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Why closed loops often integrate to zero

Mathematics · Complex Analysis · ages 20-21

Name ______________________ Date ____________

For f = u + iv with u = 3 and v = 5, what is the coefficient of dy in the imaginary part of f dz?
Answer: ______________
f(z) = 1/(z - 2) has its singularity at 2. The unit circle |z| = 1 does not enclose 2, and f is holomorphic on an open set containing the circle and its interior. What is the integral of f around the unit circle?
- 2πi
- 0
- πi
Let γ be a simple closed piecewise C1 curve traversed once positively, and let f be holomorphic on an open set containing γ and its filled interior, with continuous partials there. Green plus Cauchy-Riemann zeroes both the real and imaginary parts of the loop integral.
Circle one: True False
Write f equal u plus iv. What is the real part of f dz?
- u dx minus v dy
- u dx plus v dy
- v dx minus u dy
f is holomorphic near a closed region with simple disjoint boundaries. The outer boundary runs counterclockwise and two inner boundaries run clockwise, each once. The inner clockwise integrals are 2 and 1. What is the outer counterclockwise integral?
Answer: ______________
f is defined and holomorphic on and inside a loop except at one interior point where it blows up. The loop integral must still be zero.
Circle one: True False
Two simple circles are nested, each traversed once counterclockwise, and f is holomorphic on a neighbourhood of the closed annulus between them. The outer integral is 3 + 4i. The inner circle is traversed clockwise instead. What is the inner integral?
- 3 + 4i
- -3 - 4i
- 0
f(z) = 1/(z - 1) and γ is the circle |z| = 1 traversed once. The ordinary contour integral of f along γ is defined. True or false?
Circle one: True False
f is holomorphic near a closed region with simple disjoint boundaries. The outer boundary runs counterclockwise and two inner boundaries run clockwise, each traversed once. The inner clockwise integrals are 4 - i and -2 + 3i. The outer boundary is then traversed clockwise twice. What is the imaginary part of the resulting integral?
Answer: ______________
f has poles at 0 and 4 and is holomorphic elsewhere. For the circles A: |z - 2| = 1, B: |z - 2| = 2 and C: |z| = 5, each traversed once counterclockwise, what is the joint conclusion?
- A is 0, B is undefined, C has no zero-guarantee
- A is 0, B is 0, C is 0
- A is undefined, B is 0, C has no zero-guarantee

Answer key

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

Why closed loops often integrate to zero W1-mt_aacYFVN-HH-s1

3 · The imaginary part is v dx + u dy, so the dy coefficient is u = 3.
0 · The singularity lies outside the loop, so Cauchy's theorem applies and gives 0.
True · Each part becomes a circulation whose curl is a Cauchy-Riemann difference.
u dx minus v dy · Multiplying out (u plus iv)(dx plus i dy) gives that real part.
-3 · The sum of oriented boundary integrals is 0, so outer + 2 + 1 = 0.
False · One interior singularity removes the guarantee, so zero is no longer forced.
-3 - 4i · Outer CCW equals inner CCW at 3 + 4i, and clockwise traversal negates to -3 - 4i.
False · The pole at 1 lies on the circle, so the integrand is undefined on the path.
4 · Outer CCW is -2 - 2i, so outer clockwise twice is 4 + 4i.
A is 0, B is undefined, C has no zero-guarantee · A's filled disk contains no pole so Cauchy gives 0, B passes through both poles so the ordinary integral is undefined, and C encloses both poles so the zero-guarantee fails without forcing nonzero.