User:Frank 1729

=Proposed article on Cauchy's Integral Formula=

In mathematics, Cauchy's integral formula, named after Augustin Louis Cauchy, is a central statement in complex analysis. It expresses the remarkable fact that a holomorphic function defined on a connected, open region without holes is completely determined by its values on the boundary of the region. This formula derives its analytic significance from   the way it can be used to represent holomorphic functions using contour integrals, from which representation many important and far reaching properties of those functions can be derived, including integral formulas for their derivatives.

The theorem




Suppose U is an open, connected,  simply connected subset of the complex plane C, and f : U &rarr; C is a holomorphic function on U. Let C be a simple, closed, positively oriented contour  lying entirely inside U. Then for every point a in the interior of C we have


 * $$f(a) = {1 \over 2\pi i} \oint_C {f(z) \over z-a}\, dz. $$



Remarks on the conditions of the theorem
The domain U does not have to be simply connected as long as the contour C is null-homotopic in U, that is, continuously deformable, to a point in U, with the deformation never leaving U. Moreover, as for the Cauchy integral theorem, it is sufficient to require that f be holomorphic&mdash;that is, complex-differentiable&mdash;in the open region enclosed by the contour and continuous on its closure.

Cauchy's formula as an integral representation of holomorphic functions
The point a in Cauchy's integral formula may be any point on the open set U, so a, instead of a constant, can be thought of as a variable that assumes its values on  U. When we think of a in such terms, we may change the notation in the theorem and replace a with z and use w, for example, as the variable of integration. With that change of notation Cauchy's integral formula takes the form


 * $$f(z) = {1 \over 2\pi i} \oint_C {f(w) \over w-z}\, dw, $$

thus making it explicit that the formula provides a way of representing using a contour integral, holomorphic functions that satisfy the conditions of the theorem. It is in this form in which some authors state Cauchy's integral formula.

The role of differentiability in Cauchy's integral formula
To recover the values of some holomorphic function using the integral representation given by Cauchy's formula we need three things as input:


 * 1) A simple, closed contour C,
 * 2) the values of a function f that is continuous on C, and
 * 3) the knowledge that f is holomorphic inside C.

At first glance, in light of the integral representation form of Cauchy's formula, it might appear that we only need the first two items in the list above. Indeed, if f is an arbitrary continuous function on C, the function defined by
 * $$f(z) = {1 \over 2\pi i} \oint_C {f(w) \over w-z}\, dw $$

for all z in the interior of C is holomorphic. Haven't we now recovered f from its values on C? Has anything gone wrong?

To see what can go wrong with the approach above, consider the following example. Let C be the unit circle, traveled in the positive direction, and let f(z&#773;) = z&#773;, where z&#773; denotes the complex conjugate of z. Then, if z is a nonzero complex number inside the unit circle, we have



\begin{align} f(z)&=\frac{1}{2\pi i}\oint_C \frac{f(w)}{w-z}\, dw\\ &=\frac{1}{2\pi i}\oint_C \frac{\overline{w}}{w-z}\, dw\\ &=\frac{1}{2\pi i}\oint_C \frac{w\overline{w}}{w(w-z)}\, dw\\ &=\frac{1}{2\pi i}\oint_C \frac{\vert w\vert^2}{w(w-z)}\, dw\\ &=\frac{1}{2\pi i}\oint_C \frac{1}{w(w-z)}\, dw \quad \textrm{(because\ \textit{w}\ is\ on\ the\ unit\ circle).}\\ \end{align} $$ Since z is not equal to zero, the partial fraction decomposition of $$ \frac{1}{w(w-z)}$$ is given by
 * $$\frac{1}{w(w-z)}=\frac{1}{z}\left (\frac{1}{w-z}-\frac{1}{w}\right. )$$

By direct computation, or through the use of Cauchy's integral formula itself, we have that
 * $$\frac{1}{2\pi i}\oint_C\frac{1}{w-z}\, dw=\frac{1}{2\pi i}\oint\frac{1}{w}\, dw=1,

$$ which implies that
 * $$\frac{1}{2\pi i}\oint_C\frac{1}{z}\left(\frac{1}{w-z}-\frac{1}{w}\right)\,dw=0.$$

Thus, if z &ne; 0 then f(z) = 0. Using also a direct computation we can show that f (0) = 0. Therefore, we have recovered the function f(z) = 0 for all z in the interior of the unit circle. And the problem is now evident; we started with the continuous function f(z&#773;) = z&#773; and the unit circle as our inputs, and obtained and different function as our output (or more precisely, we extended a continuous function on the unit circle to discontinuous function on the unit disk, which is not at all what we wanted). This example clearly illustrates the need for differentiability inside the contour over which we are integrating in addition to continuity on the contour itself.

Proof
Let &epsilon; be any positive number. Since f is complex differentiable, it is continuous, so   there exists a positive number &delta; such that |f(z) - f(a)| < &epsilon; whenever |z - a| &le; &delta;. Let &Gamma; be the circle of radius &delta; centered at a. If necessary, make &delta; small enough so that &Gamma; lies entirely inside C. The deformation of contour theorem allows us to replace an integral over a simple, closed contour C in an open, though not necessarily simply connected, open set, with an integral over a closed, simple contour in the same domain, and lying inside C. Thus, this theorem implies that the integral of f(z)/(z-a) over &Gamma; has the same value as the integral of the same function over the contour C. Now, if &gamma; is a closed (not necessarily simple) contour, then the expression
 * $$ \dfrac{1}{2\pi i}\oint_{\gamma} \dfrac{1}{z-a}\,dz$$

is the winding number of &gamma; around a. If &gamma; is a circle with positive orientation and a is inside &gamma;, then the expression is equal to 1. Thus, we have



\begin{align} \frac{1}{2\pi i}\oint_{\Gamma}\frac{f(a)}{z-a} \,dz&= f(a)\cdot\frac{1}{2\pi i}\oint_{\Gamma}\frac{1}{z-a} \,dz\\ &=f(a)\cdot 1\\ &=f(a). \end{align} $$ Therefore,



\begin{align} \left\vert \dfrac{1}{2\pi i}\int_{C}\dfrac{f\left( z\right) }{z-a} \,dz-f\left( a\right) \right\vert &=\left\vert \dfrac{1}{2\pi i} \int_{\Gamma }\dfrac{f\left( z\right) }{z-a} \,dz-\dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{ f\left( a\right) }{z-a}\,dz\right\vert \\ &=\left\vert \dfrac{1}{2\pi i}\int_{\Gamma } \dfrac{f\left( z\right) -f\left( a\right) }{z-a}\,dz\right\vert \\ &=\left\vert\dfrac{1}{2\pi i}\right\vert \left\vert\int_{\Gamma } \dfrac{f\left( z\right) -f\left( a\right) }{z-a}\,dz\right \vert \\ &= \dfrac{1}{2\pi }\left\vert \int_{\Gamma } \dfrac{f\left( z\right) -f\left( a\right) }{z-a}\,dz \right\vert \end{align} $$

Using standard inequalities for contour integrals yields

\begin{align} \dfrac{1}{2\pi }\left\vert \int_{\Gamma } \dfrac{f\left( z\right) -f\left( a\right) }{z-a}\,dz \right\vert &\leq \dfrac{1}{2\pi } \int_{\Gamma } \left\vert\dfrac{f\left( z\right) -f\left( a\right) }{z-a}\right\vert\,dz \\ &= \dfrac{1}{2\pi }\int_{\Gamma }\dfrac{\left\vert f\left( z\right) -f\left( a\right) \right\vert }{\delta }\,dz \\ &\leq \dfrac{1}{2\pi }\dfrac{\epsilon }{\delta }\left( \text{circumference of } \Gamma \right) \\ &=\dfrac{1}{2\pi }\dfrac{\epsilon }{\delta }\cdot 2\pi \delta \\ &=\epsilon. \end{align} $$

We now have shown that

\left\vert \dfrac{1}{2\pi i}\int_{C}\dfrac{f\left( z\right) }{z-a} \,dz-f\left( a\right) \right\vert < \epsilon $$ for every positive number &epsilon;. This implies that

\left\vert \dfrac{1}{2\pi i}\int_{C}\dfrac{f\left( z\right) }{z-a} \,dz-f\left( a\right) \right\vert =0, $$ which in turn implies that

\dfrac{1}{2\pi i}\int_{C}\dfrac{f\left( z\right) }{z-a} \,dz-f\left( a\right) =0. $$ The theorem is now proved.

Examples
For a simple example of the application of Cauchy's integral formula, consider the contour integral


 * $$\oint_C \frac{e^z}{z-i\pi}\,dz,$$

where C is the positively oriented circle of radius 1 centered at $$i\pi$$. Since the function f(z) = ez is entire, Cauchy's integral formula tells us that


 * $$\frac{1}{2\pi i}\oint_C \frac{e^z}{z-i\pi}\,dz=f(i\pi)=e^{i\pi}.$$

Euler's formula then gives us



e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1. $$ Consequently,


 * $$\oint_C \frac{e^z}{z-i\pi}\,dz=-2\pi i.$$

There is nothing special about C being a circle. Indeed, the deformation of contour theorem tells that the value of
 * $$\frac{1}{2\pi i}\oint_C \frac{e^z}{z-i\pi}\,dz$$

is also $$-2\pi i$$ if we replace C with a any simple, positively oriented, closed contour that contains $$i\pi$$ in its interior.



For a more complicated example, consider the function


 * $$g(z)={z^2 \over z^2+2z+2}$$

and the contour described by |z| = 2, call it C.

To find out the integral of g(z) around the contour, we need to know the singularities of g(z). Observe that we can rewrite g as follows:
 * $$g(z)={z^2 \over (z-z_1)(z-z_2)}$$

where $$z_1=-1+i,$$ $$z_2=-1-i.$$

Clearly the poles become evident, their moduli are less than 2 and thus lie inside the contour and are subject to consideration by the formula. Let C1 be a small, positively oriented circle around z1 and let C2 be a small, positively oriented circle around z2. Let L1 be an arc from C1 to C2, and let L2 = -L1; that is, L2 is the contour from C2 to C1, traveled along L1. Define the contour &gamma; by


 * $$\gamma=C_1+L_1+C_2+L_2.$$

Then

\begin{align} \oint_{\gamma}g(z)\,dz&=\oint_{C_1+L_1+C_2+L_2}g(z)\,dz\\ &=\oint_{C_1}g(z)\,dz+\oint_{L_1}g(z)\,dz+\oint_{C_2}g(z)\,dz+\oint_{L_2}g(z)\,dz\\ &=\oint_{C_1}g(z)\,dz+\oint_{L_1}g(z)\,dz+\oint_{C_2}g(z)\,dz+\oint_{-L_1}g(z)\,dz\\ &=\oint_{C_1}g(z)\,dz+\oint_{L_1}g(z)\,dz+\oint_{C_2}g(z)\,dz-\oint_{L_1}g(z)\,dz\\ &=\oint_{C_1}g(z)\,dz+\oint_{C_2}g(z)\,dz. \end{align} $$ Thus, the integral of g(z) over &gamma; is the sum of the integrals of g(z) over the circles C1 and C2. The circle C can be continuously deformed into the contour &gamma; within the open set $$\mathbb C-\{z_1,z_2\}$$, so by the homotopy form of Cauchy's integral theorem, the integral of g(z) over C is the same as the integral of g(z) over &gamma;. Therefore, the integral of g(z) along C is also sum of the integrals of g(z) over the circles C1 and C2.

The function


 * $$f_1(z)=\frac{z^2}{z-z_2}$$

is holomorphic inside C1 since this contour does not contain z2. This allows us to write g in the form


 * $$g(z)={f_1(z) \over z-z_1}.$$

Now we have



\begin{align} \oint_{C_1} {g(z)}\,dz &= \oint_{C_1} {f_1(z) \over z-z_1}\, dz =2\pi i f(z_1)=(2\pi i)\frac{z_1^2}{z_1-z_2}\\ &=(2\pi i) \frac{(-1+i)^2}{-1+i-(-1-i)}=(2\pi i )\frac{-2i}{2i}=-2\pi i. \end{align} $$

To compute the integral of g over the contour C2 we proceed in the same vein and let


 * $$f_2(z)={z^2 \over z-z_1}.$$

Computations similar to those above yield



\oint_{C_2} {g(z)}\,dz = \oint_{C_2} {f_2(z) \over z-z_2}\, dz =2\pi i\cdot f(z_2)=-2\pi i. $$

The integral around the original contour C is the sum of these two integrals, and thus


 * $$\begin{align}\oint_C {z^2 \over z^2+2z+2}\,dz &= \oint_{C_1} \frac{f_1(z)}{z-z_1}\,dz + \oint_{C_2} \dfrac{f_2(z)}{z-z_2}\,dz \\

&=-2\pi i-2\pi i\\ &=-4\pi i.\end{align}$$

Consequences
One of the most important consequences of Cauchy's integral formula is   the proof that holomorphic functions are analytic. This proof shows that a holomorphic function can be expanded into a power series


 * $$\sum_{n=0}^{\infty} c_n(z-a)^n$$

at every point a in its domain. This series has a positive radius of convergence, and each coefficient cn is given by


 * $$ c_n=\oint_C \dfrac{f(z)}{(z-a)^{n+1}}\,dz,$$

and where C is a positively oriented circle with center a, and so that f is holomorphic inside C. On the other hand, the coefficients of the power series about the point a of an analytic function are given by the formula



c_n = \frac {f^{\left( n \right)}\left( a \right)} {n!} $$

These two formulas for cn show that


 * $$f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz.$$

This formula is sometimes known as Cauchy's differentiation formula. The deformation of contour theorem allows us to replace the circle C with a simple, closed contour that is inside the domain of f and on which f is holomorphic. Thus, the conditions for which Cauchy's differentiation formula holds are exactly the same as those of Cauchy's integral formula itself.

Because one may deduce from Cauchy's differentiation formula that f must be infinitely often continuously differentiable, the Cauchy integral theorem has broad implications. The fact that holomorphic functions are infinitely differentiable is used to prove Liouville's theorem, which states that every bounded entire function is constant; it is also is used to prove the residue theorem, which is a far-reaching generalization that removes the requirement that the function be analytic in the enclosed region.

The Cauchy integral theorem has no counterpart in real analysis because for a real valued function possession of a first derivative by a function will not guarantee the existence of higher order derivatives. In contrast to this, the proof of the Cauchy integral formula for $$n^{th}$$ derivatives shows that analytic functions posses derivatives of all orders.

It is known from Morera's theorem that the uniform limit of holomorphic functions is holomorphic. This can also be deduced from Cauchy's integral formula: indeed the formula also holds in the limit and the integrand, and hence the integral, can be expanded as a power series. In addition the Cauchy formulas for the higher order derivatives show that all these derivatives also converge uniformly.

Example of Cauchy's differentiation formula
Consider the contour integral
 * $$\oint_C\frac{\log(z)}{(z-2\pi i)^2}\,dz,$$

where C is any simple, closed, positively oriented contour lying in the upper complex plane, and having the point $$2\pi i$$ in its interior. The function f(z) = log(z) is the principal branch of the complex logarithm function, which is discontinuous, and thus, not differentiable, on the non positive real axis. But C avoids that axis and moves only where f is holomorphic. Hence, we can use Cauchy's differentiation formula to evaluate the integral. Recalling that $$ f'(z)=1/z$$ we get

\oint_C\frac{\log(z)}{(z-2\pi i)^2}\,dz=2\pi i f'(2\pi i)=2\pi i \cdot\frac{1}{2\pi i}=1.$$