ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The circle integral of $e^{z} / (z - 1)$ over $∣ z ∣ = 2$ is $2 π i e$

Example

If $γ (t) = 2 \exp (i t)$ for $0 \leq t \leq 2 π$ , then

$\int_{γ} \frac{e^{z}}{z - 1} d z = 2 π i e .$

Facts & Assumptions

Given: The positively oriented radius- $2$ circle $γ$ and the integrand $e^{z} / (z - 1)$ .

[L1]

The complex exponential is entire (The complex exponential is entire and its complex derivative is itself).

[L2]

If $f$ is holomorphic on $D (a, R)$ , $0 < r < R$ , $∣ z - a ∣ < r$ , and $γ (t) = a + r \exp (i t)$ for $0 \leq t \leq 2 π$ , then $\int_{γ} f (ζ) / (ζ - z) d ζ = 2 π i f (z)$ (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

Verification

technique · direct

1.1givenL1algebra

The point $1$ lies strictly inside $∣ z ∣ = 2$ , while $∣ z - 1 ∣ \geq 1$ on the circle, so the denominator has no zero on the contour; by [L1], the numerator is holomorphic on every disc.

2.1step 1.1L2∎

Apply [L2] with centre $0$ , radius $2$ , interior point $1$ , and $f (z) = e^{z}$ to obtain the displayed value $2 π i e^{1} = 2 π i e$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

Richard Howell and John Mathews, Complex Analysis, Example 6.5.3 (standard reference, not scraped)

The circle integral of ez/(z−1)ez/(z−1) over ∣z∣=2∣z∣=2 is 2πie2πie