Contour Integrals Along Parametrised Paths · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Walking a function along a path

Mathematics · Complex Analysis · ages 20-21
Name ______________________   Date ____________
  1. For γ(t) = t + it² on [0, 1], what is the imaginary part of ∫_γ z dz?

    Answer: ______________

  2. For γ(t) = t + it on [0, 1], what is γ'(t)?

    • 1 + i
    • √2
    • 1
  3. With γ(t) = t for t in [0, 3], the integral of z dz becomes the integral of t dt. What is its value?

    Answer: ______________

  4. Reversing a path negates the integral.

    Circle one:   True   False

  5. A path has two pieces of lengths 2 and 3. If |f| ≤ 5 on both, what bound does the estimation lemma give?

    Answer: ______________

  6. Let γ(t) be defined for t in [0, 2]. Which reparametrisation is guaranteed to preserve ∫_γ f(z) dz for every continuous f and any piecewise continuously differentiable γ?

    • γ(2-s) for s in [0, 2]
    • γ(2s) for s in [0, 1]
    • γ(s/2) for s in [0, 2]
  7. For γ(t) = 2e^(-it) on [0, 2π], what is ∫_γ conj(z) dz?

    • 0
    • -8πi
    • 8πi
  8. If ∫_γ f(z) dz = 0, then reversing γ gives the same value 0. True or false?

    Circle one:   True   False

  9. A path consists of a unit circle traversed twice anticlockwise with |f| ≤ 2, plus a segment of length 3 with |f| ≤ 4. Using π < 3.15, what bound does the estimation lemma give?

    Answer: ______________

  10. Let γ1 and γ2 be closed paths from z = 1 back to z = 1 with ∫_γ1 f dz = 3 + i and ∫_γ2 f dz = 2 - i. Since both are loops at the same point, they can be concatenated. A new path goes around γ1 twice, then around γ2 once in reverse. What is the integral?

    • 4 + 3i
    • 8 + i
    • 1 + 2i
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Walking a function along a path W1-mt_CaJ7ruAv8P-s1

  1. 1 · The product's imaginary part is 3t², which integrates to 1.
  2. 1 + i · Differentiate in the real parameter: γ'(t) = 1 + i.
  3. 4.5 · The antiderivative is t²/2, giving 9/2.
  4. True · Traversing backwards negates the velocity, hence the whole integral.
  5. 25 · Total length 5, so bound is 5 × 5 = 25.
  6. γ(2s) for s in [0, 1] · γ(2s) on [0,1] preserves the original oriented traversal and multiplicity, so the velocity compensates for the doubled speed.
  7. -8πi · conj(γ(t)) = 2e^(it) and γ'(t) = -2i e^(-it), so the product is -4i and the integral is -8πi.
  8. True · Reversal negates the integral, and -0 = 0, so zero stays zero.
  9. 37.2 · Circle length 2 × 2π < 12.6, segment length 3. Bound = 2 × 12.6 + 4 × 3 = 37.2.
  10. 4 + 3i · Two γ1 traversals give 2(3 + i) = 6 + 2i, and the reversed γ2 contributes -(2 - i) = -2 + i, leaving 4 + 3i.
Worksheet · LightMySky