Cauchy's integral formula states that
 |
(1)
|
where the integral is a contour integral along the contour 
enclosing the point 
.
It can be derived by considering the contour integral
 |
(2)
|
defining a path 
as an infinitesimal counterclockwise circle around the
point 
,
and defining the path 
as an arbitrary loop with a cut line (on which the forward
and reverse contributions cancel each other out) so as to go around 
. The total path is then
 |
(3)
|
so
 |
(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 
does not enclose the pole, and we are left with
 |
(5)
|
Now, let 
,
so 
.
Then
 |  |  |
(6)
|
 |  |  |
(7)
|
But we are free to allow the radius 
to shrink to 0, so
 |  |  |
(8)
|
 |  |  |
(9)
|
 |  |  |
(10)
|
 |  |  |
(11)
|
giving (1).
If multiple loops are made around the point 
, then equation (11) becomes
 |
(12)
|
where 
is the contour winding number.
A similar formula holds for the derivatives of 
,
 |  |  |
(13)
|
 |  | ![lim_(h->0)1/(2piih)[∮_gamma(f(z)dz)/(z-z_0-h)-∮_gamma(f(z)dz)/(z-z_0)]](/images/equations/CauchyIntegralFormula/Inline37.svg) |
(14)
|
 |  | ![lim_(h->0)1/(2piih)∮_gamma(f(z)[(z-z_0)-(z-z_0-h)]dz)/((z-z_0-h)(z-z_0))](/images/equations/CauchyIntegralFormula/Inline40.svg) |
(15)
|
 |  |  |
(16)
|
 |  |  |
(17)
|
Iterating again,
 |
(18)
|
Continuing the process and adding the contour winding number 
,
 |
(19)
|
