Analysis of a Complex Kind

Week 7

Lecture 1: Laurent Series

Petra Bonfert-Taylor

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 1/9

Review of Taylor Series

Recall: If f : U → C is analytic and {|z − z0 | < R} ⊂ U then f has a power series

representation
∞
X f (k ) (z0 )
f (z) = ak (z − z0 )k , where ak = , k ≥ 0.
k!
k =0

What if f is not differentiable at some point?

z
Example: f (z) = z 2 +4
is not differentiable at z = ±2i (undefined there).
f (z) = Log z not continuous on (−∞, 0], so not differentiable there.

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 2/9

Laurent Series Expansion
Theorem (Laurent Series Expansion)
If f : U → C is analytic and {r < |z − z0 | < R} ⊂ U then f has a Laurent series
expansion
∞
X a−2 a−1
f (z) = ak (z −z0 )k = · · · + +a0 +a1 (z −z0 )+a2 (z −z0 )2 +· · · ,
(z − z0 )2 z − z0
k =−∞

that converges at each point of the annulus and converges absolutely and
uniformly in each sub annulus {s ≤ |z − z0 | ≤ t}, where r < s < t < R.

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 3/9

The Coefficients ak
Note: The coefficients ak are uniquely determined by f . How do we find them?
1
Example: f (z) = is analytic in C \ {1, 2}.
(z − 1)(z − 2)
Let’s find the Laurent series in the annulus {1 < |z| < 2}.Trick:

1 (z − 1) − (z − 2) 1 1
= = −
(z − 1)(z − 2) (z − 1)(z − 2) z −2 z −1
−1 1 1
= z −
2 1− 2 z(1 − z1 )
∞ ∞
−1 X  z k 1 X 1
= −
2 2 z zk
k =0 k =0
∞ ∞ ∞ −1
X −1 k X −1 X −1 k X
= z + = z + (−1)z k .
2k +1 zk 2k +1
k =0 k =1 k =0 k =−∞

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 4/9

 1
f (z) = (z−1)(z−2)

∞ −1
1 X −1 k X
= k +1
z + (−1)z k in {1 < |z| < 2}.
(z − 1)(z − 2) 2
k =0 k =−∞

What if we choose a different annulus? f is also analytic in {2 < |z| < ∞}!

1 1 1 1 1
= − = −
(z − 1)(z − 2) z −2 z −1 z(1 − z ) z(1 − z1 )
2

∞   ∞  
1X 2 k 1X 1 k
= −
z z z z
k =0 k =0
∞ ∞ −1
X 2k −1 X 1 X
= − = (2−k −1 − 1)z k .
zk zk
k =1 k =1 k =−∞

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 5/9

 1
f (z) = (z−1)(z−2)

∞ −1
1 X −1 k X
= z + (−1)z k in {1 < |z| < 2}.
(z − 1)(z − 2) 2k +1
k =0 k =−∞
∞ ∞ −1
1 X 2k −1 X 1 X
= − = (2−k −1 − 1)z k in {2 < |z| < ∞}.
(z − 1)(z − 2) zk zk
k =1 k =1 k =−∞

What if we choose yet another annulus? f is also analytic in {0 < |z − 1| < 1}!
∞
1 1 1 X
= =− =− (z − 1)k in {0 < |z − 1| < 1}.
z −2 (z − 1) − 1 1 − (z − 1)
k =0

So
∞ ∞
1 1 X 1 X
f (z) = − =− (z−1)k − = (−1)(z−1)k in {0 < |z−1| < 1}.
z −2 z −1 z −1
k =0 k =−1

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 6/9

Another Example

sin z
Another example: is analytic in C \ {0}. What is its Laurent series, centered
z4
at 0? Recall:
∞
X (−1)k z3 z5 z7
sin z = z 2k +1 = z − + − ± ···
(2k + 1)! 3! 5! 7!
k =0

So
sin z 1 11 1 1
= 3− + z − z3 ± · · ·
z4 z 3! z 5! 7!
1 1 1
Thus a−3 = 1, a−2 = 0, a−1 = − , a0 = 0, a1 = , a2 = 0, a3 = − , . . ..
3! 5! 7!

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 7/9

Calculating ak
Recall: For a Taylor series,
∞
X
f (z) = ak (z − z0 )k , |z − z0 | < R,
k =0

f (k ) (z0 )
the ak can be calculated via ak = . How about for a Laurent series
k!
∞
X
f (z) = ak (z − z0 )k , r < |z − z0 | < R?
k =−∞

f may not be defined at z0 , so we need a new approach! Back to Taylor series for
a second:
f (k ) (z0 ) Cauchy 1
Z
f (z)
ak = = dz
k! 2πi |z−z0 |=s (z − z0 )k +1
for any s between 0 and R. One can show a similar fact for Laurent series:
Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 8/9
The Coefficients ak

Theorem
If f is analytic in {r < |z − z0 | < R}, then
∞
X
f (z) = ak (z − z0 )k ,
k =−∞

where Z
1 f (z)
ak = dz
2πi |z−z0 |=s (z − z0 )k +1
for any s between r and R and all k ∈ Z.

Note: This does not seem all that useful for finding actual values of ak , but it is
useful to estimate ak . We’ll use this when calculating integrals later.

Lecture 1: Laurent Series Analysis of a Complex Kind P. Bonfert-Taylor 9/9