corollary of Cauchy integral theorem∗
pahio†
2013-03-22 2:48:06
Theorem. Let γ be a closed contour of C not intersecting itself and
γ1 , . . . , γn likewise. Let all the contours γ1 , . . . , γn be outside each other but
insinde γ. If the closed, “holey” domain between γ and the γj s is contained in
a domain U where f is holomorphic, then
I
I
I
f (z) dz =
f (z) dz + . . . +
f (z) dz
(1)
γ
γ1
γn
where all integrals are taken with the same direction of circulation. Especially,
in the case n = 1 one has
I
I
f (z) dz =
f (z) dz.
(2)
γ
γ1
Note 1. The integrals in (1) and (2) need not necessarily vanish, since
inside a γj there may be points not belonging to U .
Note 2. When n = 0, i.e. the sum on the right hand side of (1) is empty,
it has the value 0; thus also the Cauchy integral theorem is a special case of (1).
Note 3. The theorem implies easily the residue theorem.
Proof. We prove the theorem only in the case n = 1. Other cases may be
handled analogously.
Draw two auxiliary ways connecting γ and γ1 . The integral of f taken anticlockwise around the route consisting of the upper parts of the curves and the
auxiliary ways is, by the fundamental theorem of complex analysis, equal zero.
Similarly the integral of f taken anticlockwise around the route consisting of
the lower parts of the curves and the auxiliary ways is equal zero. Thus also the
sum of both equals zero. But in the sum, the portions taken along the auxiliary
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ways are run in opposite directions and so they cancel each other. Therefore in
the sum only the portions, which are run along γ and γ1 , remain; then
I
I
f (z) dz +
f (z) dz = 0,
γ
i.e.
γ1
I
I
f (z) dz = −
f (z) dz.
γ
γ1
However, here γ is run anticlockwise and γ1 clockwise. Reversing the direction
in the right hand side of this last equation, one obtains (2). Q.E.D.
γ
γ1
Example. Calculate
I
dz
z−z
0
C
where the circle C of complex plane with centre z0 and radius R is run once
anticlockwise.
Since |z−z0 | = R we can take the parametric presentation
z−z0 = Reiϕ
Then dz = iReiϕ dϕ and
I
C
dz
=
z−z0
with
Z
0 5 ϕ < 2π.
2π
i dϕ = 2iπ.
(3)
0
By the equation (2) of the theorem, one can infer that the result (3) is true for
any continuous contour going once around the point z0 anticlockwise (cf. the
lemma of this entry).
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