You integrate f(z) = 1/z along a circular contour centered at the origin. Cauchy's theorem says the integral must be zero — true or false?
ATrue — 1/z is a complex function, and Cauchy's theorem applies to all complex functions on closed contours
BFalse — the theorem requires holomorphicity, and 1/z has a singularity at z = 0 inside the contour
CTrue — the contour is closed, which is the only condition the theorem requires
DFalse — but only because the contour must be a straight line, not a circle
Cauchy's theorem requires f to be holomorphic on a simply connected domain containing the contour. Since 1/z has a singularity at z = 0, it is not holomorphic on any domain that includes the origin. Integrating 1/z around a circle enclosing the origin gives 2πi ≠ 0 — a foundational counterexample. Options A and C represent the common misconception that the theorem applies to all complex functions.
Question 2 Multiple Choice
A function f is holomorphic everywhere except at two points inside a domain. You integrate f around a closed contour that encircles both singularities. Why does Cauchy's theorem fail?
AThe contour must be a circle for the theorem to apply
BThe domain is not simply connected — the singularities act as holes through which the contour cannot be contracted to a point
CCauchy's theorem only applies when the contour encloses no singularities and the function is holomorphic outside
DThe theorem still applies; the integral will be zero because f is holomorphic on the contour itself
Simple connectivity is the key geometric requirement: any loop in the domain must be contractible to a point without leaving the domain. Singularities inside the domain create 'holes' that prevent this contraction. The contour shape (circle, square, etc.) is irrelevant — only topology matters. Option D is wrong because holomorphicity *on* the contour is insufficient; f must be holomorphic *throughout* the interior.
Question 3 True / False
The function f(z) = z² is holomorphic on the entire complex plane (it is entire). Therefore ∮_γ z² dz = 0 for any closed contour γ in ℂ.
TTrue
FFalse
Answer: True
Because z² is entire (holomorphic everywhere with no singularities), the complex plane is a simply connected domain and Cauchy's theorem applies. The integral around any closed contour is exactly zero. This is in sharp contrast to 1/z, which has a singularity at the origin and gives a nonzero integral around contours enclosing that singularity.
Question 4 True / False
Cauchy's theorem states that the integral of any complex function around any closed contour equals zero.
TTrue
FFalse
Answer: False
This overstates the theorem. Two conditions are required: the function must be *holomorphic* (complex-differentiable throughout the relevant region) and the domain must be *simply connected* (no holes). The classic counterexample is ∮ (1/z) dz around a circle enclosing the origin, which equals 2πi, not zero — because 1/z is not holomorphic at z = 0.
Question 5 Short Answer
Explain why the simply connected condition in Cauchy's theorem is not merely a technical formality. Use the example of f(z) = 1/z to illustrate what goes wrong when it fails.
Think about your answer, then reveal below.
Model answer: Simple connectivity guarantees that any closed loop can be continuously shrunk to a point without leaving the domain — meaning there are no 'holes.' For 1/z on the punctured plane ℂ \ {0}, the origin is a singularity that creates a hole: a loop enclosing the origin cannot be contracted past it. The proof of Cauchy's theorem uses Green's theorem and the Cauchy-Riemann equations, which require the integrand to be well-defined and holomorphic everywhere inside the contour. When a singularity is enclosed, those conditions fail at the singularity, and the integral picks up a nonzero contribution — exactly 2πi per enclosed pole of 1/z. This failure is precisely what makes residue theory possible: nonzero contour integrals encode information about the singularities enclosed.
Simple connectivity is the topological condition that makes the proof go through. Without it, the homotopy argument breaks down: you cannot deform the contour to a point without crossing the singularity. The fact that ∮ (1/z) dz = 2πi is not an exception to be dismissed — it is the gateway to the entire residue theorem and complex integration methods used across physics and engineering.