ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The circle integral of $\cos z / (z - 1)^{3}$ over $∣ z ∣ = 2$ is $- π i \cos 1$

Example

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

$\int_{γ} \frac{\cos z}{(z - 1)^{3}} d z = - π i \cos 1.$

Facts & Assumptions

Given: The positively oriented radius- $2$ circle and the displayed integrand.

[L1]

If $f$ is holomorphic on $D (a, R)$ , $0 < r < R$ , $z \in D (a, r)$ , $n \in N$ , and $γ (t) = a + r \exp (i t)$ for $0 \leq t \leq 2 π$ , then $\int_{γ} f (ζ) / (ζ - z)^{n + 1} d ζ = (2 π i / n!) f^{(n)} (z)$ (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L2]

The complex cosine is entire, with $(\cos z)^{'} = - \sin z$ and $(\sin z)^{'} = \cos z$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives).

Verification

technique · direct

1.1givenL2algebra

The point $1$ lies strictly inside the radius- $2$ circle and the denominator is nonzero on it; by [L2], $f (z) = \cos z$ is entire and $f^{''} (1) = - \cos 1$ .

2.1step 1.1L1algebra∎

Apply [L1] with $n = 2$ and $z = 1$ : the integral is $(2 π i / 2!) f^{''} (1) = - π i \cos 1$ .

Depends on

Used by

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The circle integral of cos⁡z/(z−1)3cosz/(z−1)3 over ∣z∣=2∣z∣=2 is −πicos⁡1−πicos1