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Cauchy's Theorem and Deforming a Contour

Prove that an analytic function integrates to zero around a closed loop it is analytic inside, and use that to slide contours freely.

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What a learner can do afterwards

  • Derive Cauchy's theorem from Green's theorem plus the Cauchy-Riemann equations
  • Deform a contour past a region of analyticity without changing the integral
  • Explain what fails when a singularity lies inside the loop

1 · Read

Cauchy's theorem says an analytic function integrates to zero around a closed loop it is analytic inside. Split f into u plus iv and the loop integral becomes two real circulations. Green plus Cauchy-Riemann zeroes both curls, and multiplying out shows the real part is u dx minus v dy.

Try it together

z squared is analytic everywhere, so its integral around any closed loop is 0. But 1 divided by z blows up at the origin, so its unit circle integral is 2 pi i, not zero. The missing analyticity at one interior point is the whole story.

You may slide a contour across any region where f stays analytic without changing the integral. The old and new paths differ by loops with zero integral. This freedom replaces awkward contours with friendly circles: a preserved value like 3 plus 4 stays 7 on the deformed loop.

Good to know

A single interior singularity breaks the spell. Never apply the theorem after spotting a blowup inside, and never deform across one. The unit circle with 1 divided by z is the classic nonzero counterexample to keep in mind.

Analytic inside means zero on the loop and free sliding nearby, while one singularity inside changes everything.

2 · Watch

Take it off screen

Print a worksheetA4 with an answer key page for grown-ups. No screen, no internet.

Where it sits

Then practise

8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.

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Cauchy's Theorem and Deforming a Contour · Mathematics, ages 20 to 21 · LightMySky