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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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Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

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Any individual member, a community, or a respected organization.

What are tags?

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