A function f is analytic everywhere on ℂ\{0, 1}. A student wants to integrate f along two different paths from point A to point B, both staying within the domain. Can they assume the two integrals are equal?
AYes — since f is analytic on the domain, path-independence is guaranteed by Cauchy's theorem
BNo — the domain has holes at 0 and 1, so it is not simply connected, and path-independence is not guaranteed
CYes — both paths stay within the domain, so they can be continuously deformed into each other
DNo — complex integration is never path-independent regardless of the domain
Path-independence for analytic functions requires a simply connected domain. ℂ\{0, 1} has two holes (the removed points), so it is not simply connected. A path that winds around the puncture at 0 cannot be continuously deformed into one that does not — the hole blocks the deformation. For functions like 1/z or 1/(z-1), integrating around these holes gives nonzero results. Option C is wrong because paths that wind differently around holes cannot be continuously deformed into each other even though both stay in the domain.
Question 2 Multiple Choice
The annulus A = {z : 1 < |z| < 2} is connected but not simply connected. What does 'not simply connected' mean geometrically for this region?
AThe annulus has two separate components that cannot be connected by a path within the region
BThe annulus is one piece, but a loop circling the inner hole cannot be continuously shrunk to a point while staying inside the annulus
CThe annulus has infinitely many connected components, one for each point removed from the disk
DThe outer and inner boundary circles each form separate connected components
Connected means the region is in one piece: any two points can be joined by a path inside A. The annulus satisfies this — you can travel from any point to any other without leaving A. Simply connected additionally requires no holes: every closed loop can be contracted to a point within the region. A loop winding around the inner hole of the annulus cannot be contracted to a point without crossing that hole. The annulus is the canonical example of a connected-but-not-simply-connected domain.
Question 3 True / False
The integral ∮_C (1/z) dz = 2πi for a circle C centered at the origin demonstrates that ℂ\{0} is not simply connected.
TTrue
FFalse
Answer: True
If ℂ\{0} were simply connected, Cauchy's theorem would guarantee that the integral of any analytic function around any closed curve equals zero. The fact that ∮ 1/z dz = 2πi ≠ 0 for a loop around the origin reveals that something is topologically non-trivial — the loop cannot be contracted to a point, because the origin (a hole) is inside it. The function 1/z 'detects' the hole through integration; the nonzero residue is a topological signature.
Question 4 True / False
A connected region in ℂ is automatically simply connected.
TTrue
FFalse
Answer: False
Connectivity only requires that the region is in one piece — any two points can be joined by a path inside the region. Simple connectivity is a stronger condition requiring additionally that every closed loop can be contracted to a point, i.e., that the region has no holes. The annulus {1 < |z| < 2} and the punctured plane ℂ\{0} are both connected (one piece) but not simply connected (they have holes). The implication goes one way: simply connected implies connected, but connected does not imply simply connected.
Question 5 Short Answer
Why is simple connectivity, rather than mere connectivity, the condition required to guarantee that analytic functions have path-independent integrals?
Think about your answer, then reveal below.
Model answer: In a connected but not simply connected domain, two paths from A to B can wind differently around holes in the domain. If the function has a singularity at such a hole, the contribution from winding around it (the residue) causes the two integrals to differ. Simple connectivity rules out holes entirely: any two paths from A to B can be continuously deformed into each other without leaving the domain, and for analytic functions this deformation preserves the integral value. Connectivity alone (one piece) is insufficient because it does not prevent paths from winding differently around holes; simple connectivity removes the holes that would allow such winding.
The geometric intuition is that simple connectivity means there is 'nothing inside' any closed curve — no singularity the curve can encircle. Cauchy's theorem formalizes this: ∮_C f(z) dz = 0 for every closed curve in a simply connected domain where f is analytic. The punctured plane and annulus show the consequences of failure: 1/z has a pole at the origin, and integrating around it gives 2πi rather than 0, because the hole allows the curve to detect the singularity.