# Cauchy's theorem

Here we follow standard texts, such as Spiegel (1964) or Levinson and Redheffer (1970). 

If $C$ is a closed contour (Figure 1.), and the complex valued function $f(z)$ is an analytic function of the complex variable $z$ inside the region bounded by, and on $C$ then

$\int _{C}f(z)\;dz=0.$ ## Proof

If we substitute for $f(z)=u(x,y)+iv(x,y)$ and $dz=dx+idy$ $\int _{C}f(z)\;dz=\int _{C}(u(x,y)+iv(x,y))\;(dx+idy)=\int _{C}u(x,y)\;dx-\int _{C}v(x,y)\;dy+i\left(\int _{C}u(x,y)\;dy+\int _{C}v(x,y)\;dx\right).$ By Green's theorem, for any two functions $P(x,y)$ and $Q(x,y)$ such that ${\frac {\partial P}{\partial y}}$ and ${\frac {\partial Q}{\partial x}}$ exist in a two dimensional region ${\mathcal {R}}$ bounded by a curve $C$ $\int _{\mathcal {R}}\int \left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\;dx\;dy=\int _{C}P\;dx+\int _{C}Q\;dy.$ If we apply Green's theorem to the real and complex terms of the integral above, we have (identifying the real and imaginary parts of $f(x)$ with $P$ and $Q$ where appropriate, we have

$\int _{C}f(z)\;dz=-\left(\int _{\mathcal {R}}\int \left({\frac {\partial v(x,y)}{\partial x}}+{\frac {\partial u(x,y)}{\partial y}}\right)\;dx\;dy\right)+i\left(\int _{\mathcal {R}}\int \left({\frac {\partial u(x,y)}{\partial x}}-{\frac {\partial v(x,y)}{\partial y}}\right)\;dx\;dy\right).$ Because $f(z)$ is analytic inside ${\mathcal {R}}$ its real and imaginary parts must satisfy the Cauchy-Riemann equations

${\frac {\partial u(x,y)}{\partial x}}={\frac {\partial v(x,y)}{\partial y}}\quad \quad {\mbox{and}}\quad \quad {\frac {\partial v(x,y)}{\partial x}}=-{\frac {\partial u(x,y)}{\partial y}}.$ Thus the real and imaginary parts vanish independently showing that

$\int _{C}f(z)\;dz=0.$ We note that the shape of $C$ is quite general. It may have any shape, as long as it does not cross itself, and may have any finite number of corners, where the function describing the curve is continuous, but not differentiable.

The extension of Cauchy's theorem to a region with any finite number of holes is called the Cauchy-Goursat theorem.

# Cauchy Goursat theorem

If a complex valued function $f(z)$ is analytic in a region of the complex plane bounded by a simple closed curve $C_{1}$ , except possibly on any number of finite subdomains (holes) bounded by simple closed curves $C_{k}$ for $k>1$ then Cauchy's theorem holds in that region bounded by $C_{1}$ and all of the curves $C_{k}$ .

## Proof:

Consider a region bounded by a simple closed curve $C_{1}$ with a hole bounded by $C_{2}.$ (See Figure 2.) We may connect the two regions with a cut long the curve $[a,b].$ The integral over the full boundary of the shaded region, where $f(z)$ is analytic is given by

$\int _{C_{1}+[a,b]+C_{2}+[b,a]}f(z)\;dz=\left\{\oint _{C_{1}}+\int _{a}^{b}+\oint _{C_{2}^{-}}+\int _{b}^{a}\right\}f(z)\;dz=0$ where the notation $C_{2}^{-}$ indicates that the integration path is in the clockwise (negative) direction in the complex plane.

Because

$\int _{a}^{b}f(z)\;dz=-\int _{b}^{a}f(z)\;dz$ we have

$\int _{C_{1}}f(z)\;dz=\int _{C_{2}^{-}}f(z)\;dz$ Reversing the direction of integration on the integral on the right hand side yields

$\int _{C_{1}}f(z)\;dz=\int _{C_{2}}f(z)\;dz.$ Thus the integrals over the integration contours $C_{1}$ and $C_{2}$ are equivalent. Because $f(z)$ need not be analytic in the interior of $C_{2}$ these integrals are not necessarily zero.

We may have any finite number of holes in our domain, and the sum of the integrals over the curves $C_{k}$ bounding these holes is equivalent to the integral over the bounding contour $C_{1}.$ (See Figure 3.)

Another way of interpreting this result is that we may continuously deform the countour $C_{1}$ to any other closed simple curve $C_{2}$ enclosing the same region. Again, there is no restriction on the shape of the contours, only that they are connected, and that they have at most a finite number of corners.