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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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On a positively oriented circle about a, the integral of (z-a)^m is zero for every integer m except -1, and is 2 pi i for m=-1

Statement

Let aC, r>0, and γ(t)=a+rexp(it) for 0t2π. For every integer m, γ(za)mdz={2πi,m=1,0,m1.

Facts & Assumptions

Given: The positively oriented circle γ and an integer m.

[L1]

On a piecewise-C1 contour, the Riemann–Stieltjes integral agrees with f(γ(t))γ(t)dt (For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals).

[L2]

Negative integer powers are defined exactly for nonzero complex bases (Integer powers in the complex field).

[L3]

The complex exponential is entire with derivative itself and satisfies exp(z+w)=expzexpw (The complex exponential is entire and its complex derivative is itself, exp(z+w)=expzexpw, and the complex exponential extends the real exponential).

[L4]

For real x,y, exp(x+iy)=ex(cosy+isiny) and exp(x+iy)=ex; in particular eiπ+1=0 (exp(x+iy)=ex(cosy+isiny), exp(x+iy)=ex, and eiπ+1=0).

[L5]

If a real function G is differentiable on [a,b] and G is integrable, then abG=G(b)G(a) (The second fundamental theorem: if G is differentiable on [a,b] with G=f and f is integrable, then abf=G(b)G(a)).

[L6]

Proof

technique · cases
1.1

Since r>0, γ(t)a0, so all integer powers in [L2] are defined. By [L1] and [L3], the integrand becomes irm+1exp(i(m+1)t).

L1L2L3algebra
2.1

If m=1, the expression in step 1.1 is the constant i, whose integral from 0 to 2π is 2πi.

assume-case exceptionalstep 1.1algebra
2.2

If m1, an antiderivative is rm+1exp(i(m+1)t)/(m+1) by [L3]. Apply the real theorem [L5] to its two components using [L6]; the complex integral is the endpoint difference rm+1(exp(2πi(m+1))1)/(m+1). Write k:=m+1, a nonzero integer. For k>0 the addition law in [L3] gives exp(2πik)=exp(iπ)2k, and exp(iπ)=1 by [L4], so exp(2πik)=(1)2k=1; for k<0 the addition law gives exp(2πik)exp(2πik)=exp(0)=1 with exp(2πik)=1 by the previous case, so again exp(2πik)=1. The endpoint difference is therefore 0.

assume-case regularstep 1.1L3L4L5L6algebra
3.1

The integer cases m=1 and m1 are exhaustive, proving the formula.

step 2.1step 2.2cases-exhaustive

Depends on

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