November 24, 2010
Lecturer Dmitri Zaitsev
Michaelmas Term 2010
Course 2325 2010 Complex Analysis I
Sheet 3
Due: at the end of the lecture
Exercise 1
(i) Show that (ez )0 = ez . (Hint. Differentiate in the direction of the x-axis.)
Solution. Since
f 0 (z0 ) = lim
z→z0
=
f (z0 + h) − f (z0 )
f (z) − f (z0 )
= lim
=
h→0
z − z0
h
f (z0 + h) − f (z0 )
= ∂x f (z0 )
h∈IR, h→0
h
lim
holds for any holomorphic f (z), we have
(ez )0 = ∂x ex (cosy + isiny) = ex (cosy + isiny) = ez .
Exercise 2
(iii) Use (ii) to show that the conclusion of Cauchy’s theorem does not hold for f (z) = z̄.
Solution. It follows from (ii) that
Z
z̄ dz 6=
γ1
Z
z̄ dz
γ2
whenever γj corresponds to some yj with y1 6= y2 . Form a closed path C by following
γ1 from −1 to 1 and then γ2 with reversed orientation from 1 to −1. Then
Z
z̄ dz =
C
Z
z̄ dz −
γ1
Z
z̄ dz 6= 0.
γ2
Thus we have a closed path for which the conclusion of Cauchy’s theorem does not hold
with f (z) = z̄.
Exercise 3
(i) Calculate
R
γ
f (z) dz, where f (z) =
Solution.
Z
γ
1
dz =
z
Z
1
z
and γ(t) = eit , 0 ≤ t ≤ 2π, is the unit circle.
2π
0
ieit dz
= 2πi.
eit
(ii) Use (i) to show that f (z) does not have an antiderivative in its domain of definition.
Solution. It f had antiderivative, its integral would be zero over any closed path,
which is not the case.
(iii) Use Exercise 1 (ii) to give an example of a domain Ω, where f does have an
antiderivative.
Solution. Take Ω = C \ {0}, the domain of definition of f (z) = 1z , then (ii) shows
that f does not have antiderivative in Ω.
(iv) Does f (z) =
1
zn
have an antiderivative, where n ≥ 2 is an integer?
Solution. Yes, take F (z) =
Justify your answer.
1
(−n+1)z n−1 .