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Fundamental Theorem of Algebra, Analysis: Cauchy's Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: A module containing theorems and exercises leading up to Cauchy's theorem.

We begin with a simple observation connecting differentiability of a function of a complex variable to a relation among of partial derivatives of the real and imaginary parts of the function. Actually, we have already visited this point in (Reference).

Theorem 1: Cauchy-Riemann equations

Let f=u+ivf=u+iv be a complex-valued function of a complex variable z=x+iy(x,y),z=x+iy(x,y), and suppose ff is differentiable, as a function of a complex variable, at the point c=(a,b).c=(a,b). Then the following two partial differential equations, known as the Cauchy-Riemann Equations, hold:

t i a l u t i a l x ( a , b ) = t i a l v t i a l y ( a , b ) , t i a l u t i a l x ( a , b ) = t i a l v t i a l y ( a , b ) ,
(1)

and

t i a l u t i a l y ( a , b ) = - t i a l v t i a l x ( a , b ) . t i a l u t i a l y ( a , b ) = - t i a l v t i a l x ( a , b ) .
(2)

Proof

We know that

f ' ( c ) = lim h 0 f ( c + h ) - f ( c ) h , f ' ( c ) = lim h 0 f ( c + h ) - f ( c ) h ,
(3)

and this limit is taken as the complex number hh approaches 0. We simply examine this limit for real hh's approaching 0 and then for purely imaginary hh's approaching 0. For real hh's, we have

f ' ( c ) = f ' ( a + i b ) = lim h 0 f ( a + h + i b ) - f ( a + i b ) h = lim h 0 u ( a + h , b ) + i v ( a + h , b ) - u ( a , b ) - i v ( a , b ) h = lim h 0 u ( a + h , b ) - u ( a , b ) h + i lim h 0 v ( a + h , b ) - v ( a , b ) h = t i a l u t i a l x ( a , b ) + i t i a l v t i a l x ( a , b ) . f ' ( c ) = f ' ( a + i b ) = lim h 0 f ( a + h + i b ) - f ( a + i b ) h = lim h 0 u ( a + h , b ) + i v ( a + h , b ) - u ( a , b ) - i v ( a , b ) h = lim h 0 u ( a + h , b ) - u ( a , b ) h + i lim h 0 v ( a + h , b ) - v ( a , b ) h = t i a l u t i a l x ( a , b ) + i t i a l v t i a l x ( a , b ) .
(4)

For purely imaginary hh's, which we write as h=ik,h=ik, we have

f ' ( c ) = f ' ( a + i b ) = lim k 0 f ( a + i ( b + k ) ) - f ( a + i b ) i k = lim k 0 u ( a , b + k ) + i v ( a , b + k ) - u ( a , b ) - i v ( a , b ) i k = - i lim k 0 u ( a , b + k ) - u ( a , b ) k + v ( a , b + k ) - v ( a , b ) k = - i t i a l u t i a l y ( a , b ) + t i a l v t i a l y ( a , b ) . f ' ( c ) = f ' ( a + i b ) = lim k 0 f ( a + i ( b + k ) ) - f ( a + i b ) i k = lim k 0 u ( a , b + k ) + i v ( a , b + k ) - u ( a , b ) - i v ( a , b ) i k = - i lim k 0 u ( a , b + k ) - u ( a , b ) k + v ( a , b + k ) - v ( a , b ) k = - i t i a l u t i a l y ( a , b ) + t i a l v t i a l y ( a , b ) .
(5)

Equating the real and imaginary parts of these two equivalent expressions for f'(c)f'(c) gives the Cauchy-Riemann equations.

As an immediate corollary of this theorem, together with Green's Theorem ((Reference)), we get the following result, which is a special case of what is known as Cauchy's Theorem.

Corollary 1

Let SS be a piecewise smooth geometric set whose boundary CSCS has finite length. Suppose ff is a complex-valued function that is continuous on SS and differentiable at each point of the interior S0S0 of S.S. Then the contour integral CSf(ζ)dζ=0.CSf(ζ)dζ=0.

Exercise 1

  1. Prove the preceding corollary. See (Reference).
  2. Suppose f=u+ivf=u+iv is a differentiable, complex-valued function on an open disk Br(c)Br(c) in C,C, and assume that the real part uu is a constant function. Prove that ff is a constant function. Derive the same result assuming that vv is a constant function.
  3. Suppose ff and gg are two differentiable, complex-valued functions on an open disk Br(c)Br(c) in C.C. Show that, if the real part of ff is equal to the real part of g,g, then there exists a constant kk such that f(z)=g(z)+k,f(z)=g(z)+k, for all zBr(c).zBr(c).

For future computational purposes, we give the following implications of the Cauchy-Riemann equations. As with Theorem 1, this next theorem mixes the notions of differentiability of a function of a complex variable and the partial derivatives of its real and imaginary parts.

Theorem 2

Let f=u+ivf=u+iv be a complex-valued function of a complex variable, and suppose that ff is differentiable at the point c=(a,b).c=(a,b). Let AA be the 2×22×2 matrix

A = ( u x ( a , b ) v x ( a , b ) u y ( a , b ) v y ( a , b ) ) . A = ( u x ( a , b ) v x ( a , b ) u y ( a , b ) v y ( a , b ) ) .
(6)

Then:

  1. | f ' ( c ) | 2 = det ( A ) . | f ' ( c ) | 2 = det ( A ) .
  2. The two vectors
    V1=(ux(a,b),uy(a,b))andV2=(vx(a,b),vy(a,b))V1=(ux(a,b),uy(a,b))andV2=(vx(a,b),vy(a,b))
    (7)
    are linearly independent vectors in R2R2 if and only if f'(c)0.f'(c)0.
  3. The vectors
    V3=(ux(a,b),vx(a,b))andV4=(uy(a,b),vy(a,b))V3=(ux(a,b),vx(a,b))andV4=(uy(a,b),vy(a,b))
    (8)
    are linearly independent vectors in R2R2 if and only if f'(c)0.f'(c)0.

Proof

Using the Cauchy-Riemann equations, we see that the determinant of the matrix AA is given by

det A = u x ( a , b ) v y ( a , b ) - u y ( a , b ) v x ( a , b ) = ( u x ( a , b ) ) 2 + ( v x ( a , b ) ) 2 = ( u x ( a , b ) + i v x ( a , b ) ) ( u x ( a , b ) - i v x ( a , b ) ) = f ' ( c ) f ' ( c ) ¯ = | f ' ( c ) | 2 , det A = u x ( a , b ) v y ( a , b ) - u y ( a , b ) v x ( a , b ) = ( u x ( a , b ) ) 2 + ( v x ( a , b ) ) 2 = ( u x ( a , b ) + i v x ( a , b ) ) ( u x ( a , b ) - i v x ( a , b ) ) = f ' ( c ) f ' ( c ) ¯ = | f ' ( c ) | 2 ,
(9)

proving part (1).

The vectors V1V1 and V2V2 are the columns of the matrix A,A, and so, from elementary linear algebra, we see that they are linearly independent if and only if the determinant of AA is nonzero. Hence, part (2) follows from part (1). Similarly, part (3) is a consequence of part (1).

It may come as no surprise that the contour integral of a function ff around the boundary of a geometric set SS is not necessarily 0 if the function ff is not differentiable at each point in the interior of S.S. However, it is exactly these kinds of contour integrals that will occupy our attention in the rest of this chapter, and we shouldn't jump to any conclusions.

Exercise 2

Let cc be a point in C,C, and let SS be the geometric set that is a closed disk B¯r(c).B¯r(c). Let φφ be the parameterization of the boundary CrCr of SS given by φ(t)=c+reitφ(t)=c+reit for t[0,2π].t[0,2π]. For each integer nZ,nZ, define fn(z)=(z-c)n.fn(z)=(z-c)n.

  1. Show that Crfn(ζdζ=0Crfn(ζdζ=0 for all n-1.n-1.
  2. Show that
    Crf-1(ζ)dζ=Cr1ζ-cdζ=2πi.Crf-1(ζ)dζ=Cr1ζ-cdζ=2πi.
    (10)

There is a remarkable result about contour integrals of certain functions that aren't differentiable everywhere within a geometric set, and it is what has been called the Fundamental Theorem of Analysis, or Cauchy's Theorem. This theorem has many general statements, but we present one here that is quite broad and certainly adequate for our purposes.

Theorem 3: Cauchy's Theorem, Fundamental Theorem of Analysis

Let SS be a piecewise smooth geometric set whose boundary CSCS has finite length, and let S^S0S^S0 be a piecewise smooth geometric set, whose boundary CS^CS^ also is of finite length. Suppose ff is continuous on SS^0˜,SS^0˜, i.e., at every point zz that is in SS but not in S^0,S^0, and assume that ff is differentiable on S0S^˜,S0S^˜, i.e., at every point zz in S0S0 but not in S^.S^. (We think of these sets as being the points “between” the boundary curves of these geometric sets.) Then the two contour integrals CSf(ζ)dζCSf(ζ)dζ and CS^f(ζ)dζCS^f(ζ)dζ are equal.

Proof

Let the geometric set SS be determined by the interval [a,b][a,b] and the two bounding functions uu and l,l, and let the geometric set S^S^ be determined by the subinterval [a^,b^][a^,b^] of [a,b][a,b] and the two bounding functions u^u^ and l^.l^. Because S^S0,S^S0, we know that u^(t)<u(t)u^(t)<u(t) and l(t)<l^(t)l(t)<l^(t) for all t[a^,b^].t[a^,b^]. We define four geometric sets S1,...,S4S1,...,S4 as follows:

  1.  S1S1 is determined by the interval [a,a^][a,a^] and the two bounding functions uu and ll restricted to that interval.
  2.  S2S2 is determined by the interval [a^,b^][a^,b^] and the two bounding functions uu and u^u^ restricted to that interval.
  3.  S3S3 is determined by the interval [a^,b^][a^,b^] and the two bounding functions l^l^ and ll restricted to that interval.
  4.  S4S4 is determined by the interval [b^,b][b^,b] and the two bounding functions uu and ll restricted to that interval.

Observe that the five sets S^,S1,...,S4S^,S1,...,S4 constitute a partition of the geometric set S.S. The corollary to Theorem 1 applies to each of the four geometric sets S1,...,S4.S1,...,S4. Hence, the contour integral of ff around each of the four boundaries of these geometric sets is 0. So, by (Reference),

C S f ( ζ ) d ζ = C S ^ f ( ζ ) d ζ + k = 1 4 C S k f ( ζ ) d ζ = C S ^ f ( ζ ) d ζ , C S f ( ζ ) d ζ = C S ^ f ( ζ ) d ζ + k = 1 4 C S k f ( ζ ) d ζ = C S ^ f ( ζ ) d ζ ,
(11)

as desired.

Exercise 3

  1. Draw a picture of the five geometric sets in the proof above and justify the claim that the sum of the four contour integrals around the geometric sets S1,...,S4S1,...,S4 is the integral around CSCS minus the integral around CS^.CS^.
  2. Let S1,...,SnS1,...,Sn be pairwise disjoint, piecewise smooth geometric sets, each having a boundary of finite length, and each contained in a piecewise smooth geometric set SS whose boundary also has finite length. Prove that the SkSk's are some of the elements of a partition {S˜l}{S˜l} of S,S, each of which is piecewise smooth and has a boundary of finite length. Show that, by reindexing, S1,...,SnS1,...,Sn can be chosen to be the first nn elements of the partition {S^l}.{S^l}. HINT: Just carefully adjust the proof of (Reference).
  3. Suppose SS is a piecewise smooth geometric set whose boundary has finite length, and let S1,...,SnS1,...,Sn be a partition of SS for which each SkSk is piecewise smooth and has a boundary CSkCSk of finite length. Suppose ff is continuous on each of the boundaries CSkCSk of the SkSk's as well as the boundary CSCS of S,S, and assume that ff is continuous on each of the SkSk's, for 1km,1km, and differentiable at each point of their interiors. Prove that
    CSf(ζ)dζ=k=m+1nCSkf(ζ)dζ.CSf(ζ)dζ=k=m+1nCSkf(ζ)dζ.
    (12)
  4. Prove the following generalization of the Cauchy Theorem: Let S1,...,SnS1,...,Sn be pairwise disjoint, piecewise smooth geometric sets whose boundaries have finite length, all contained in the interior of a piecewise smooth geometric set SS whose boundary also has finite length. Suppose ff is continuous at each point of SS that is not in the interior of any of the SkSk's, and that ff is differentiable at each point of S0S0 that is not an element of any of the SkSk's. Prove that
    CSf(ζ)dζ=k=1nCSkf(ζ)dζ.CSf(ζ)dζ=k=1nCSkf(ζ)dζ.
    (13)

Perhaps the main application of Theorem 3 is what's called the Cauchy Integral Formula. It may not appear to be useful at first glance, but we will be able to use it over and over throughout this chapter. In addition to its theoretical uses, it is the basis for a technique for actually evaluating contour integrals, line integrals, as well as ordinary integrals.

Theorem 4: Cauchy Integral Formula

Let SS be a piecewise smooth geometric set whose boundary CSCS has finite length, and let ff be a continuous function on SS that is differentiable on the interior S0S0 of S.S. Then, for any point zS0,zS0, we have

f ( z ) = 1 2 π i C S f ( ζ ) ζ - z d ζ . f ( z ) = 1 2 π i C S f ( ζ ) ζ - z d ζ .
(14)

REMARK This theorem is an initial glimpse at how differentiable functions of a complex variable are remarkably different from differentiable functions of a real variable. Indeed, Cauchy's Integral Formula shows that the values of a differentiable function ff at all points in the interior of a geometric set SS are completely determined by the values of that function on the boundary of the set. The analogous thing for a function of a real variable would be to say that all the values of a differentiable function ff at points in the open interval (a,b)(a,b) are completely determined by its values at the endpoints aa and b.b. This is patently absurd for functions of a real variable, so there surely is something marvelous going on for differentiable functions of a complex variable.

Proof

Let rr be any positive number such that B¯r(z)B¯r(z) is contained in the interior S0S0 of S,S, and note that the close disk B¯r(z)B¯r(z) is a piecewise smooth geometric set S^S^ contained in S0.S0. We will write CrCr instead of CS^CS^ for the boundary of this disk, and we will use as a parameterization of the curve CrCr the function φ:[0,2π]Crφ:[0,2π]Cr given by φ(t)=z+reit.φ(t)=z+reit. Now the function g(ζ)=f(ζ)/(ζ-z)g(ζ)=f(ζ)/(ζ-z) is continuous on SS^0˜SS^0˜ and differentiable on S0S^˜,S0S^˜, so that Theorem 3 applies to the function g.g. Hence

1 2 π i C S f ( ζ ) ζ - z d ζ = 1 2 π i C S g ( ζ ) d ζ = 1 2 π i C R g ( ζ ) , d ζ = 1 2 π i C r f ( ζ ) ζ - z d ζ = 1 2 π i 0 2 π f ( z + r e i t ) z + r e i t - z i r e i t d t = 1 2 π 0 2 π f ( z + r e i t ) d t . 1 2 π i C S f ( ζ ) ζ - z d ζ = 1 2 π i C S g ( ζ ) d ζ = 1 2 π i C R g ( ζ ) , d ζ = 1 2 π i C r f ( ζ ) ζ - z d ζ = 1 2 π i 0 2 π f ( z + r e i t ) z + r e i t - z i r e i t d t = 1 2 π 0 2 π f ( z + r e i t ) d t .
(15)

Since the equality established above is valid, independent of r,r, we may take the limit as rr goes to 0, and the equality will persist. We can evaluate such a limit by replacing the rr by 1/n,1/n, in which case we would be evaluating

lim n 1 2 π 0 2 π f ( z + 1 n e i t ) d t = lim n 1 2 π 0 2 π f n ( t ) d t , lim n 1 2 π 0 2 π f ( z + 1 n e i t ) d t = lim n 1 2 π 0 2 π f n ( t ) d t ,
(16)

where fn(t)=f(z+frac1neit).fn(t)=f(z+frac1neit). Finally, because the function ff is continuous at the point z,z, it follows that the sequence {fn}{fn} converges uniformly to the constant function f(z)f(z) on the interval [0,2π].[0,2π]. So, by Theorem 5.6, we have that

lim n 1 2 π 0 2 π f n ( t ) d t = 1 2 π 0 2 π f ( z ) d t = f ( z ) . lim n 1 2 π 0 2 π f n ( t ) d t = 1 2 π 0 2 π f ( z ) d t = f ( z ) .
(17)

Therefore,

1 2 π i C S f ( ζ ζ - z d ζ = lim r 0 1 2 π 0 2 π f ( z + r e i t ) d t = f ( z ) , 1 2 π i C S f ( ζ ζ - z d ζ = lim r 0 1 2 π 0 2 π f ( z + r e i t ) d t = f ( z ) ,
(18)

and the theorem is proved.

The next exercise gives two simple but strong consequences of the Cauchy Integral Formula, and it would be wise to spend a few minutes deriving other similar results.

Exercise 4

  1. Let SS and ff be as in the preceding theorem, and assume that f(z)=0f(z)=0 for every point on the boundary CSCS of S.S. Prove that f(z)=0f(z)=0 for every zS.zS.
  2. Let SS be as in part (a), and suppose that ff and gg are two continuous functions on S,S, both differentiable on S0,S0, and such that f(ζ)=g(ζ)f(ζ)=g(ζ) for every point on the boundary of S.S. Prove that f(z)=g(z)f(z)=g(z) for all zS.zS.

The preceding exercise shows that two differentiable functions of a complex variable are equal everywhere on a piecewise smooth geometric set SS if they agree on the boundary of the set. More is true. We will see below in the Identity Theorem that they are equal everywhere on a piecewise smooth geometric set SS if they agree just along a single convergent sequence in the interior of S.S.

Combining part (b) of Exercise 3, (Reference), and Theorem 3, we obtain the following corollary:

Corollary 2

Let S1,...,SnS1,...,Sn be pairwise disjoint, piecewise smooth geometric sets whose boundaries have finite length, all contained in the interior of a piecewise smooth geometric set SS whose boundary has finite length. Suppose ff is continuous at each point of SS that is not in the interior of any of the SkSk's, and that ff is differentiable at each point of S0S0 that is not an element of any of the SkSk's. Then, for any zS0zS0 that is not an element of any of the SkSk's, we have

f ( z ) = 1 2 π i C S f ( ζ ) ζ - z d ζ - k = 1 n C S k f ( ζ ) ζ - z d ζ . f ( z ) = 1 2 π i C S f ( ζ ) ζ - z d ζ - k = 1 n C S k f ( ζ ) ζ - z d ζ .
(19)

Proof

Let r>0r>0 be such that B¯r(z)B¯r(z) is disjoint from all the SkSk's. By part (b) of Exercise 3, let T1,...,TmT1,...,Tm be a partition of SS such that Tk=SkTk=Sk for 1kn,1kn, and Tn+1=B¯r(z).Tn+1=B¯r(z). By (Reference), we know that

C S f ( ζ ) ζ - z d ζ = k = 1 m C T k f ( ζ ) ζ - z d ζ . C S f ( ζ ) ζ - z d ζ = k = 1 m C T k f ( ζ ) ζ - z d ζ .
(20)

From the Cauchy Integral Formula, we know that

C T n + 1 f ( ζ ) ζ - z d ζ = 2 π i f ( z ) . C T n + 1 f ( ζ ) ζ - z d ζ = 2 π i f ( z ) .
(21)

Also, since f(ζ)/(ζ-z)f(ζ)/(ζ-z) is differentiable at each point of the interior of the sets TkTk for k>n+1,k>n+1, we have from Theorem 2 that for all k>n+1k>n+1

C t k f ( ζ ) ζ - z d ζ = 0 . C t k f ( ζ ) ζ - z d ζ = 0 .
(22)

Therefore,

C S f ( ζ ) ζ - z d ζ = k = 1 n C S k f ( ζ ) ζ - z d ζ + 2 π i f ( z ) , C S f ( ζ ) ζ - z d ζ = k = 1 n C S k f ( ζ ) ζ - z d ζ + 2 π i f ( z ) ,
(23)

which completes the proof.

Exercise 5

Suppose SS is a piecewise smooth geometric set whose boundary has finite length, and let c1,...,cnc1,...,cn be points in S0.S0. Suppose ff is a complex-valued function that is continuous at every point of SS except the CkCk's and differentiable at every point of S0S0 except the ckck's. Let r1,...,rnr1,...,rn be positive numbers such that the disks {B¯Rk(ck)}{B¯Rk(ck)} are pairwise disjoint and all contained in S0.S0.

  1. Prove that
    CSf(ζ)dζ,=k=1nCkf(ζ)dζCSf(ζ)dζ,=k=1nCkf(ζ)dζ
    (24)
    where CkCk denotes the boundary of the disk B¯rk(ck).B¯rk(ck).
  2. For any zS0zS0 that is not in any of the closed disks B¯rk(ck),B¯rk(ck), show that
    CSf(ζ)ζ-zdζ=2πif(z)+k=1nCkf(ζ)ζ-zdζ.CSf(ζ)ζ-zdζ=2πif(z)+k=1nCkf(ζ)ζ-zdζ.
    (25)
  3. (c) Specialize part (b) to the case where S=B¯r(c),S=B¯r(c), and ff is analytic at each point of Br(c)Br(c) except at the central point c.c. For each zczc in Br(c),Br(c), and any 0<δ<|z-c|,0<δ<|z-c|, derive the formula
    f(z)=12πiCrf(ζ)ζ-zdζ-12πiCδf(ζ)ζ-zdζ.f(z)=12πiCrf(ζ)ζ-zdζ-12πiCδf(ζ)ζ-zdζ.
    (26)

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