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Chapter 1, Exercise 5
A set Ω is said to be pathwise connected if any two points in Ω can be joined
by a (piecewise-smooth) curve contained entirely in Ω. The purpose of this
exercise is to prove that an open set Ω is pathwise connected if and only if Ω is
connected.
Part (a)
Suppose first that Ω is open and pathwise connected, and that it can be written
as Ω = Ω 1 ∪ Ω 2 where Ω 1 and Ω 2 are disjoint, non-empty open sets. Chhose
two points w 1 ∈ Ω 1 and w 2 ∈ Ω 2 and let γ denote a curve in Ω joining w 1 to
w 2. Consider a Parameterization z : [0, 1] → Ω of this curve with z(0) = w 1 and
z(1) = w 2 , and let
t
∗ = sup 0 ≤t≤ 1
{t : z(s) ∈ Ω 1 for all 0 ≤ s < t}.
Arrive at a contradiction by considering the point z(t
∗ )
Solution
As suggested, we consider the point z(t
∗ ). We ask the question: which of Ω 1
and Ω 2 contains this point? Evidently, this point is not in Ω 1 : if z(t
∗ ) is in Ω 1 ,
then, because Ω 1 is open, there is an open ball B containing z(t
∗ ). Since z is
continuous, it follows that z
− 1 (B) is open as a subset of [0, 1]. Thus (assuming
t
∗ < 1) z
− 1 (Ω 1 ) contains points to the right of t
∗ , which is impossible. If
t
∗ = 1, then there is a sequence of points in Ω 1 that converges to z(1) ∈ Ω 2 ,
contradicting the assumption that Ω 2 is open.
If we assume instead that z(t ∗ ) ∈ Ω 2 , we recognize that z(t) ∈ Ω 2 if and
only if t > t ∗
. Thus t ∗ is the infimum of all values of t such that z(t) ∈ Ω 2 ,
and we can use the same argument as in the previous paragraph to conclude
z(t ∗ ) ∈/ Ω 2. Since z(t ∗ ) ∈ Ω 1 ∪ Ω 2 , this is a contradiction.
0.1 Part b
Conversely, suppose that Ω is open and connected. Fix a point w ∈ Ω and let
Ω 1 ⊂ Ω denote the set of all points that can be joined to w by a curve contained
in Ω. Also, let Ω 2 ⊂ Ω denote the set of all points that cannot be joined to w
by a curve in Ω. Prove that both Ω 1 , Ω 2 are open, disjoin, and their union is
Ω. Finally, since Ω 1 is nonempty (why?) conclude that Ω = Ω 1 as desired.
0.1.1 Solution
Evidently Ω 1 ∪ Ω 2 = Ω and Ω 1 is disjoint from Ω 2. The only thing that remains
to be shown is that both Ω 1 and Ω 2 are open.
Let w 1 ∈ Ω 1. Because Ω is open, Ω contains an open ball B centered at w 1.
It is obvious that if w
∗ ∈ B, then there is a path z
∗ connecting w 1 and w
∗
. Let
z 1 be a curve joining w to w 1. Then consider the curve defined by
z(t) =
z(2t) if 0 ≤ t < 1 / 2
z(2t − 1) if 1/ 2 ≤ t ≤ 1
Then z is a continuous, piecewise smooth curve that connects w to w ∗
. It follows
that B ⊂ Ω 1 and that Ω 1 is open.
Now, let w 2 ∈ Ω 2. Because Ω is open Ω contains an open ball B centered at
w 2. Let w ∗ ∈ B. If there were a curve z 2 that connected w to w ∗ , then we could,
as in the previous paragraph, find a curve connecting w to w 2 by concatenating
the path from w to w ∗ and the path from w ∗ to w 2. Thus w 2 ∈ Ω 1 , which is a
contradiction.
Thus Ω can be written as Ω 1 ∪ Ω 2 for disjoint open sets Ω 1 and Ω 2. Since
Ω is connected, either Ω 1 = Ω or Ω 2 = Ω. But w ∈ Ω 1 , so Ω 1 is nonempty and
therefore Ω 1 = Ω.
Chapter 1, Exercise 7
The family of mappings introduced here plays an important role in complex
analysis. These mappings, sometimes called Blaschke factors, will reappear in
the various applications in later chapters.
Part a
Let z, w be two complex numbers such that ¯zω 6 = 1. Prove that
w − z
1 − wz¯
< 1 if |z| < 1 and |w| < 1
and also that (^) ∣ ∣ ∣ ∣
w − z
1 − wz¯
= 1 if |z| = 1 or |w| = 1.
[Hint: Why can we assume that z is real? It then suffices to prove that
(r − w)(r − w¯) ≤ (1 − rw)(1 − r w¯)
with equality for appropriate r and |w|.]
Solution
(i) and (iii) directly follow from part (a) of the problem except for the holomor-
phy, which is clear except when z w¯ = 1. This can only happen if |w| = 1 and
z =
1 w
. It is seen that F has a removable singularity at z =
1 w with value w.
(ii) follows by plugging in: the numerator is clearly 0 when z = w, and plugging
in z = 0 makes the numerator equal to w and the denominator equal to 1. All
that remains to be seen is that F is bijective on D. Consider F ◦ F (z). This is
w − w−z 1 − wz¯
1 − w¯
w−z 1 − wz¯
We simplify this:
w −
w−z 1 − wz¯
1 − w¯
w−z 1 − wz¯
w −
w−z 1 − wz¯
w¯(w−z) 1 − wz¯
w(1 − wz¯ ) − (w − z)
q − wz¯ − w¯(w − z)
w − |w| 2 z − w + z
1 − wz¯ − |w| 2
z(1 − |w| 2 )
1 − |w| 2
= z
so the function F is an involution and therefore bijective on D.
Chapter 1, Exercise 13
Suppose that f is holomorphic in an open set Ω. Prove that in any one of the
following cases:
- Re(f ) is constant;
- Im(f ) is constant;
- |f | is constant; one can conclude that f is constant.
Solution
Suppose that Re(f ) is constant. Then f (x, y) = a + iv(x, y) for z = x + iy.
Then we consider the PDEs from the Cauchy-Riemann equations:
∂u
∂x
∂v
∂y
So
∂v ∂y is zero, and thus v depends only on x and
∂u
∂y
∂v
∂x
so
∂v ∂x is zero and thus v depends only on y. Since v cannot depend on
either x or y, it follows that v is constant.
The same argument works if Im(f ) is constant. Alternatively, if Im(f ) is
constant, then Re(if ) is constant and so if , and thus f , is constant.
Now suppose |f | is constant. Writing f (z) = u(x, y) = iv(x, y) for z =
x + iy, we then have that u(x, y) 2
- v(x, y) 2 is constant. In particular,
this implies that
∂u ∂x
∂v ∂x
and that
∂u ∂y
∂v ∂y
Thus we can again use
the Cauchy-Riemann equations:
∂u
∂x
∂v
∂y
∂v
∂x
and ∂u
∂x
∂u
∂y
∂v
∂x
So we get
∂u ∂x is equal to both
∂v ∂x and −
∂v ∂x , showing that both are equal
to zero, and by the same logic as before, f is constant.
