Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
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Chapter 15<br />
Analytic Continuation along a<br />
Curve<br />
Example. Let<br />
and<br />
so that<br />
Let<br />
D1 := {z ∈ C : Rez > 0},<br />
D2 := {z ∈ C : Imz > Rez},<br />
D3 := {z ∈ C : Imz < −Rez},<br />
D1 ∪D2 ∪D3 = C\{0}.<br />
g: C\{0} → C, z ↦→ 1<br />
z ,<br />
and let f1 = Log|D1, so that f1 is an antiderivative <strong>of</strong> g on D1. Since D2 is simply<br />
connected, g also has anantiderivative onD2; since (f ′ 1 −f′ 2 )|D1∩D2= (g−g)|D1∩D2≡ 0,<br />
it follows that (f1−f2)|D1∩D2 is constant, and by altering f2 by an additive constant,<br />
we can achieve that f1|D1∩D2= f2|D1∩D2. In the same fashion, we can find an antiderivative<br />
f3 <strong>of</strong> g on D3 such that f2|D2∩D3= f3|D2∩D3. However, f1|D1∩D3�= f3|D1∩D3<br />
because otherwise, we would have an antiderivative <strong>of</strong> g on all <strong>of</strong> C\{0}, which we<br />
know to be impossible.<br />
Since f ′ 1 − f′ 3 |D1∩D3= g − gD1∩D3 ≡ 0, however, there exists c ∈ C such that<br />
f3(z) = f1(z)+cforz ∈ D1∩D2. Weclaimthatc = 2πi. Toseethis, letz1,z2,z3 ∈ ∂D<br />
be such that z1 ∈ D1∩D3, z2 ∈ D2∩D1, and z3 ∈ D3∩D2. Let γz1,z2, γz2,z3, and γz3,z1<br />
be the arc segments <strong>of</strong> ∂D from z1 to z2, from z2 to z3, and from z3 to z1, respectively.<br />
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