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Section VI.1, Problem 1
Want to get everything to the form
1 1 −w
where we know that |w| < 1.
a. For 0 < |z| < 1, we have
f (z) =
z
1 − z
z
∑^ ∞
k=
z
k
∑^ ∞
k=− 1
−z
k .
For |z| > 1, we have
f (z) =
z^2
1 − (1/z)
z^2
∞ ∑
k=
z
)k
− 2 ∑
k=−∞
z
k .
b.
0 ≤ |z| < 1 : f (z) = 1 −
1 − (−z)
∞ ∑
k=
(−z)
k = −1 +
∞ ∑
k=
k+ z
k .
|z| > 1 : f (z) = 1 −
z
1 − (− 1 /z)
z
∑^ ∞
k=
z
)k
∑^ −^1
k=−∞
k z
k .
c. Do partial fractions; i.e. write
(z
2 − 1)(z
2 − 4)
a
z − 1
b
z + 1
c
z − 2
d
z + 2
then cross-multiply to get
1 = a(z +1)(z −2)(z +2)+b(z −1)(z −2)(z +2)+c(z −1)(z +1)(z +2)+d(z −1)(z +1)(z −2).
Successively trying the values 1, − 1 , 2, and −2, we have
1 = − 6 a, 1 = 6b, 1 = 12c, 1 = − 12 d.
Thus
f (z) =
z − 1
z + 1
z − 2
z + 2
but it’s probably easier to recombine the pairs to get
f (z) =
1 − z
2
4 − z
2
1
0 ≤ |z| < 1 : f (z) =
1 − (z
2 )
1 − (z
2 /4)
∞ ∑
k=
(z
2 )
k
∞ ∑
k=
z
2
)k
∞ ∑
k=
k )
z
2 k
1 ≤ |z| < 2 : f (z) =
3 z^2
1 − (1/z^2 )
1 − (z^2 /4)
3 z
2
∑^ ∞
k=
z
2
)k
∑^ ∞
k=
z
2
)k
∑^ −^1
k=−∞
z
2 k
∑^ ∞
k=
k )
z
2 k
|z| > 2 : f (z) =
3 z^2
1 − (1/z^2 )
3 z^2
1 − (4/z^2 )
3 z
2
∞ ∑
k=
z
2
)k
3 z
2
∞ ∑
k=
z
2
)k
∑^ −^1
k=−∞
−k− 1
z
2 k
Section VI.1, Problem 2
Let w = z − (−1) = z + 1 so I don’t have to write as much. Note that the distance from
−1 to
1 2
is
3 2
so we are trying to find power series that converge at |w| =
3 2
a. (Using partial fractions in the middle.)
f (z) =
w
2 − 3 w + 2
(w − 2)(w − 1)
w − 2
w − 1
1 − (w/2)
w
1 − (1/w)
∞ ∑
k=
w
)k
w
∞ ∑
k=
w
)k
− 1 ∑
k=−∞
−(z − (−1))
k
∞ ∑
k=
k+
(z − (−1))
k
When we expanded out the geometric series on the second-to-last line, we assumed 1 < |w| <
b.
f (z) =
w + 2
w
w
= 2w
− 1
and this Laurent expansion converges on |w| > 0.
Section VII.1, Problem 3
b. Simple poles at ±1, residues are
e±^1 2(±1)
using Rule 3, so the value of the integral is
2 πi
e − e
− 1
= πi
e − e
− 1
d. sin z has simple zeros at every multiple of π; the only one that’s inside our contour is
at 0, but this is cancelled by the zero on top, hence the function has is analytic inside the
contour and the integral is zero.
f. tan z has a pole at every zero of its denominator, cos z; i.e. at every odd multiple of
π 2
The pole at z = 0 is cancelled by the zero of tan z there and so the only pole inside the
contour is at z =
π 2
. Using Rule 3, we calculate the residue as
sin z z d dz
cos z
z
π
as z →
π 2
, then the value of the integral is
2 πi
π
= − 4 i.
Section VII.1, Problem 6
a. Draw a picture of the parallelogram; it looks like a rectangle of width
1 2
that’s been
stretched so the sides are at 45
◦ .
The denominator has a simple zero at z = 0, so use Rule 3 to calculate the residue:
e
πi(z− 1 /2)^2
d dz
(1 − e
− 2 πiz )
e
πi(z− 1 /2)^2
2 πie−^2 πiz^
e
πi/ 4
2 πi
at z = 0,
and so the value of the integral is
e
πi/ 4
1 + i √ 2
b. For the top side of DR, use an M-L estimate: Parameterize by z = x + (1 + i)R where x
ranges from
1 2
to −
1 2
. The modulus of the integral on the top side (for R large enough) is
e
πi(x+R+iR− 1 /2)^2
1 − e−^2 πi(x+R+iR)
∣e
πi(x+R+iR− 1 /2)^2
|e−^2 πi(x+R+iR)| − 1
e
− 2 πR(x+R− 1 /2)
e^2 πR^ − 1
e^2 πR^ − 1
and the length of the integral is 1 thus
∣ ∣ ∣ ∣
top of DR
e
2 πR − 1
as R → ∞.
For the bottom side of DR, also use an M-L estimate: Parameterize by z = x − (1 + i)R
where x ranges from −
1 2
to
1 2
. The modulus of the integral on the bottom side (for R large
enough) is
e
πi(x−R−iR− 1 /2)^2
1 − e
− 2 πi(x−R−iR)
∣e
πi(x−R−iR− 1 /2)^2
1 − |e
− 2 πi(x−R−iR) |
e
− 2 πR(−x+R+1/2)
1 − e
− 2 πR
e
− 2 πR^2
and the length of the integral is 1 thus ∣ ∣ ∣ ∣
bottom of DR
≤ 2 e
− 2 πR^2 → 0
as R → ∞.
On the right side we parameterize by z =
1 2
- (1 + i)t where t ranges from −R to R. On
the left side we parameterize by z =
1 2
- (1 + i)t where t ranges from R to −R. Thus
∫
right side of DR
left side of DR
∫ R
−R
e
− 2 πt^2
1 + e−^2 πi(1+i)t^
(1 + i)dt −
∫ R
−R
−e
− 2 πt^2 − 2 πi(1+i)t
1 + e−^2 πi(1+i)t^
(1 + i)dt
= (1 + i)
R
−R
e
− 2 πt^2 1 +^ e
− 2 πi(1+i)t
1 + e
− 2 πi(1+i)t
dt
= (1 + i)
∫ R
−R
e
− 2 πt^2 dt
c. We have that
(1 + i)
−∞
e
− 2 πt^2 dt =
1 + i √ 2
and a u-substitution gets the result.
Section VII.2, Problem 1
Formula (2.4) where the poles are at ±ai with residues (Rule 3)
± 1 2 ai
(so only one is in the
upper half-plane). It would be a good idea to be able to do this problem without the use
of formula 2.4, but that derivation is exactly the same as the integral they evaluate at the
start of the section.
Section VII.2, Problem 4
The integrand is a rational function (quotient of two polynomials) and the degree of the
numerator is 0, the denominator, 4 so we can use formula (2.4). The (simple) poles are at
the fourth roots of −1: e
(πi+2πik)/ 4 with k = 0, 1 , 2 , 3, but only the first two are in the upper
half-plane and these have residues (Rule 3)
1 4 eπi/^4
1 4 e^3 πi/^4
. Then formula (2.4) gives that the
integral is
2 πi
4 eπi/^4
4 e^3 πi/^4
πi
e
−πi/ 4
− 3 πi/ 4
π √ 2
Section VII.2, Problem 6
Simple poles at − 1 ± i and ± 2 i. The ones in the upper half-plane are −1 + i and 2i. It is
likely easiest to calculate the residues by Rule 1 (instead of Rule 3), and they are
−1 + i
((−1 + i) − (− 1 − i))((−1 + i)^2 + 4)
−1 + i
2 i(− 2 i + 4)
1 + 3i
2 i
((2i)^2 + 2(2i) + 2)(2i − (− 2 i))
2 i
4 i(4i − 2)
− 1 − 2 i
and so the value of the integral is
2 πi
1 + 3i
− 1 − 2 i
−π
