Exercise 8.25

Exercise 25: Suppose γ 1 and γ 2 are curves as in Exercise 23, and

| γ 1 ( t ) − γ 2 ( t ) | < | γ 1 ( t ) | ( a ≤ t ≤ b ) .

Prove that ( γ 1 ) = ( γ 2 ) .

Answers

Following the hint, let γ = γ 2 ∕ γ 1 . This is well-defined since γ 1 is nowhere 0, and so γ is also a continuous differentiable closed curve defined on [ a , b ] which is nowhere 0, so that we can define ( γ ) . By the condition on γ 1 and γ 2 we have for all t ∈ [ a , b ]

| 1 − γ ( t ) | = | γ 1 ( t ) − γ 2 ( t ) | | γ 1 ( t ) | < | γ 1 ( t ) | | γ 1 ( t ) | = 1 .

Hence the range of γ does not intersect the negative real axis, so by Exercise 24 we have

0 = ( γ ) = 1 2 πi ∫ a b γ ′ ( t ) γ ( t ) dt = 1 2 πi ∫ a b γ 1 ( t ) γ 2 ′ ( t ) − γ 2 ( t ) γ 1 ′ ( t ) γ 1 2 ( t ) ⋅ γ 1 ( t ) γ 2 ( t ) dt = 1 2 πi ∫ a b ( γ 2 ′ ( t ) γ 2 ( t ) − γ 1 ′ ( t ) γ 1 ) ) dt = ( γ 2 ) − ( γ 1 )

so that ( γ 1 ) = ( γ 2 ) .

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2023-08-07 00:00
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