Math 411: Honours Complex Variables - University of Alberta

1. Given a < b < c and two curves γ1 : [a,b] → C and γ2 : [b,c] → C with
γ1(b) = γ2(b), the concatenation of γ1 and γ2 is the curve
�
γ1(t), t ∈ [a,b],
γ1 ⊕γ2: [a,c] → C, t ↦→
γ2(t), t ∈ [b,c].
If γ1 and γ2 are piecewise smooth, then so is γ1 ⊕γ2, and we have
�
γ1⊕γ2
�
f =
for each continuous f: {γ1}∪{γ2} → C.
γ1
�
f +
2. For any curve γ: [a,b] → C, the reversed curve is defined as
γ − : [a,b] → C, t ↦→ γ(a+b−t).
If γ is piecewise smooth, then so is γ− , and we have
� �
f = − f
for each continuous f: {γ} → C.
γ −
3. We denote the straight line segment {z0 +t(z −z0) : t ∈ [0,1]} by [z0,z].
Proposition 4.2 (Locally Constant vs. Connectivity). Let D ⊂ C be open. Then
the following are equivalent:
(i) D is connected;
(ii) every locally constant function f: D → C is constant;
(iii) for any z,w ∈ D, there exists a piecewise smooth curve γ: [a,b] → D such that
γ(a) = z and γ(b) = w.
Proof. (iii) =⇒ (ii): Let f: D → C be a locally constant function, and let z,w ∈ D.
Let γ: [a,b] → D be a piecewise smooth curve with γ(a) = z and γ(b) = w. Since f
is locally constant, the function
γ
γ2
[a,b] → C, t ↦→ f(γ(t))
is differentiable with zero derivative and therefore constant. It follows that f(z) =
f(γ(a)) = f(γ(b)) = f(w).
(ii) =⇒ (i): Suppose that D is not connected. Then there exist non-empty open
sets U,V ⊂ C with U ∩V = ∅ and U ∪V = D. Define
f: D → C, z ↦→
f
� 0, z ∈ U,
1, z ∈ V.
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