7.1: Cauchy's Theorem

Introduction

Our main goal is a better understanding of the partial fraction expansion of a given transfer function. With respect to the example that closed the discussion of complex differentiation, see the equation - In this equation, we found

$$(zI-B)^{-1} = \frac{1}{z-\lambda_{1}}P_{1}+\frac{1}{(z-\lambda_{1})^2}D_{1}+\frac{1}{z-\lambda_{2}}P_{2} \nonumber$$

where the $P_{j}$ and $D_{j}$ enjoy the amazing properties

$$\begin{align*} BP_{1} &= P_{1}B \\[4pt] &= \lambda_{1}P_{1}+D_{1} \end{align*}$$

and

$$BP_{2} = P_{2}B = \lambda_{2}P_{2} \nonumber$$

$$P_{1}+P_{2} = I \nonumber$$

$$P_{1}^{2} = P_{1} \nonumber$$

$$P_{2}^{2} = P_{2} \nonumber$$

and

$$D_{1}^{2} = 0 \nonumber$$

$$P_{1}D_{1} = D_{1}P_{1} \nonumber$$

$$= D_{1} \nonumber$$

and

$$P_{2}D_{1} = D_{1}P_{2} = 0$$

In order to show that this always happens, i.e., that it is not a quirk produced by the particular $B$, we require a few additional tools from the theory of complex variables. In particular, we need the fact that partial fraction expansions may be carried out through complex integration.

Integration of Complex Functions Over Complex Curves

We shall be integrating complex functions over complex curves. Such a curve is parameterized by one complex valued or, equivalently, two real valued, function(s) of a real parameter (typically denoted by $t$). More precisely,

$$C \equiv \{z(t) = x(t)+iy(t) | a \le t \le b\} \nonumber$$

For example, if $x(t) = y(t) = t$ while $a = 0$ and $b = 1$, then $C$ is the line segment joining $0+i0$ to $1+i$.

We now define

$$\int f(z) dz = \equiv \int_{a}^{b} f(z(t)) z'(t) dt \nonumber$$

For example, if $C = \{t+it | 0 \le t \le 1\}$ as above and $f(z) = z$ then

$$\int z dz = \int_{0}^{1} (t+it) (1+i) dt = \int_{0}^{1} t-t+i2t dt = i \nonumber$$

while if $C$ is the unit circle $\{e^{it} | 0 \le t \le 2\pi\}$ then

$$\int z dz =\in_{0}^{2\pi} e^{it}ie^{it} dt = i \int_{0}^{2\pi} e^{i2t} dt = i \int_{0}^{2\pi} \cos(2t)+i \sin(2t) dt = 0 \nonumber$$

Remaining with the unit circle but now integrating $f(z) = \frac{1}{z}$ we find

$$\int z^{-1} dz = \int_{0}^{2 \pi} e^{-(it)}ie^{it} dt = 2\pi i \nonumber$$

We generalize this calculation to arbitrary (integer) powers over arbitrary circles. More precisely, for integer mm and fixed complex $(z-a)^{m}$ over

$$C(a,r) \equiv \{a+re^{it} | 0 \le t \le 2\pi\} \nonumber$$

the circle of radius $r$ centered at $a$

$$\int (z-a)^{m} dz = \int_{0}^{2\pi} (a+re^{it}-a)^{m} rie^{it} dt \nonumber$$

$$= ir^{m+1} \int_{0}^{2\pi} e^{i(m+1)t} dt \nonumber$$

$$\int (z-a)^{m} dz = ir^{m+1} \int_{0}^{2\pi} \cos((m+1)t)+i \sin((m+1)t) dt = \left \{ \begin{array}{cc} {2\pi i}&{\text{if} m = -1}\\ {0}&{\text{otherwise}} \end{array} \right . \nonumber$$

When integrating more general functions it is often convenient to express the integral in terms of its real and imaginary parts. More precisely

$$\begin{align*} \int f(z) dz &=\int u(x,y)+iv(x,y) dx+i \int u(x,y)+iv(x,y) dy \\[4pt] &=\int u(x,y)dx - \int v(x,y) dy+i \int v(x,y)dx+i \int u(x,y) dy \\[4pt] &=\int_{a}^{b} u(x(t), y(t))x'(t)-v(x(t), y(t))y'(t) dt+i \int_{a}^{b} u(x(t), y(t))y'(t)+v(x(t), y(t))x'(t) dt \end{align*}$$

The second line should invoke memories of:

Green's Theorem

If $C$ is a closed curve and $M$ and $N$ are continuously differentiable real-valued functions on $C_{in}$, the region enclosed by $C$, then

$$\int M dx+ \int N dy = \iint \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} dxdy \nonumber$$

Applying this to the situation above, we find, so long as $C$ is closed, that

$$\int f(z) dz = -\iint \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} dxdy+ i \iint \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} dxdy \nonumber$$

At first glance it appears that Green's Theorem only serves to muddy the waters. Recalling the Cauchy-Riemann equations however we find that each of these double integrals is in fact identically zero! In brief, we have proven:

Cauchy's Theorem

If $f$ is differentiable on and in the closed curve $C$ then $\int f(z) dz = 0$.

Strictly speaking, in order to invoke Green's Theorem we require not only that ff be differentiable but that its derivative in fact be continuous. This however is simply a limitation of our simple mode of proof; Cauchy's Theorem is true as stated.

This theorem, together with $C(a,r) \equiv \{a+re^{it} | 0 \le t \le 2\pi\}$, permits us to integrate every proper rational function. More precisely, if $q = \frac{f}{g}$ where $f$ is a polynomial of degree at most $m-1$ and $g$ is an mth degree polynomial with h distinct zeros at $\{\lambda_{j} | j = \{1, \cdots, h\}\}$ with respective multiplicities of $\{m_{j} | j = \{1, \cdots, h\}\}$ we found that

$$q(z) = \sum_{j = 1}^{h} \sum_{k = 1}^{m_{j}} \frac{q_{j,k}}{(z-\lambda_{j})^{k}} \nonumber$$

Observe now that if we choose $r_{j}$ so small that $\lambda_{j}$ is the only zero of $g$ encircled by $C_{j} \equiv C(\lambda_{j}, r_{j})$ then by Cauchy's Theorem

$$\int q(z) dz = \sum_{k = 1}^{m_{j}} q_{j,k} \int \frac{1}{(z-\lambda_{j})^k} dz \nonumber$$

In Equation we found that each, save the first, of the integrals under the sum is in fact zero. Hence,

$$\int q(z) dz = 2\pi i q_{j,1} \nonumber$$

With $q_{j,1}$ in hand, say from this equation or residue, one may view Equation as a means for computing the indicated integral. The opposite reading, i.e., that the integral is a convenient means of expressing $q_{j,1}$, will prove just as useful. With that in mind, we note that the remaining residues may be computed as integrals of the product of q and the appropriate factor. More precisely,

$$\int q(z)(z-\lambda_{j})^{k-1} dz = 2\pi i q_{j,k} \nonumber$$

One may be led to believe that the precision of this result is due to the very special choice of curve and function. We shall see ...