Class2 16
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jarchowsilvia26Class2 16
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jarchowsilvia26Math 752 Spring 2011
Class 2/16
1 Laurent series development
Theorem 1. Let f be analytic in the annulus R1 < |z − a| < R2 . Then
∞
X
f (z) = an (z − a)n
n=−∞
where the series is absolutely and uniformly convergent in every r1 < |z −
a| < r2 with R1 < r1 < r2 < R2 and
Z
1 f (z)
an = dz
2πi γ (z − a)n+1
where γ is the circle |z − a| = r with r ∈ (R1 , R2 ). This series is unique.
Proof. Fix r1 , r2 with R1 < r1 < r2 < R2 . Denote by γ1 and γ2 the two
circles traced counterclockwise with radius r1 and r2 respectively, and note
that they are homotopic in the annulus.
The integrand of the integral that is proposed to equal an is analytic in
the annulus, and hence its value is independent of r1 and r2 .
Connect the two annuli by a radial line segment λ missing z (traced
outwards), and consider the path
γ = γ2 − λ − γ1 + λ.
This path is homotopic to zero, and has winding number 1 around z.
Moreover, it has winding number zero about every point z with |z − a| < r1
or |z − a| > r2 . Hence Cauchy’s theorem applies and gives
Z
1 f (w)
f (z) = dw
2πi γ w − z
Z Z
1 f (w) 1 f (w)
= dw − dw
2πi γ2 w − z 2πi γ1 w − z
=: f2 (z) + f1 (z).
1
Crucial fact: Since f is uniformly continuous on γ1 and on γ2 , the func-
tion f1 is analytic in B(a, r2 ) and f2 is analytic in C\B(a, r1 ). We expand
f2 in a power series to get
∞
X
f2 (z) = an (z − a)n
n=0
with an as in the statement of the theorem. (Note that in general f 6= f2 ,
though.) We would like to expand f1 into a power series in (z − a)−1 . To
do so, we define
g(z) = f1 (a + z −1 ).
We note from the definition of f1 that
sup{|f (z)| : z ∈ {γ1 }} · r1
|f1 (z)| ≤
d(z, {γ1 })
and the denominator goes to ∞ with z. Hence, by setting g(0) = 0 we
obtain that g is analytic in B(0, 1/R1 ). Its power series expansion gives
∞
X
f1 (a + z −1 ) = Bn z n ,
n=1
and the substitution w = a + z −1 gives
∞
X
f (w) = Bn (w − a)−n .
n=1
We have (with 1/r < 1/R1 )
f (a + z −1 )
Z
1
Bn = dz,
2πi |z|=1/r z n+1
and as before (note that the change of direction cancels the sign from the
substitution)
|f (w)|
Z
1 dw
Bn = −n−1
,
2πi |w−a|=r (w − a) (w − a)2
hence the formula for a−n := Bn follows. Uniqueness follows from the fact
that if we start with an absolutely and uniformly convergent Laurent series,
then the representation for the coefficients follows with an interchange of
integration and summation.
2
Corollary 1. z = a isolated singularity of f and f (z) = n an (z − a)n its
P
Laurent expansion in the annulus 0 < |z − a| < R (some R > 0). Then
a) a is removable iff an = 0 for n ≤ −1,
b) a is a pole of order m iff a−m 6= 0 and an = 0 for n < −m,
c) a is essential singularity if infinitely many an with negative index are
non-zero.