SOLUTIONS OF HW6
MINGFENG ZHAO
March 25, 2011
1. [Problem 1, in Page 127] Prove that a function w(z) which is continuous on a closed arc l is
uniformly continuous on l, i.e, that for every > 0, there exists a ρ > 0 such that |w(z1 ) − w(z2 )| < whenever z1 and z2 are points of the arc l and |z1 − z2 | < ρ .
Proof. Let γ : [0, 1] → C a closed curve, that is, γ is continuous on [0, 1, and γ(0) = γ(1). Denote
Γ = γ([0, 1]). By the assumption, we know w : Γ → C is continuous.
Assume w is not uniformly continuous on Γ, that is, there exists some 0 > 0, such that for any
δ > 0, there exists z1,δ , z2,δ ∈ Γ such that |z1,δ − z2,δ | < δ, but |w(z1,δ ) − w(z2,δ )| ≥ 0 .
Take δ =
1
n,
then for all n ≥ 1, there exits z1,n , z2,n ∈ Γ such that |z1,n − z2,n | <
1
n,
but
|w(z1,n ) − w(z2,n )| ≥ 0 . Since Γ = γ([0, 1]), so we can find t1,n , t2,n ∈ [0, 1] such that z1,n =
γ(t1,n ), z2,n = γ(t2,n ). So we get
|w ◦ γ(t1,n ) − w ◦ γ(t2,n )| = |w(z1,δ ) − w(z2,δ )|
≥ 0 .
∞
For {t1,n }∞
n=1 ⊂ [0, 1], since [0, 1] is compact, then there exists a subsequence {t1,nk }k=1 such
that t1,nk → t1 ∈ [0, 1], as k → ∞. For {t2,nk }∞
k=1 ⊂ [0, 1], since [0, 1] is compact, then there exists a
∞
subsequence {t2,nkl }∞
l=1 such that t2,nkl → t2 ∈ [0, 1], as l → ∞. Consider the subsequences {t1,nkl }l=1
and {t2,nkl }∞
l=1 , we know
lim t1,nkl
= t1
lim t2,nkl
= t2
l→∞
l→∞
1
2
MINGFENG ZHAO
Since γ is continuous on [0, 1], then we have
lim γ(t1,nkl )
l→∞
=
lim z1,nkl
l→∞
= γ(t1 )
lim γ(t2,nkl )
l→∞
=
lim z2,nkl
l→∞
= γ(t2 )
Since w is continuous, then
lim w ◦ γ(t1,nkl )
= w ◦ γ(t1 )
lim w ◦ γ(t2,nkl )
= w ◦ γ(t2 )
l→∞
l→∞
lim |w ◦ γ(t1,nkl ) − w ◦ γ(t2,nkl )| =
l→∞
lim |w ◦ γ(t1 ) − w ◦ γ(t2 )|
l→∞
= |w ◦ γ(t1 ) − w ◦ γ(t2 )|
≥ 0
> 0.
On the other hand, we know
|z1,nkl − z2,nkl | <
1
,
nkl
which implies that liml→∞ |z1,nkl − z2,nkl | = 0, that is
lim z1,nkl = lim z1,nkl .
l→∞
l→∞
So γ(t1 ) = γ(t2 ), which implies that |w ◦ γ(t1 ) − w ◦ γ(t2 ) = 0. Contradiction.
Therefore, w is uniformly continuous on Γ.
SOLUTIONS OF HW6
3
2. [Problem 4, in Page 127] Show that the value of the integral (8.2) is independent of the parametric
representation (8.1) of the arc l, provided the conditions for the existence of the integral are fulfilled.
Proof. We assume γ : [a, b] → C be a curve, and Γ = γ([a, b]), and f : Γ → C is complex function.
And
R
γ
f (z)dz exists, that is, there exists A ∈ C such that for any > 0, there exits δ > 0 such that
for any partition P = {a = t0 < t1 < · · · < tn = b} of [a, b] such that
∆(P) = max (tk − tk−1 ) < δ.
1≤k≤n
For any 1 ≤ k ≤ n, we take any sk ∈ [tk−1 , tk ], We have
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A < .
k=1
In this case, we denote
Z
A=
f (z)dz
γ,[a,b]
Let γ̃ : [c, d] → C be another reparametric representation, that is, there exits a function φ : [c, d] →
[a, b] such that
a. φ is continuous on [c, d];
b. φ(c) = a, φ(d) = b;
c. φ is increasing on [c, d]. (Sometimes we easily assume φ0 ≥ 0);
d. γ̃(t) = γ ◦ φ(t) for all t ∈ [c, d].
So our question is that we need to show
i.
R
ii.
R
γ̃,[c,d]
f (z)dz exists;
γ̃,[c,d]
f (z)dz = A.
More precisely, we need to show that for any > 0, there exits δ̃ > 0 such that for any partition
P = {c = r0 < r1 < · · · < rn = d} of [c, d] such that
∆(P) = max (rk − rk−1 ) < δ̃.
1≤k≤n
4
MINGFENG ZHAO
For any 1 ≤ k ≤ n, we take any qk ∈ [rk−1 , rk ], We have
n
X
f (γ̃(qk ))(γ̃(rk ) − γ̃(rk−1 )) − A < .
k=1
Since φ : [c, d] → [a, b] is continuous, then φ is uniformly continuous on [c, d]. For the above δ > 0,
there exists δ̃ such that, whenever |r − r0 | < δ̃ for r, r0 ∈ [c, d], we have
|φ(r) − φ(r0 )| < δ.
So for any partition P = {c = r0 < r1 < · · · < rn = d} of [c, d] such that
∆(P) = max (rk − rk−1 ) < δ̃.
1≤k≤n
For any 1 ≤ k ≤ n, we take any qk ∈ [rk−1 , rk ]. We just let tk = φ(rk ) for all 0 ≤ k ≤ n, and
sk = φ(qk ) for all 1 ≤ k ≤ n. Since |rk − rk−1 | = rk − rk−1 < δ̃ for all 1 ≤ k ≤ n, then
tk − tk−1 = |tk − tk−1 | < δ.
So we have
n
X
f (γ̃(qk ))(γ̃(rk ) − γ̃(rk−1 )) − A
k=1
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A
= k=1
<
.
Therefore, we complete the proof.
3. [Problem 5, in Page 127] Prove the rules of computation (1) to (5) for line integrals in Section 8.7.
Proof. 1)a. We assume γ : [a, b] → C, Γ = γ([a, b]), f : Γ → C, and
R
γ
f (z)dz exists, and
R
γ
f (z)dz =
A ∈ C. So we have for any > 0, there exits δ > 0 such that for any partition P = {a = t0 < t1 <
· · · < tn = b} of [a, b] such that
∆(P) = max (tk − tk−1 ) < δ.
1≤k≤n
SOLUTIONS OF HW6
5
For any 1 ≤ k ≤ n, we take any sk ∈ [tk−1 , tk ], We have
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A < .
k=1
Hence, for any c ∈ C, we have
n
X
cf (γ(sk ))(γ(tk ) − γ(tk−1 )) − cA =
k=1
<
By the definition of line integral, we know
R
γ
n
X
|c| f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A
k=1
c
cf (z)dz exists, and
R
b. We assume γ : [a, b] → C, Γ = γ([a, b]), f : Γ → C, g : Γ → C and
exists, and
R
γ
f (z)dz = A ∈ C, and
R
γ
cf (z)dz = cA = c
γ
R
γ
R
γ
f (z)dz exists, and
∆(P) = max (tk − tk−1 ) < δ.
1≤k≤n
For any 1 ≤ k ≤ n, we take any sk ∈ [tk−1 , tk ], We have
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A < .
k=1
And
n
X
g(γ(sk ))(γ(tk ) − γ(tk−1 )) − B < .
k=1
Hence we have
n
X
(f + g)(γ(sk ))(γ(tk ) − γ(tk−1 )) − (A + B)
k=1
n
n
X
X
= f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A +
g(γ(sk ))(γ(tk ) − γ(tk−1 )) − B k=1
k=1
n
n
X
X
g(γ(sk ))(γ(tk ) − γ(tk−1 )) − B ≤ f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A + <
2.
R
γ
g(z)dz
g(z)dz = B ∈ C. So we have for any > 0, there exits δ > 0
such that for any partition P = {a = t0 < t1 < · · · < tn = b} of [a, b] such that
k=1
f (z)dz.
k=1
6
MINGFENG ZHAO
By the definition of line integral, we know
R
γ
f (z)dz +
R
γ
R
γ
(f (z)+g(z))dz exists, and
R
γ
(f (z)+g(z))dz = A+B =
g(z)dz.
2). Assume the curve γ1 : [a, b] → C, γ2 : [b, c] → C, and γ : [a, c] → C such that γ1 (b) = γ2 (b), and




 γ1 (t) if t ∈ [a, b]
γ(t) =



 γ (t) if t ∈ [b, c]
2
Denote Γ1 = γ1 ([a, b]), Γ2 = γ2 ([b, c]), and Γ = γ([a, c]) = Γ1 ∪ Γ2 . And f : Γ → C such that
R
γ1
f (z)dz and
R
γ2
f (z)dz exist, and
R
γ1
f (z)dz = A ∈ C and
R
γ2
f (z)dz = B ∈ C. So we have for any
> 0, there exits δ > 0 such that for any partition P1 = {a = t0 < t1 < · · · < tm = b} of [a, b], and
partition P2 = {b = r0 < r1 < · · · < rn = c} of [b, c], such that
∆(P1 ) = max (tk − tk−1 ) < δ.
1≤k≤m
and
∆(P2 ) = max (rk − rk−1 ) < δ.
1≤k≤n
For any 1 ≤ k ≤ m, we take any sk ∈ [tk−1 , tk ], and for any 1 ≤ l ≤ n, we take any ql ∈ [rl−1 , rl ].
We have
m
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A < .
k=1
And
n
X
f (γ(qk ))(γ(rk ) − γ(rk−1 )) − B < .
k=1
So for the above > 0, δ > 0, we consider any partition P = {a = t0 < t1 < · · · < tn } such that
∆(P) = max (tk − tk−1 ) < δ.
1≤k≤n
Assume ti−1 ≤ b ≤ ti for some 1 ≤ i ≤ n − 1. Then P1 = {a = t0 < t1 < · · · ti−1 ≤ b} is a partition
of [a, b] such that
∆(P1 ) = max
max
1≤k≤i−1
(tk − tk−1 ), b − ti−1
< δ.
SOLUTIONS OF HW6
7
Figure 1. a = t0 < t1 < · · · < ti−1 ≤ b ≤ ti < · · · < tn = c
And P2 = {b = ti < ti+1 < · · · < ti−1 ≤ c} is a partition of [b, c] such that
∆(P2 ) = max
max
i+1≤k≤n
(tk − tk−1 ), ti − b
< δ.
Then we have
=
m
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − (A + B)
k=1
i−1
n
X
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A − B f (γ(sk ))(γ(tk ) − γ(tk−1 )) + f (γ(si ))(γ(ti ) − γ(ti−1 )) +
k=1
=
k=i+1
i−1
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) + f (γ(si ))(γ(b) − γ(ti−1 )) + f (γ(si ))(γ(ti ) − γ(b))
k=1
n
X
+
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A − B k=i+1
≤
i−1
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) + f (γ(si ))(γ(b) − γ(ti−1 )) − A
k=1
n
X
+
f (γ(sk ))(γ(tk ) − γ(tk−1 )) + f (γ(si ))(γ(ti ) − γ(b)) − B k=i+1
≤
+
=
2.
8
MINGFENG ZHAO
By the definition of line integral, we know
R
γ2
R
γ
f (z)dz exists, and
R
γ
f (z)dz = A + B =
R
γ1
f (z)dz +
f (z)dz.
3) In fact, this is the definition of
Ra
b
f (z)dz = −
Rb
a
f (z)dz.
4). We assume γ : [a, b] → C, Γ = γ([a, b]), f : Γ → C, and
R
γ
f (z)dz exists, and
R
γ
f (z)dz = A ∈ C,
and assume M = supz∈Γ |f (z)|. So we have for any > 0, there exits δ > 0 such that for any partition
P = {a = t0 < t1 < · · · < tn = b} of [a, b] such that
∆(P) = max (tk − tk−1 ) < δ.
1≤k≤n
For any 1 ≤ k ≤ n, we take any sk ∈ [tk−1 , tk ], We have
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A < .
k=1
So we have
|A|
n
n
X
X
f (γ(sk ))(γ(tk ) − γ(tk−1 ))
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A + ≤ k=1
k=1
<
+
n
X
|f (γ(sk ))||γ(tk ) − γ(tk−1 ))|
k=1
≤ +M
n
X
|γ(tk ) − γ(tk−1 ))|
k=1
≤ + ML
By the definition of the length
Take → 0, we get |A| ≤ M L.
5). We assume γ : [a, b] → C, Γ = γ([a, b]), f : Γ → C, and
R
γ
f (z)dz exists, and
R
γ
f (z)dz =
a + ib ∈ C, a, b ∈ R. So we have for any > 0, there exits δ > 0 such that for any partition
P = {a = t0 < t1 < · · · < tn = b} of [a, b] such that
∆(P) = max (tk − tk−1 ) < δ.
1≤k≤n
SOLUTIONS OF HW6
9
For any 1 ≤ k ≤ n, we take any sk ∈ [tk−1 , tk ], We have
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 )) − A < .
(1)
k=1
We let γ(t) = x(t) + iy(t) for all t ∈ [a, b], and f (z) = f (x, y) = u(x, y) + iv(x, y) for all z ∈ Γ. So
we have
n
X
f (γ(sk ))(γ(tk ) − γ(tk−1 ))
k=1
=
n
X
[u(x(sk ), y(sk )) + iv(x(sk ), y(sk ))](x(tk ) + iy(tk ) − x(tk−1 ) − iy(tk−1 ))
k=1
=
n
X
u(x(sk ), y(sk ))(x(tk ) − x(tk−1 )) −
k=1
+i
n
X
v(x(sk ), y(sk ))(y(tk ) − y(tk−1 ))
k=1
n
X
v(x(sk ), y(sk ))(x(tk ) − x(tk−1 )) +
k=1
!
u(x(sk ), y(sk ))(y(tk ) − y(tk−1 ))
k=1
Z
−→
n
X
Z
(udx − vdy) + i
γ
as ∆(P) → 0.
(vdx + udy)
γ
By (1), we get
Z
Z
γ
Z
(udx − vdy) + i
f (z)dz =
γ
(vdx + udy).
γ
4. [Problem 7, in Page 127] Evaluate
Z
t
|z|dz
−t
for the following paths of integration:
a. A straight line segment;
b. The arc |z| = 1, Re z ≥ 0;
c. The arc |z| = 1, Re z ≤ 0.
Proof. a. In this case, we denote the curve γ : [−1, 1] → C by γ(t) = it.
10
MINGFENG ZHAO
Figure 2. A straight line segment
Then
Z
Z
|z|dz
1
|t|d(it)
=
−1
γ
Z
=
1
|t|dt
i
−1
Z
=
2i
1
tdt
0
=
i.
b. In this case, we denote the curve γ : − π2 , π2 → C by γ(t) = eit .
Then
Z
π
2
Z
|z|dz
|eit |d(eit )
=
−π
2
γ
π
2
Z
ieit dt
=
−π
2
Z
=
π
2
i
(cos t + i sin t)dt
−π
2
Z
=
π
2
−π
2
=
Z
π
2
cos tdt −
i
2i.
sin tdt
−π
2
SOLUTIONS OF HW6
Figure 3. The arc |z| = 1, Re z ≥ 0;
c. In this case, we denote the curve γ : − π2 , π2 → C by γ(t) = − cos t + i sin t.
Figure 4. The arc |z| = 1, Re z ≤ 0;
Then
Z
Z
|z|dz
π
2
| − cos t + i sin t|d(− cos t + i sin t)
=
−π
2
γ
Z
π
2
=
(sin t + i cos t)dt
−π
2
Z
π
2
=
Z
π
2
sin tdt + i
−π
2
cos tdt
−π
2
11
12
MINGFENG ZHAO
=
2i.
5. [Problem 2, in Page 127] If w(z) is continuous in a domain G, then it is uniformly continuous on
every compact (closed and bounded) subset G0 of G, i.e, for every > 0, there is a corresponding
ρ > 0, such that |w(z1 ) − w(z2 )| < provided that |z1 − z2 | < ρ and z1 , z2 ∈ G0 .
Proof. Let G0 be any compact subset of G. If w is not uniformly continuous on G0 , that is, there
exists 0 > 0 such that for any δ > 0, there exits z1,δ , z2,δ ∈ G0 such that |z1,δ − z2,δ | < δ, but
|w(z1,δ ) − w(z2,δ )| ≥ 0 .
Take δ =
1
n,
then for all n ≥ 1, there exits z1,n , z2,n ∈ Γ such that |z1,n − z2,n | <
1
n,
but
|w(z1,n ) − w(z2,n )| ≥ 0 .
0
0
∞
Consider sequences {z1,n }∞
n=1 ⊂ G , since G is compact, then there exists a subsequence {z1,nk }k=1
such that
lim z1,nk = z1 ∈ C.
k→∞
0
0
Consider sequences {z2,nk }∞
k=1 ⊂ G , since G is compact, then there exists a subsequence
{z2,nkl }∞
l=1 such that
lim z2,nkl = z2 ∈ C.
l→∞
Since w is continuous on G0 , then we have
lim w(z1,nkl )
=
w(z1 )
lim w(z2,nkl )
=
w(z2 )
l→∞
l→∞
Since |w(z1,nkl ) − w(z2,nkl )| ≥ 0 for all l ≥ 1, then we have |w(z1 ) − w(z2 )| ≥ 0 .
On the other hand, since |z1,nkl − z2,nkl | <
1
nkl
which implies that w(z1 ) = w(z2 ), contradiction.
for all l ≥ 1, when taking l → ∞, we have z1 = z2 ,
SOLUTIONS OF HW6
13
6. [Problem 9, in Page 127] Let the function w(z) be continuous outside the circle |z| = r0 , and let
limr→∞ rM (r) = 0, where M (r) is the maximum of |w(z)| on Kr : |z| = r > r0 . Show that
Z
lim
w(z)dz = 0.
r→∞
Kr
Proof. For any r > r0 , by the property 4), in Page 114, we have
Z
Kr
w(z)dz ≤
Z
|w(z)||dz|
Kr
Z
≤
M (r)|dz|
Kr
=
M (r)L(Kr )
=
2πrM (r)
Since limr→∞ rM (r) = 0, then we get
Z
lim r→∞
Kr
w(z)dz ≤ 0.
So we have
Z
w(z)dz = 0.
lim
r→∞
Kr
7.
[Problem 10, in Page 128] Let the function w(z) be continuous for 0 < |z| < r0 , and let
limr→0 rM (r) = 0, where M (r) is the maximum of |w(z)| on Kr : |z| = r < r0 . Show that
Z
lim
r→0
w(z)dz = 0.
Kr
Proof. For any 0 < r < r0 , by the property 4), in Page 114, we have
Z
Kr
w(z)dz ≤
Z
|w(z)||dz|
Kr
Z
≤
M (r)|dz|
Kr
14
MINGFENG ZHAO
=
M (r)L(Kr )
=
2πrM (r)
Since limr→0 rM (r) = 0, then we get
Z
lim r→0
Kr
w(z)dz ≤ 0.
So we have
Z
lim
r→0
w(z)dz = 0.
Kr
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu