# SOLUTIONS TO HOMEWORK ASSIGNMENT # 5

```SOLUTIONS TO HOMEWORK ASSIGNMENT # 5
1. Use Cauchy’s Integral Theorem to evaluate the following integrals.
Z
z
(a)
dz, where C is the positively oriented circle |z − 2| = 2.
3
C z +1
Z
z
(b)
dz, where C is the circle |z| = 3, oriented in the clockwise direction.
2
C z +z−2
Z
cos π(z − 1)
(c)
dz, where C is the circle |z| = π ori2
2
2
2 2
2
C z(z + 16)(z − 16)(z + 25) (z − 25)
ented positively.
Solution:
(a) Suppose the partial fraction decomposition of
z3
z3
z
is
+1
B
C
z
A
+
=
+
πı/3
+1
z+1 z−e
z − e−πı/3
Z
A
dz = 0 since z = −1 is exterior to C. The other 2 roots are interior
C z+1
to C. Next we compute B and C :
z(z − eπı/3 ) eπı/3
e−πı/3
=
B =
=
z3 + 1
3e2πı/3
3
z=eπı/3
−πı/3 −πı/3
e
eπı/3
z(z − e
)
=
C =
=
z3 + 1
3e−2πı/3
3
z=e−πı/3
Notice that
Therefore
Z
C
z3
z
2πı
dz = 2πı(B + C) =
+1
3
z
2/3
1/3
(b) The partial fraction decomposition is 2
=
+
. Since both roots
z + z −2
z
+2 z−1
Z
z
2 1
are interior to C we have
dz = 2πı
+
= 2πı.
2
3 3
C z +z−2
1
(c) The partial fraction decomposition of
will
2
2
z(z + 16)(z − 16)(z 2 + 25)2 (z 2 − 25)2
1
A
= + other terms, where the
have the form
2
2
2
2
2
2
z(z + 16)(z − 16)(z + 25) (z − 25)
z
other terms contribute 0 to the integral. Why is this?
z
Now we compute A =
= −10−8 . Therefore
z(z 2 + 16)(z 2 − 16)(z 2 + 25)2 (z 2 − 25)2 z=0
Z
Z
cos π(z − 1)
cos π(z − 1)
dz = A
dz = 2πı &times; 10−8 .
2
2
2
2
2
2
z(z
+
16)(z
−
16)(z
+
25)
(z
−
25)
z
C
C
Question Where did the minus sign go?
1
2. Let C be a simple closed contour and let D be its interior. Suppose f (z) and g(z) are
analytic in D and on C.
(a) Show that f (z) = g(z) ∀z ∈ D if f (z) = g(z) ∀z ∈ C.
Z
Z
f 0 (ζ)
f (ζ)
dζ ∀z ∈ D.
(b) Show that
dζ =
2
C ζ −z
C (ζ − z)
Solution:
(a) By the Cauchy Integral Theorem:
1
for any z ∈ D, f (z) =
2πı
Z
C
1
f (ζ)
dζ =
ζ −z
2πı
Z
C
g(ζ)
dζ = g(z)
ζ −z
(b) Apply the Cauchy Integral Theorem twice:
Z
Z
1
f (ζ)
1
f (ζ)
0
f (z) =
dz =⇒ f (z) =
dz
2πı C ζ − z
2πı C (ζ − z)2
Z
1
f 0 (ζ)
0
dz
f (z) =
2πı C ζ − z
3. Suppose f (z) is an entire function such that Im(f (z)) is bounded. Show that f (z) is
a constant.
Solution: Let f (z) = u(x, y) + ıv(x, y). Then apply Liouville’s theorem to the entire
function g(z) = eıf (z) = e−v+ıu :
|g(z)| = = e−v+ıu = e−v = e−Im(f (z)) is bounded =⇒ f (z) is a constant.
Q
sj
4. Suppose P (z) = j=k
is a polynomial in factored form and C is a positively
j=1 (z − rj )
oriented simple closed contour such that rZ1 , . . . , rn are in the interior of C and the rest
P 0 (z)
of the roots are exterior to C. Show that
dz = 2πı(s1 + &middot; &middot; &middot; + sn ).
C P (z)
j=k
X
P 0 (z)
P 0 (z)
sj
is
=
, and
Solution: The partial fraction decomposition of
P (z)
P (z)
z − rj
j=1
therefore
Z
C
j=k
X
P 0 (z)
dz =
P (z)
j=1
Z
5. Evaluate
C
Z
C
j=n
X
sj
dz = 2πı
sj since the other roots are exterior to C
z − rj
j=1
eız
dz, where C is the circle x2 + (y − 1)2 = 1, oriented positively.
(z 2 + 1)3
2
eız
fails to be analytic only at z = &plusmn;ı. First we deter(z 2 + 1)3
1
.
mine the partial fraction decomposition of 2
(z + 1)3
Solution: The integrand
(z 2
A1
A2
A3
B1
B2
B3
1
=
+
+
+
+
+
3
2
3
2
+ 1)
z − ı (z − ı)
(z − ı)
z + ı (z + ı)
(z + ı)3
1 d2
1 d2
3
(z − ı)3 1
=
A1 =
=− ı
2
2
3
2
3
2 dz
(z + 1)
2 dz
(z + ı)
16
z=ı
z=ı
3
d
3
(z − ı)
1
d
A2 =
=
=−
2
3
3
dz (z + 1)
dz (z + ı)
16
z=ı
z=ı
3 ı
(z − ı) 1
=
=
A3 =
(z 2 + 1)3 z=ı (z + ı)3 z=ı 8
eız
We don’t need to compute the other numerators since
is analytic on and within
(z + ı)n
Z
eız
C for any integer, and therefore
dz = 0. Finally
n
C (z + ı)
Z
Z
Z
Z
eız
eız
eız
eız
dz
=
A
dz
+
A
dz
dz
+
A
1
3
2
2
3
2
3
C z−ı
C (z − ı)
C (z + 1)
C (z − ı)
d
d2
= 2πıA1 eız |z=ı +2πıA2 (eız ) |z=ı +πıA3 2 (eız ) |z=ı
dz
dz
3π 3π
π
7π
=
+
+
=
8e
8e 8e
8e
3
```