👁 Image

Cauchy Integral Formula

👁 CauchysIntegralFormula

Cauchy's integral formula states that

👁 f(z_0)=1/(2pii)∮_gamma(f(z)dz)/(z-z_0),

(1)

where the integral is a contour integral along the contour 👁 gamma
enclosing the point 👁 z_0
.

It can be derived by considering the contour integral

👁 ∮_gamma(f(z)dz)/(z-z_0),

(2)

defining a path 👁 gamma_r
as an infinitesimal counterclockwise circle around the point 👁 z_0
, and defining the path 👁 gamma_0
as an arbitrary loop with a cut line (on which the forward and reverse contributions cancel each other out) so as to go around 👁 z_0
. The total path is then

👁 gamma=gamma_0+gamma_r,

(3)

👁 ∮_gamma(f(z)dz)/(z-z_0)=∮_(gamma_0)(f(z)dz)/(z-z_0)+∮_(gamma_r)(f(z)dz)/(z-z_0).

(4)

From the Cauchy integral theorem, the contour integral along any path not enclosing a pole is 0. Therefore, the first term in the above equation is 0 since 👁 gamma_0
does not enclose the pole, and we are left with

👁 ∮_gamma(f(z)dz)/(z-z_0)=∮_(gamma_r)(f(z)dz)/(z-z_0).

(5)

Now, let 👁 z=z_0+re^(itheta)
, so 👁 dz=ire^(itheta)dtheta
. Then

👁 ∮_gamma(f(z)dz)/(z-z_0)	👁 =	👁 ∮_(gamma_r)(f(z_0+re^(itheta)))/(re^(itheta))ire^(itheta)dtheta	(6)
👁 Image	👁 =	👁 ∮_(gamma_r)f(z_0+re^(itheta))idtheta.	(7)

But we are free to allow the radius 👁 r
to shrink to 0, so

👁 ∮_gamma(f(z)dz)/(z-z_0)	👁 =	👁 lim_(r->0)∮_(gamma_r)f(z_0+re^(itheta))idtheta	(8)
👁 Image	👁 =	👁 ∮_(gamma_r)f(z_0)idtheta	(9)
👁 Image	👁 =	👁 if(z_0)∮_(gamma_r)dtheta	(10)
👁 Image	👁 =	👁 2piif(z_0),	(11)

giving (1).

If multiple loops are made around the point 👁 z_0
, then equation (11) becomes

👁 n(gamma,z_0)f(z_0)=1/(2pii)∮_gamma(f(z)dz)/(z-z_0),

(12)

where 👁 n(gamma,z_0)
is the contour winding number.

A similar formula holds for the derivatives of 👁 f(z)
,

👁 f^'(z_0)	👁 =	👁 lim_(h->0)(f(z_0+h)-f(z_0))/h	(13)
👁 Image	👁 =	👁 lim_(h->0)1/(2piih)[∮_gamma(f(z)dz)/(z-z_0-h)-∮_gamma(f(z)dz)/(z-z_0)]	(14)
👁 Image	👁 =	👁 lim_(h->0)1/(2piih)∮_gamma(f(z)[(z-z_0)-(z-z_0-h)]dz)/((z-z_0-h)(z-z_0))	(15)
👁 Image	👁 =	👁 lim_(h->0)1/(2piih)∮_gamma(hf(z)dz)/((z-z_0-h)(z-z_0))	(16)
👁 Image	👁 =	👁 1/(2pii)∮_gamma(f(z)dz)/((z-z_0)^2).	(17)

Iterating again,

👁 f^('')(z_0)=2/(2pii)∮_gamma(f(z)dz)/((z-z_0)^3).

(18)

Continuing the process and adding the contour winding number 👁 n
,

👁 n(gamma,z_0)f^((r))(z_0)=(r!)/(2pii)∮_gamma(f(z)dz)/((z-z_0)^(r+1)).

(19)

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References

Arfken, G. "Cauchy's Integral Formula." §6.4 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 371-376, 1985.Kaplan, W. "Cauchy's Integral Formula." §9.9 in Advanced Calculus, 4th ed. Reading, MA: Addison-Wesley, pp. 598-599, 1991.Knopp, K. "Cauchy's Integral Formulas." Ch. 5 in Theory of Functions Parts I and II, Two Volumes Bound as One, Part I. New York: Dover, pp. 61-66, 1996.Krantz, S. G. "The Cauchy Integral Theorem and Formula." §2.3 in Handbook of Complex Variables. Boston, MA: Birkhäuser, pp. 26-29, 1999.Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 367-372, 1953.Woods, F. S. "Cauchy's Theorem." §146 in Advanced Calculus: A Course Arranged with Special Reference to the Needs of Students of Applied Mathematics. Boston, MA: Ginn, pp. 352-353, 1926.

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Cauchy Integral Formula

Cite this as:

Weisstein, Eric W. "Cauchy Integral Formula." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/CauchyIntegralFormula.html

URL: https://mathworld.wolfram.com/CauchyIntegralFormula.html