The document discusses three functions and their integrals over the unit circle, showing that the integrals evaluate to zero. It employs parameterization and properties of odd functions to conclude that the integrals for various forms of f(z) result in cancellation. Specifically, it demonstrates this for f(z) = z³ - 1 + 3i, f(z) = z² + 1/(z - 4), and f(z) = sin(z)(z² - 25)(z² + 9).

1. Show that
C
f(z)dz = 0, where f is the given function and C is the unit circle |z| = 1.
a) f(z) = z3
− 1 + 3i
b) f(z) = z2
+
1
z − 4
c) f(z) =
sin z
(z2 − 25) (z2 + 9)
Solution by Mikołaj Hajduk: The points a) and b) can be solved with use of parametrization z(t) = eit
where t ∈ [0, 2π] and dz = ieit
dt.
Ad. a)
C
f(z)dz =
C
(z3
− 1 + 3i)dz =
2π
0
((eit
)3
− 1 + 3i)ieit
dt =
=
2π
0
e3it
ieit
+ (−1 + 3i)ieit
dt =
2π
0
ie4it
dt +
2π
0
(−1 + 3i)ieit
dt =
= i
2π
0
e4it
dt + (−1 + 3i)i
2π
0
eit
dt = i
1
4i
e4it
2π
0
+ (−1 + 3i)i
1
i
eit
2π
0
=
=
1
4
e4it
2π
0
+ (−1 + 3i)eit
2π
0
=
1
4
(e4i∗2π
− e4i∗0
) + (−1 + 3i)(ei∗2π
− ei∗0
) =
c 2015/09/17 02:27:23, Mikołaj Hajduk 1 / 4 next

2. =
1
4
((e2iπ
)4
− e0
) + (−1 + 3i)(e2iπ
− e0
) =
1
4
(1 − 1) + (−1 + 3i)(1 − 1) = 0
Ad. b)
C
f(z)dz =
C
z2
+
1
z − 4
dz =
2π
0
(eit
)2
+
1
eit − 4
ieit
dt =
=
2π
0
ie2it
eit
dt +
2π
0
ieit
eit − 4
dt = i
2π
0
e3it
dt +
2π
0
(eit
− 4)
eit − 4
dt =
= i
1
3i
e3it
2π
0
+ ln |eit
− 4|
2π
0
=
1
3
(e3i∗2π
− e3i∗0
) + (ln |ei∗2π
− 4| − ln |ei∗0
− 4|) =
=
1
3
((e2iπ
)3
− e0
) + (ln |1 − 4| − ln |1 − 4|) =
1
3
(1 − 1) + (ln 3 − ln 3) = 0
Ad. c)
Let’s notice that the function f is odd in its domain because
sin(−z)
def
=
ei(−z)
− e−i(−z)
2i
=
e−iz
− eiz
2i
= −
eiz
− e−iz
2i
def
= − sin(z)
and hence
f(−z) =
sin(−z)
((−z)2 − 25) ((−z)2 + 9)
=
− sin z
(z2 − 25) (z2 + 9)
= −f(z)
Let CU and CB denote the circle arcs corresponding to the upper and bottom halves of the circle C. We have
then
C
f(z)dz =
CU
f(z)dz +
CB
f(z)dz
c 2015/09/17 02:27:23, Mikołaj Hajduk 2 / 4 next

3. The function h : CU −→ CB defined as h(z) = −z is a bijection between points of the arcs CU and CB that
transforms the beginning point of the arc CU into the beginning point of the arc CB and the ending point of
the arc CU into the ending point of the arc CB. The arc CB is an image of the arc CU under the function h:
CB = h(CU)
c 2015/09/17 02:27:23, Mikołaj Hajduk 3 / 4 next

4. therefore, bearing in mind that f(z) is odd and h−1
(z) = −z, we get
CB
f(z)dz =
h(CU)
f(z)dz =
CU
f(h−1
(z))dz =
CU
f(−z)dz =
CU
−f(z)dz = −
CU
f(z)dz
Hence
C
f(z)dz =
CU
f(z)dz +
CB
f(z)dz =
CU
f(z)dz −
CU
f(z)dz = 0
c 2015/09/17 02:27:23, Mikołaj Hajduk 4 / 4