ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

1.1

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.

givenL1algebra

2.1

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

step 1.1L2∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 83 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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