Cauchy's theorem

"Figure 1: a simple closed contour C in the complex z plane"

Here we follow standard texts, such as Spiegel (1964)[1] or Levinson and Redheffer (1970). [2]

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

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

Proof

If we substitute for ${\displaystyle f(z)=u(x,y)+iv(x,y)}$ and ${\displaystyle dz=dx+idy}$

${\displaystyle \int _{C}f(z)\;dz=\int _{C}(u(x,y)+iv(x,y))\;(dx+idy)=\int _{C}u(x,y)\;dx-\int _{C}v(x,y)\;dy+i\left(\int _{C}u(x,y)\;dy+\int _{C}v(x,y)\;dx\right).}$

By Green's theorem, for any two functions ${\displaystyle P(x,y)}$ and ${\displaystyle Q(x,y)}$ such that ${\displaystyle {\frac {\partial P}{\partial y}}}$ and ${\displaystyle {\frac {\partial Q}{\partial x}}}$ exist in a two dimensional region ${\displaystyle {\mathcal {R}}}$ bounded by a curve ${\displaystyle C}$

${\displaystyle \int _{\mathcal {R}}\int \left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\;dx\;dy=\int _{C}P\;dx+\int _{C}Q\;dy.}$

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 ${\displaystyle f(x)}$ with ${\displaystyle P}$ and ${\displaystyle Q}$ where appropriate, we have

${\displaystyle \int _{C}f(z)\;dz=-\left(\int _{\mathcal {R}}\int \left({\frac {\partial v(x,y)}{\partial x}}+{\frac {\partial u(x,y)}{\partial y}}\right)\;dx\;dy\right)+i\left(\int _{\mathcal {R}}\int \left({\frac {\partial u(x,y)}{\partial x}}-{\frac {\partial v(x,y)}{\partial y}}\right)\;dx\;dy\right).}$

Because ${\displaystyle f(z)}$ is analytic inside ${\displaystyle {\mathcal {R}}}$ its real and imaginary parts must satisfy the Cauchy-Riemann equations

${\displaystyle {\frac {\partial u(x,y)}{\partial x}}={\frac {\partial v(x,y)}{\partial y}}\quad \quad {\mbox{and}}\quad \quad {\frac {\partial v(x,y)}{\partial x}}=-{\frac {\partial u(x,y)}{\partial y}}.}$

Thus the real and imaginary parts vanish independently showing that

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

We note that the shape of ${\displaystyle C}$ 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

"Figure 2: f(z) is analytic in the shaded region"

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

Proof:

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

${\displaystyle \int _{C_{1}+[a,b]+C_{2}+[b,a]}f(z)\;dz=\left\{\oint _{C_{1}}+\int _{a}^{b}+\oint _{C_{2}^{-}}+\int _{b}^{a}\right\}f(z)\;dz=0}$

where the notation ${\displaystyle C_{2}^{-}}$ indicates that the integration path is in the clockwise (negative) direction in the complex plane.

"Figure 3: equivalent contour integrals"

Because

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

we have

${\displaystyle \int _{C_{1}}f(z)\;dz=\int _{C_{2}^{-}}f(z)\;dz}$

Reversing the direction of integration on the integral on the right hand side yields

${\displaystyle \int _{C_{1}}f(z)\;dz=\int _{C_{2}}f(z)\;dz.}$

Thus the integrals over the integration contours ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ are equivalent. Because ${\displaystyle f(z)}$ need not be analytic in the interior of ${\displaystyle C_{2}}$ 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 ${\displaystyle C_{k}}$ bounding these holes is equivalent to the integral over the bounding contour ${\displaystyle C_{1}.}$ (See Figure 3.)

Another way of interpreting this result is that we may continuously deform the countour ${\displaystyle C_{1}}$ to any other closed simple curve ${\displaystyle C_{2}}$ 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.