ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The iterated Cauchy formula computed for $z_{0} z_{1}$ on a bidisc

Example

Let $f (z) = z_{0} z_{1}$ on the unit bidisc, let $C (t) = \frac{1}{2} \exp (i t)$ on $[0, 2 π]$ , and let $z = (z_{0}, z_{1})$ satisfy $∣ z_{0} ∣ < 1 / 2$ and $∣ z_{1} ∣ < 1 / 2$ . Then

$\frac{1}{(2 π i)^{2}} \int_{C} ⁣ \int_{C} \frac{ζ_{0} ζ_{1}}{(ζ_{0} - z_{0}) (ζ_{1} - z_{1})} d ζ_{1} d ζ_{0} = z_{0} z_{1} .$

The computation is done in the stated order: the inner integral is taken in $ζ_{1}$ first, with $ζ_{0}$ held fixed, and only then is the outer integral taken in $ζ_{0}$ .

Facts & Assumptions

Given: The function $f (z) = z_{0} z_{1}$ , the circle $C (t) = \frac{1}{2} \exp (i t)$ , and a point $z$ with $∣ z_{0} ∣ < 1 / 2$ and $∣ z_{1} ∣ < 1 / 2$ .

[L1]

The iterated Cauchy formula on a polydisc represents a continuous separately holomorphic function by successive one-variable contour integrals (The iterated Cauchy integral formula on a polydisc).

[L2]

The one-variable Cauchy integral formula on a circle gives $h (w) = (2 π i)^{- 1} \int_{γ} h (ζ) (ζ - w)^{- 1} d ζ$ for $∣ w ∣$ strictly inside the circle (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

Verification

technique · direct

1.1

Fix $ζ_{0}$ on the circle $C$ . As a function of $ζ_{1}$ , the integrand is $ζ_{0} ζ_{1} / ((ζ_{0} - z_{0}) (ζ_{1} - z_{1})) = ζ_{0} (ζ_{0} - z_{0})^{- 1} \cdot ζ_{1} (ζ_{1} - z_{1})^{- 1}$ , where the scalar factor in front is constant in $ζ_{1}$ .

givenalgebra

2.1

Apply [L2] to the holomorphic function $ζ_{1} \mapsto ζ_{1}$ on the disc $∣ ζ_{1} ∣ < 1$ : then $(2 π i)^{- 1} \int_{C} ζ_{1} (ζ_{1} - z_{1})^{- 1} d ζ_{1} = z_{1}$ , so the inner integral equals $2 π i \cdot ζ_{0} z_{1} / (ζ_{0} - z_{0})$ .

step 1.1L2

3.1

Substituting step 2.1 into the outer integral gives $\frac{1}{2 π i} \int_{C} ζ_{0} z_{1} (ζ_{0} - z_{0})^{- 1} d ζ_{0} = z_{1} \cdot \frac{1}{2 π i} \int_{C} ζ_{0} (ζ_{0} - z_{0})^{- 1} d ζ_{0}$ , and [L2] applied again yields the value $z_{1} z_{0}$ .

step 2.1L2

4.1

This matches the value $f (z) = z_{0} z_{1}$ , exactly as [L1] predicts.

step 3.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Tasty Bits of Several Complex Variables, v4.4, §1.2 (standard reference, not scraped)

The iterated Cauchy formula computed for z0z1z0​z1​ on a bidisc