--- title: "State and prove Cauchy integral theorem" subject: "Applied Mathematics" chapter: 1 key: "semester_4/applied_mathematics:cauchy-integral-theorem-prove" source: https://www.bctengineer.com/q/semester_4/applied_mathematics/cauchy-integral-theorem-prove site: BCT Engineering --- # State and prove Cauchy integral theorem Applied Mathematics · Chapter 1: Complex Analysis - Asked in: 2074 Bhadra, 5 marks; 2068 Magh, 5 marks; 2069 Poush (asked 3 times) - Video: [Cauchy Integral Theorem](https://www.youtube.com/watch?v=-d9-V7-2lPE) — IGNITED MINDS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/cauchy-integral-theorem-prove - Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1 ## Question State and prove Cauchy integral theorem. ## Short answer If f is analytic inside and on a simple closed contour C, then ∮ f(z)dz=0. Write f=u+iv, split the integral, apply Green's theorem, and use the Cauchy–Riemann equations u_x=v_y, u_y=−v_x to make both double integrals vanish. ## Answer **Statement (Cauchy's integral theorem).** If f(z) is analytic at all points inside and on a simple closed contour C, then ∮_C f(z) dz = 0. #### Proof (using Green's theorem) Assume, as in Cauchy's original proof, that f′(z) is continuous in the region R bounded by C. Let f(z) = u(x, y) + i v(x, y) and dz = dx + i dy. Then ```text ∮ f(z) dz = ∮ (u + iv)(dx + i dy) = ∮ (u dx − v dy) + i ∮ (v dx + u dy) ``` Green's theorem: ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy. Apply it to each integral. ```text ∮ (u dx − v dy) = ∬_R ( −∂v/∂x − ∂u/∂y ) dx dy ∮ (v dx + u dy) = ∬_R ( ∂u/∂x − ∂v/∂y ) dx dy ``` Since f(z) is analytic, u and v satisfy the Cauchy–Riemann equations in R: ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x. So both integrands vanish identically: −v_x − u_y = −v_x + v_x = 0 and u_x − v_y = 0. Hence ∮_C f(z) dz = 0 + i·0 = 0. ∎ **Remark.** Goursat later proved the theorem without assuming that f′ is continuous. The theorem also gives path independence: if f is analytic in a simply connected region, ∫ f(z) dz between two points does not depend on the path joining them. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Cauchy's integral theorem says that for f analytic inside and on a simple closed contour C: A. ∮_C f(z) dz = 2πi f(0) B. ∮_C f(z) dz = f(a) C. ∮_C f(z) dz = 0 D. ∮_C f(z) dz = 2πi Answer: C. With no singularities inside, the contour integral vanishes. (Key point: Statement) 2. Which theorem converts the contour integral into a double integral in the proof? A. Residue theorem B. Green's theorem C. Taylor's theorem D. Stokes' theorem in 3D Answer: B. Green's theorem turns ∮(P dx + Q dy) into ∬(Q_x − P_y) dx dy. (Key point: Tool used) 3. Why do the double integrals vanish? A. The area of R is zero B. The Cauchy–Riemann equations make each integrand zero C. u and v are constants D. C is a circle Answer: B. With u_x = v_y and u_y = −v_x, both integrands are 0. (Key point: Why zero) 4. What extra assumption does the Green's theorem proof make that Goursat later removed? A. f is real B. f′(z) is continuous in R C. f is entire D. C is a circle Answer: B. Goursat proved the theorem without assuming continuity of f′. 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