a power series centred at z0 = 3 and can
apply the Ratio Test to see that
a (z − z )n+1 4+i n+1 (z − 3)n+1 n+1
0
lim = lim 6 4+i n
n
n
n→∞
n→∞
an (z − z0 )
(z − 3)
6
4 + i
= lim · (z − 3)
n→∞
6
√
17
|z − 3| = L.
=
6
By the Ratio Test, the series converges absolutely if L < 1 and diverges if L > 1.
In this manner, we see that the radius of convergence of the power series is given
by R = √617 , since
√
17
|z − 3| < 1
6
⇔
6
|z − 3| < √ .
17
Another type of question we may encounter is one in which we are asked to
determine the power series of a given function.
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Example 3. Find the power series expansion of the function f (z) =
about
2z − 3
the point z0 = 0 and determine the radius of convergence. To accomplish this,
we observe that f can be written in the form of a convergent geometric series as
follows:
n
∞
∞ ∞ n
X
1X 2
2n n
1
1
1
1 X 2z
n
=− ·
=−
=−
z =−
z .
2z − 3
3 1 − (2z/3)
3 n=0 3
3 n=0 3
3n+1
n=0
The series expansion holds only if |2z/3| < 1, that is, |z| < 3/2, so the radius of
convergence of the series is 23 .
2
Some natural questions to pose at this point are: What happens if we are asked
to find the power series representation of a function f which cannot be expressed
as a Geometric Series? Is there a general method by which we can find the power
series representation of an arbitrary function? How do we even know that a function
actually admits a power series representation?
We give a partial answer to these questions in the next subsection.
6.2
Taylor Series
THEOREM 1. (Taylor Series) Let f be holomorphic on the disc D = {z ∈ C :
|z − z0 | < R}. Then f is differentiable arbitrarily many times at z0 and for all
z ∈ D,
∞
X
f (n) (z0 )
f (z) =
(z − z0 )n ,
(1)
n!
n=0
the series on the right, called the Taylor series for f at z0 , converges to f (z).
∞
X
an (z − z0 )n
Moreover, this is the unique representation of f as a power series
n=0
on D.
Proof. We shall prove this theorem later in the lectures, using complex integration
theory.
Note that Theorem 1 says that any holomorphic function (i.e. once differentiable)
can be written as a power series, and by applying the Proposition on power series
from the previous Lecture we can deduce that
0
f (z) =
f 00 (z) =
∞
X
n=1
∞
X
n=2
..
.
1
f (n) (z0 )(z − z0 )n−1
(n − 1)!
1
f (n) (z0 )(z − z0 )n−2
(n − 2)!
for all z ∈ D.
This is very different from the real case, where a function can be differentiable once
without being differentiable twice (can you think of an example?). Indeed, a real
everywhere differentiable function need not be equal to its Taylor series. for example,
consider the function
(
e−1/x x > 0
f (x) =
.
(2)
0
x≤0
3
This function satisfies f (n) (0) = 0 for all n ≥ 0 so its Taylor series must be 0 + 0 ·
x + 0 · x2 + · · · = 0. But f itself is not equal to its Taylor series.
The complex version of Taylor’s Theorem, Theorem 1 tells us that once we know
f (z0 ), f 0 (z0 ), f 00 (z0 ), . . . , f (n) (z0 ), . . ., (at only the point z0 ), we know the value of
P
f (n)
n
f (z) everywhere in the disc of convergence of the Taylor series ∞
n=0 n! (z − z0 ) .
Now we give some examples of Taylor Series:
∞
X
1 n
z
z by definition. This series converges absolutely for all
Example 4. e =
n!
n=0
z ∈ C (by the Ratio Test), so R = ∞, i.e. the radius of convergence is infinite.
Hence, ez is differentiable ofr all z ∈ C and has derivative
∞
∞
X
1 n−1 X 1 n
(e ) =
nz
=
z = ez .
n!
n!
n=1
n=0
z 0
dn (ez )
= ez which evaluated at z = 0 gives value 1. Hence, the Taylor series
dz n
∞
X
1 n
z
for e about z0 = 0 is
z (which we could also have deduced from the
n!
n=0
uniqueness statement in Theorem 1
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Example 5. Let f (z) =
. what is the power series of this function about
2z − 3
the point z0 = 2? We need to find a Taylor series expansion of the function of the
∞
X
form
an (z − 2)n . The first method is to use the formula for the coefficients
Now
n=0
∞
X
f (n) (2)
(z − 2)n . Subsequently, we will
n!
n=0
need to compute the radius of convergence R.
A second, alternative method is to rewrite the function f as a geometric series.
Observe that
in the Taylor Series Theorem to find
f (z) =
=
=
=
=
1
1
1
= ·
2z − 3
2 z − 23
1
1
·
2 (z − 2) + 12
1
2(z − 2) + 1
1
1 − (−2(z − 2))
∞
∞
X
X
n
(−2(z − 2)) =
(−2)n (z − 2)n .
n=0
n=0
4
This series is absolutely convergent for z satisfying
1
1
|2(z − 2)| < 1 ⇔ |z − 2| <
⇒R= .
2
2
Thus, the function f (z) is represented by the above power series on the disc of
convergence given by
1
D = {z ∈ C : |z − 2| < }.
2
1
3
Notice that f (z) = 2z−3 is undefined at z = 2 , so the series and function only
agree up to, but not including the boundary of the disc D.
6.3
Laurent Series
Taylor series exist for functions which are holomorphic on discs. Holomorphic functions on certain other types of regions also have series representations.
∞
X
1
z n for |z| < 1. What
Example 6. Let f (z) =
. We have a power series
1−z
n=0
can we do for |z| > 1? In this case, | z1 | < 1, so we can try to form a series in z1
instead:
1
1
1
=
· 1
1−z
z z −1
1
1
= − ·
z 1 − z1
∞ n
1X 1
= −
z n=0 z
n+1 X
∞
∞
X
1
=
−
=
−1 · z −n .
z
n=0
n=1
This series is absolutely convergent for all z with |z| > 1, but divergent for all z
with |z| < 1.
Definition 1. Let A be an annulus centred at z0 with inner radius R1 and outer
radius R2 , 0 ≤ R1 < R2 ≤ ∞:
A = {z ∈ C : R1 < |z − z0 | < R2 }.
A series given by
∞
X
n
an (z − z0 ) +
n=0
∞
X
bn (z − z0 )−n
n=1
converging to f (z) at every z ∈ A is called a Laurent series for f on A.
5
(3)
(4)
Remark 1. We sometimes write
∞
X
an (z − z0 )n to denote the Laurent series,
n=−∞
however, we must be careful to remember that this denotes the sum of two series.
THEOREM 2. Let f be holomorphic on the annulus A centred at z0 . The f (z)
has a unique Laurent series expansion
∞
X
n
an (z − z0 ) +
n=0
∞
X
bn (z − z0 )−n
(5)
n=1
converging absolutely to f (z) at every point z in A.
Proof. We shall prove this in a subsequent section, using complex integration.
There are formulae for the coefficients in Laurent series involving integrals1 .
∞
X
1
(−1) · z −n on
Example 7. We have now seen that
has Laurent series
1−z
n=1
∞
X
z n on the disc
the annulus A = {z : 1 < |z| < ∞} and has Laurent series
n=0
A = D = {z : 0 ≤ |z| < 1}. Note that the latter Laurent series is just the Taylor
series when the domain is a disc.
∞
X
1
1 −n
z
Example 8. e =
z (by definition) on A = {z : 0 < |z| < ∞}, so this is
n!
n=0
its Laurent series there, by uniqueness.
1
Example 9. What is the Laurent series of f (z) =
on the annulus
(z − 1)(z − 2)
A = {z : 1 < |z| < 2}? Observe that we may write f (z) using a partial fraction
expansion as
1
1
1
=
−
(z − 2)(z − 1)
z−2 z−1
−1
1
=
z −
2(1 − 2 ) z(1 − z1 )
∞
∞
1 X z n X −n
= − ·
−
z
2 n=0 2
n=1
which is its Laurent series. Notice that the first sum only exists if |z/2| < 1
(equivalently, if |z| < 2)and the second sum only exists if |1/z| < 1 (equivalently,
if |z| > 1), so the last line is valid only on the annular region 1 < |z| < 2, as
desired.
1
However, these formulae are, in general, not practical for computations.
6
1
on the annulus
(z − 1)(z − 2)
A = {z : 0 < |z − 1| < 1}? Next, notice that we can write f (z) as
Example 10. What is the Laurent series of f (z) =
1
1
1
1
1
=
·
=
·
(z − 2)(z − 1)
1−z 2−z
1 − z 1 − (z − 1)
∞
1 X
=
(z − 1)n
1 − z n=0
=
∞
X
−(z − 1)n
n=−1
∞
X
1
= −
+
−(z − 1)n .
z − 1 n=0
This is the Laurent series for f on A.
In the final example, note that in the notation of Theorem 2, b1 = −1, bn = 0
for all n ≥ 2. Note also that we are considering a Laurent series expansion in a
punctured disc
D0 = {z ∈ C : 0 < |z − z0 | < r}.
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