Key to Complex Analysis Homework 2
September 27, 2011
Chapter 3 (in Greene & Krantz’s book)
4. Use Morera’s theorem to give another proof of Theorem 3.5.1: If {fj } is a
sequence of holomorphic functions on a domain U and if the sequence converges
unifromly on compact subsets of U to a limit function f , then f is holomorphic
on U .
Proof. Let γ be a closed piecewise C 1 curve in U , then by the generalized
Cauchy’s theorem, we have that
Z
fn dz = 0, for each n ∈ N.
γ
Then, we have that
Z
Z
f dz = fn − f dz γ
γ
Z
≤
|fn − f ||dz|
γ
By the given condition that fn converges uniformly to f on a compact subset
of U which contains the curve γ and its interior , for each > 0, we can choose
an N ∈ N, such that
|fn − f | < , n > N.
Hence, it follows that
Z
Z
f dz ≤
|fn − f ||dz| < · length(γ),
γ
n > N.
γ
Letting → 0, we conclude
Z
f dz = 0.
γ
Since our γ is arbitratily chosen, by the Morerea’s theorem, we know that f is
holomorphic on U .
1
2
10. Find the complex power series expansion for z 2 /(1 − z 2 )3 about 0 and
determine the radius of convergence(do not use Taylor’s formula).
Proof. We use the formula, for |z| < 1,
∞
X
1
=
zn.
1 − z n=0
By differentiating above, we get
∞
X
1
=
nz n−1 .
(1 − z)2
n=1
By differentiating above again, we get
∞
X
2
n(n − 1)z n−2 .
=
(1 − z)3
n=2
Replacing z with z 2 , so for |z| < 1,
∞
X
1
n(n − 1) 2(n−2)
=
z
.
2
3
(1 − z )
2
n=2
Hence,
z 2 /(1 − z 2 )3 =
∞
X
n(n − 1) 2(n−1)
z
2
n=2
with the radius of convergence being one.
2
19. Prove the case of Lemma 3.2.6 where
lim sup |ak |1/k = 0.
k→+∞
P∞
Proof. We prove that series k=0 ak z k converges for each (fixed) z ∈ C. For
each fixed z, from lim supk→+∞ |ak |1/k = 0, we know that, for k > N , |ak |1/k <
1
2|z| . Hence
∞
N
N
X
X
X
|ak ||z|k ≤
|ak ||z|k +
|1/2|k < +∞.
k=0
P∞
k=0
k=0
k
Therefore, k=0 |ak ||z| converges for each (fixed) z ∈ C, and thus
converges for each (fixed) z ∈ C.
2
P∞
k=0
ak z k
2
20. Find the power series expansion for each of the following holomorphic functions about the given point (i) by using the statement of Theorem 3.3.1 and
(ii) by using the proof of Theorem 3.3.1. Determine the disc of convergence of
each series.
(a) f(z)=1/(1+2z),
P=0.
Proof. (i) By the statement of Theorem 3.3.1, we know that the coefficients of
power series are given by
1 ∂kf
(0).
ak =
k! ∂z k
Hence, calculating the complex derivative of the given function f (z) = 1/(1 +
2z), we have
∂kf
2k k!
k
=
(−1)
, k = 0, 1, 2, · · ·
∂z k
(1 + 2z)k+1
So the coefficents are
ak =
2k k!
1
(−1)k
= (−1)k 2k ,
k!
(1 + 2 · 0)k+1
k = 0, 1, 2, · · ·
(ii)However, by the proof of Theorem 3.3.1, from Cauchy’s integral formula, for
r < 1/2,
I
1
1
f (z) =
dζ.
2πi |ζ|=r (1 + 2ζ)(ζ − z)
For |ζ| = r < 1/2 and |z| < r (so |z/ζ| < 1) ,
∞ n
∞ X
1
1X z
1
1
=
=
=
zn.
n+1
ζ −z
ζ(1 − z/ζ)
ζ n=0 ζ
ζ
n=0
Hence, for |z| < 1/2,
1
f (z) =
2πi
I
|ζ|=r
I
∞
∞
X
X
1
1
1
n
dζ =
dζz =
an z n ,
n+1
(1 + 2ζ)(ζ − z)
2πi
(1
+
2ζ)ζ
|ζ|=r
n=0
n=0
where, obviously
an =
1
2πi
I
|ζ|=r
1
dζ.
(1 + 2ζ)ζ n+1
2
21. Prove that the function
f (z) =
∞
X
j=0
3
j
2−j z (2
)
is holomorphic on D(0, 1) and continuous on D̄(0, 1). Prove that if w is a
(2N )th root of unity, then limr→1− |f 0 (rw)| = +∞. Deduce that f cannot be
the restriction to D(0, 1) of a holomorphic function defined on a connected open
set that is strictly larger than D(0, 1).
Proof. First of all, we know that if |z| ≤ 1, then
j
|2−j z (2 ) | < 2−j
Also, the series
∞
X
2−j
j=0
converges.
Pn
j
Thus, by the Weierstrass M-test, we know thatfn (z) = j=0 2−j z (2 ) converges
uniformly to f on D̄(0, 1). This implies that f is holomorphic on D(0, 1), as
each fn is holomorphic on D(0, 1); and continuous on D̄(0, 1), as each fn is
continuous on D̄(0, 1).
Next, from Corollary 3.5.2, on D(0, 1),
f 0 (z) =
n
n
X
∂ −j (2j ) X (2j −1)
2 z
z
,
=
∂z
j=0
j=0
so, obviously, at z = 1,
lim |f 0 (r)| = +∞.
r→1−
This means that f can not be extended holomorphically across the point z = 1
(or there is no open neighborhood of z = 1 on which f is holomorphic, i.e. the
series still convergent). We now show that this not only true at z = 1, but also
at almost every points on the circle |z| = 1, i.e. the convergence of the series
P∞ −j (2j )
can not be extended across at almost every point on the circle
j=0 2 z
|z| = 1.
To do so, consider the (2N )th root of unity (note that when N getting bigger,
2πi
the root of N -units spread out on the circle |z| = 1), denoted as w = e 2N , and,
for 0 < r < 1,
X
∞
j
|f 0 (rw)| = (rw)(2 −1) j=0
X
∞
2πi
j
= (re 2N )(2 −1) j=0
4
Note that, for each j ≥ N , we actually have that
2πi
j
j
w2 = (e 2N )2 = 1
which implies after Nth term, the series becomes
N
N +1
w−1 r2 −1 + r2 −1 + · · · .
Hence
∞
X
j
0
(2 −1) |f (rw)| = (rw)
j=0
∞
N −1
X
X
j
j
(2 −1) (2 −1) ≥
(rw)
(rw)
−
j=N
j=0
−1
N
NX
j
N +1
(2 −1) −1
2 −1
2
−1
(rw)
= |w|
r
+r
+ ··· − j=0
−1
NX
1 2N
(2j −1) 2N +1
=
(rw)
r +r
+ ··· − |w|r
j=0
−1
NX
N
j
1 2N
2
2
(2 −1) =
r + (r ) + · · · − (rw)
.
|w|r
j=0
Using the fact that, for each positive integer m,
N
N
N
N
N
N
r2 + (r2 )2 + · · · > r2 + (r2 )2 + · · · + (r2 )2m ≥ m(r2 )2m ,
N
N
where in above, we used the fact that, for 0 < r < 1, r2 > (r2 )2 > · · · , and
the RHS contains m terms. Hence
N −1
X
(2j −1) 0
2N 2m
|f (rw)| ≥ m(r ) − (rw)
.
j=0
By letting r → 1, we get
NX
−1
j
0
(2 −1) lim inf
|f (rw)| ≥ m − (w)
r→1−
j=0
and then by letting m → +∞, it gives
lim inf
|f 0 (rw)| = +∞.
−
r→1
Hence
lim |f 0 (rw)| = +∞.
r→1−
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