Mathematics 3
Prof. F. Brock
Homework 8, solutions
1. Evaluate the complex line integral
Z
f (z) dz
γ
in the following cases.
(a) f (z) = z 3 , γ: part of the parabola x = y 2 , that connects the points 0 and
1 + i in the complex plane.
(b) f (z) = z + 2z, γ: the circular arc |z| = 2, −π/2 ≤ arg z ≤ π/2.
Solution:
(a) An antiderivative of f (z) = z 3 is F (z) = z 4 /4. Since f is holomorphic on
C, we find that
Z
(1 + i)4
z 3 dz = F (1 + i) − F (0) =
= −1.
4
γ
(b) We parametrize γ by z = 2eit , (−π/2 ≤ t ≤ π/2), so that dz = 2i eit dt,
and hence
Z π/2
Z π/2
Z
it
−it
it
(4i e2it + 8i) dt
(2e + 4e )2ie dt =
(z + 2z) dz =
−π/2
−π/2
γ
π/2
= 2(eπi − e−πi ) + 8πi = 8πi.
2e2it + 8it =
−π/2
2.(a) Show the identities
sin z = sin x cosh y + i cos x sinh y, cos z = cos x cosh y − i sin x sinh y.
(b) Evaluate the values cos(2 − i), Log (2 − 3i) and (3 + 4i)i .
Solution:
to (b): There holds
eiz =
∞
X
(iz)n
n=0
n!
=
∞
X
k=0
(−1)k
∞
X
z 2k
z 2k+1
+i
(−1)k
= cos z + i sin z.
(2k)!
(2k
+
1)!
k=0
Setting z = −iy, and z = iy, we obtain ey = cos(iy) − i sin(iy), and e−y =
cos(iy) + i sin(iy), respectively. This implies
cosh y = cos(iy), sinh y = −i sin(iy), and i sinh y = sin(iy).
Now the Theorems of Addition for trigonometric functions give
sin z = sin(x + iy) = sin x cos(iy) + cos x sin(iy) = sin x cosh y + i cos x sinh y
and
cos z = cos(x+iy) = cos x cos(iy)−sin x sin(iy) = cos x cosh y −i sin x sinh y.
to (a): Using (b) we have
cos(2 − i) = cos 2 cosh 1 + i sin 2 sinh 1.
By the definition of the Logarithm it follows that
Log (2 − 3i) = log(2 − 3i) + i 2kπ = ln |2 − 3i| + i(arg(2 − 3i) + 2kπ)
√
3
= ln 13 + i(arctan + 2kπ), (k ∈ Z).
2
Finally we have
4
(3 + 4i)i = ei log(3+4i) = ei (ln 5+i arctan 3
4
4
= ei ln 5−arctan 3 = e− arctan 3 (cos(ln 5) + i sin(ln 5)).
3. Find the Taylor expansions of the following functions:
(a) e2z−1 , into powers of z + 2.
1
, into powers of z.
(b)
(z − 1)2 (z + 2)
Solution:
(a) There holds
e
2z−1
2(z+2)−5
=e
−5 2(z+2)
=e e
=e
−5
∞
X
2n
(z + 2)n .
n!
n=0
(b) Using partial fractions: Since
1
d
=
2
(1 − z)
dz
=
∞
X
n=0
1
1−z
d
=
dz
(n + 1) z n ,
∞
X
n=0
!
z
n
=
∞
X
n=1
n z n−1
we obtain
1
1 1
1 1
1
1
=
−
+
2
(z − 1) (z + 2)
9 z + 2 9 z − 1 3 (z − 1)2
1 1 1
1
3
=
+
+
9 2 1 + z2
1 − z (1 − z)2
!
∞
∞
∞
X
1 1 X z n X n
−
+
z +3
(n + 1) z n
=
9 2 n=0
2
n=0
n=0
∞ n
X
(−1)
1
=
+ 1 + 3(n + 1) z n
n+1
9 n=0
2
∞
1 X (−1)n
=
+ 3n + 4 z n .
9 n=0
2n+1
Using Cauchy’s product formula: As above we have
∞
X
1
=
(n + 1) z n
(z − 1)2
n=0
∞
and
X (−1)n
1
=
zn.
z + 2 n=0 2n+1
By Cauchy’s product formula we obtain finally
∞ X
n
k
X
1
n−k (−1)
=
(n
−
k
+
1)
z
zk
k+1
(z − 1)2 (z + 2)
2
n=0 k=0
=
∞
X
n=0
Since
n
X
n
n
n + 1 X (−1)k 1 X (−1)k
−
k
2 k=0 2k
2 k=0
2k
wk =
k=0
wn+1 − 1
w−1
and
n
X
k=0
n
kw
k
∂ X k
∂
= w
w =w
∂w k=0
∂w
=
wn+1 − 1
w−1
n wn+2 − (n + 1) wn+1 + w
(w − 1)2
!
zn.
we find, for w = −1/2
n
n
n + 1 X (−1)k 1 X (−1)k
−
k
2 k=0 2k
2 k=0
2k
(−1)n+2
2
n + 1 (−1)n+1
(−1)n+1 1
n
−1 −
− (n + 1) n+1 −
= −
3
2n+1
9
2n+2
2
2
n+2
n+2
n+2
1
(−1)
(−1)
(−1)
=
(3 n + 3)
+ 3n + 3 − n
− (2n + 2) n+1 + 1
9
2n+1
2n+1
2
n
1 (−1)
+ 3n + 4 ,
=
9 2n+1
that is,
∞
1
1X
=
(z − 1)2 (z + 2)
9 n=0
(−1)n
+ 3n + 4
2n+1
zn.
4. Evaluate the complex line integral
Z
cos z
dz
2
γ z(z − 2)
in the following cases. The closed curves below are oriented counterclockwise.
(a) γ: the circle |z| = 1.
(b) γ: the circle |z − 2| = 1.
(c) γ: the circle |z − 2i| = 1.
Hint: Use Cauchy’s integral theorem and Cauchy’s integral formula.
Solution: Decomposition into partial fractions gives
11 1 1
1
1
1
=
−
+
,
2
z(z − 2)
4 z 4 z − 2 2 (z − 2)2
that is,
Z
cos z
dz
2
γ z(z − 2)
Z
Z
Z
1
cos z
1
cos z
1
cos z
=
dz −
dz +
dz
4 γ z
4 γ z−2
2 γ (z − 2)2
1
1
1
=:
I1 − I2 + I3 .
4
4
2
(a) The integrands of I2 and I3 are holomorphic for |z| ≤ 1. Hence both
integrals are = 0 by Cauchy’s integral theorem. Further, Cauchy’s integral
formula gives I1 = 2 π i cos 0 = 2 π i, hence we have that
Z
cos z
πi
I=
dz =
.
2
2
γ z(z − 2)
I
=
(b) The integrand of I1 is holomorphic for |z − 2| ≤ 1, and hence I1 = 0.
Further, Cauchy’s integral formulas give I2 = 2 π i cos 2 and I3 = −2 π i sin 2,
hence
Z
cos z
iπ
I=
cos 2 − i π sin 2.
dz = −
2
2
γ z(z − 2)
(c) All three integrands are holomorphic for |z − 2i| ≤ 1, so that I1 = I2 =
I3 = 0 by Cauchy’s integral theorem. Hence
Z
cos z
I=
dz = 0.
2
γ z(z − 2)