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Course by: Steven J. Cox. E-mail the author

# Cauchy's Theorem

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

Summary: In this module we begin by discussing integrations of complex functions over complex curves and end with 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 this equation - In this equation, we found zIB-1=1z λ 1 P 1 +1z λ 1 2 D 1 +1z λ 2 P 2 z I B 1 z λ 1 P 1 1 z λ 1 2 D 1 1 z λ 2 P 2 where the P j P j and D j D j enjoy the amazing properties

1. B P 1 = P 1 B= λ 1 P 1 + D 1 B P 1 P 1 B λ 1 P 1 D 1
(1)
and B P 2 = P 2 B= λ 2 P 2 B P 2 P 2 B λ 2 P 2
2. P 1 + P 2 =I P 1 P 2 I
(2)
P 1 2= P 1 P 1 2 P 1 P 2 2= P 2 P 2 2 P 2 and D 1 2=0 D 1 2 0
3. P 1 D 1 = D 1 P 1 = D 1 P 1 D 1 D 1 P 1 D 1
(3)
and P 2 D 1 = D 1 P 2 =0 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 B in this equation, 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 t). More precisely, C zt=xt+iyt atb C z t x t y t a t b For example, if xt=yt=t x t y t t while a=0 a 0 and b=1 b 1 , then C C is the line segment joining 0+i0 0 0 to 1+i 1 .

We now define fzdzabfztztdt z C f z t a b f z t z t For example, if C= t+it 0t1 C t t 0 t 1 as above and fz=z f z z then zdz=01(t+it)(1+i)dt=01tt+i2tdt=i z C z t 0 1 t t 1 t 0 1 t t 2 t while if C C is the unit circle eit 0t2π t 0 t 2 then zdz=02πeitieitdt=i02πei2tdt=i02πcos2t+isin2tdt=0 z C z t 0 2 t t t 0 2 2 t t 0 2 2 t 2 t 0

Remaining with the unit circle but now integrating fz=1z f z 1 z we find z-1dz=02πe(it)ieitdt=2πi z C z t 0 2 t t 2

We generalize this calculation to arbitrary (integer) powers over arbitrary circles. More precisely, for integer m m and fixed complex a a we integrate zam z a m over Car a+reit 0t2π C a r a r t 0 t 2 the circle of radius r r centered at a a. We find

zamdz=02πa+reitamrieitdt=irm+102πei(m+1)tdt z C a r z a m t 0 2 a r t a m r t r m 1 t 0 2 m 1 t
(4)
zamdz=irm+102πcos(m+1)t+isin(m+1)tdt={2πi  if  m=-10  otherwise   z C a r z a m r m 1 t 0 2 m 1 t m 1 t 2 m -1 0

When integrating more general functions it is often convenient to express the integral in terms of its real and imaginary parts. More precisely fzdz=uxy+ivxydx+iuxy+ivxydy z C f z x C u x y v x y y C u x y v x y fzdz=uxydxvxydy+ivxydx+iuxydy z C f z x C u x y y C v x y x C v x y y C u x y fzdz=abuxtytxtvxtytytdt+iabuxtytyt+vxtytxtdt z C f z t a b u x t y t x t v x t y t y t t a b u x t y t y t v x t y t x t

The second line should invoke memories of:

### Theorem 1: Green's Theorem

If C C is a closed curve and M M and N N are continuously differentiable real-valued functions on C in C in , the region enclosed by C C, then Mdx+Ndy=NxMydxdy x C M y C N y x C in x N y M

Applying this to the situation above, we find, so long as C C is closed, that fzdz=vx+uydxdy+iux+vydxdy z C f z y x C in x v y u y x C in x u y v

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:

### Theorem 2: Cauchy's Theorem

If f f is differentiable on and in the closed curve C C then fzdz=0 z C f z 0 .

Strictly speaking, in order to invoke Green's Theorem we require not only that f f 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 Equation 4, permits us to integrate every proper rational function. More precisely, if q=fg q f g where f f is a polynomial of degree at most m1 m 1 and gg is an mmth degree polynomial with hh distinct zeros at λ j j=1h λ j j 1 h with respective multiplicities of m j j=1h m j j 1 h we found that

qz= j =1h k =1 m j q j , k z λ j k q z j 1 h k 1 m j q j , k z λ j k
(5)
Observe now that if we choose r j r j so small that λ j λ j is the only zero of gg encircled by C j C λ j r j C j C λ j r j then by Cauchy's Theorem qzd z = k =1 m j q j , k 1z λ j kd z z C j q z k 1 m j q j , k z C j 1 z λ j k In Equation 4 we found that each, save the first, of the integrals under the sum is in fact zero. Hence,
qzd z =2πi q j , 1 z C j q z 2 q j , 1
(6)
With q j , 1 q j , 1 in hand, say from this equation or residue, one may view Equation 6 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 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 q and the appropriate factor. More precisely,
qzz λ j k1d z =2πi q j , k z C j q z z λ j k 1 2 q j , k
(7)
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 ...

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