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Fundamental Theorem of Algebra, Analysis: Basic Applications of the Cauchy Integral Formula

Module by: Lawrence Baggett. E-mail the author

Summary: As a major application of the Cauchy Integral Formula, let us show the much alluded to remarkable fact that a function that is a differentiable function of a complex variable on an open set UU is actually expandable in a Taylor series around every point in U,U, i.e., is an analytic function on U.U.

As a major application of the Cauchy Integral Formula, let us show the much alluded to remarkable fact that a function that is a differentiable function of a complex variable on an open set UU is actually expandable in a Taylor series around every point in U,U, i.e., is an analytic function on U.U.

Theorem 1

Suppose ff is a differentiable function of a complex variable on an open set U⊆C,U⊆C, and let cc be an element of U.U. Then ff is expandable in a Taylor series around c.c. In fact, for any r>0r>0 for which B¯r(c)⊆U,B¯r(c)⊆U, we have

f ( z ) = ∑ n = 0 ∞ a n ( z - c ) n f ( z ) = ∑ n = 0 ∞ a n ( z - c ) n
(1)

for all z∈Br(c).z∈Br(c).

Proof

Choose an r>0r>0 such that the closed disk B¯r(c)⊆U,B¯r(c)⊆U, and write CrCr for the boundary of this disk. Note that, for all points ζζ on the curve Cr,Cr, and any fixed point zz in the open disk Br(c),Br(c), we have that |z-c|<r=|ζ-c|,|z-c|<r=|ζ-c|, whence |z-c|/|ζ-c|=|z-c|/r<1.|z-c|/|ζ-c|=|z-c|/r<1. Therefore the geometric series

∑ n = 0 ∞ z - c ζ - c n converges to 1 1 - z - c ζ - c . ∑ n = 0 ∞ z - c ζ - c n converges to 1 1 - z - c ζ - c .
(2)

Moreover, by the Weierstrass MM-Test, as functions of the variable ζ,ζ, this infinite series converges uniformly on the curve Cr.Cr. We will use this in the calculation below. Now, according to (Reference), we have that

f ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c + c - z d ζ = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) ( 1 - z - c ζ - c ) d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c ∑ n = 0 ∞ z - c ζ - c n d ζ = 1 2 π i ∫ C r ∑ n = 0 ∞ f ( ζ ) ( ζ - c ) n + 1 ( z - c ) n d ζ = 1 2 π i ∑ n = 0 ∞ ∫ C r f ( ζ ) ( ζ - c ) n + 1 ( z - c ) n d ζ = ∑ n = 0 ∞ a n ( z - c ) n , f ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c + c - z d ζ = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) ( 1 - z - c ζ - c ) d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c ∑ n = 0 ∞ z - c ζ - c n d ζ = 1 2 π i ∫ C r ∑ n = 0 ∞ f ( ζ ) ( ζ - c ) n + 1 ( z - c ) n d ζ = 1 2 π i ∑ n = 0 ∞ ∫ C r f ( ζ ) ( ζ - c ) n + 1 ( z - c ) n d ζ = ∑ n = 0 ∞ a n ( z - c ) n ,
(3)

where we are able to bring the summation sign outside the integral by part (3) of (Reference), and where

a n = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) n + 1 d ζ . a n = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) n + 1 d ζ .
(4)

This proves that ff is expandable in a Taylor series around the point c,c, as desired.

Using what we know about the relationship between the coefficients of a Taylor series and the derivatives of the function, together with the Cauchy Integral Theorem, we obtain the following formulas for the derivatives of a differentiable function ff of a complex variable. These are sometimes also called the Cauchy Integral Formulas.

Corollary 1

Suppose ff is a differentiable function of a complex variable on an open set U,U, and let cc be an element of U.U. Then ff is infinitely differentiable at c,c, and

f ( n ) ( c ) = n ! 2 π i ∫ C s f ( ζ ) ( ζ - c ) n + 1 d ζ , f ( n ) ( c ) = n ! 2 π i ∫ C s f ( ζ ) ( ζ - c ) n + 1 d ζ ,
(5)

for any piecewise smooth geometric set S⊆US⊆U whose boundary CSCS has finite length, and for which cc belongs to the interior S0S0 of S.S.

Exercise 1

  1. Prove the preceding corollary.
  2. Let f,U,f,U, and cc be as in Theorem 1. Show that the radius of convergence rr of the Taylor series expansion of ff around cc is at least as large as the supremum of all ss for which Bs(c)⊆U.Bs(c)⊆U.
  3. Conclude that the radius of convergence of the Taylor series expansion of a differentiable function of a complex variable is as large as possible. That is, if ff is differentiable on a disk Br(c),Br(c), then the Taylor series expansion of ff around cc converges on all of Br(c).Br(c).
  4. Consider the real-valued function of a real variable given by f(x)=1/(1+x2).f(x)=1/(1+x2). Show that ff is differentiable at each real number x.x. Show that ff is expandable in a Taylor series around 0,0, but show that the radius of convergence of this Taylor series is equal to 1. Does this contradict part (c)?
  5. Let ff be the complex-valued function of a complex variable given by f(z)=1/(1+z2).f(z)=1/(1+z2). We have just replaced the real variable xx of part (d) by a complex variable z.z. Explain the apparent contradiction that parts (c) and (d) present in connection with this function.

Exercise 2

  1. Let SS be a piecewise smooth geometric set whose boundary CSCS has finite length, and let ff be a continuous function on the curve CS.CS. Define a function FF on S0S0 by
    F(z)=∫CSf(ζ)ζ-zdζ.F(z)=∫CSf(ζ)ζ-zdζ.
    (6)
    Prove that FF is expandable in a Taylor series around each point c∈S0.c∈S0. Show in fact that F(z)=∑an(z-c)nF(z)=∑an(z-c)n for all zz in a disk Br(c)⊆S0,Br(c)⊆S0, where
    an=n!2πi∫CSf(ζ)(ζ-c)n+1dζ.an=n!2πi∫CSf(ζ)(ζ-c)n+1dζ.
    (7)
    HINT: Mimic the proof of Theorem 1.
  2. Let ff and FF be as in part (a). Is FF defined on the boundary CSCS of S?S? If zz belongs to the boundary CS,CS, and z=limzn,z=limzn, where each zn∈S0,zn∈S0, Does the sequence {F(zn)}{F(zn)} converge, and, if so, does it converge to f(z)?f(z)?
  3. Let SS be the closed unit disk B¯1(0),B¯1(0), and let ff be defined on the boundary C1C1 of this disk by f(z)=z¯,f(z)=z¯, i.e., f(x+iy)=x-iy.f(x+iy)=x-iy. Work out the function FF of part (a), and then re-think about part (b).
  4. Let ff and FF be as in part (a). If, in addition, ff is continuous on all of SS and differentiable on S0,S0, show that F(z)=2πif(z)F(z)=2πif(z) for all z∈S0.z∈S0. Think about this “magic” constant 2πi.2πi. Review the proof of the Cauchy Integral Formula to understand where this constant comes from.

(Reference) and (Reference) constitute what we called the “identity theorem” for functions that are expandable in a Taylor series around a point c.c. An even stronger result than that is actually true for functions of a complex variable.

Theorem 2: Identity Theorem

Let ff be a continuous complex-valued function on a piecewise smooth geometric set S,S, and assume that ff is differentiable on the interior S0S0 of S.S. Suppose {zk}{zk} is a sequence of distinct points in S0S0 that converges to a point cc in S0.S0. If f(zk)=0f(zk)=0 for every K,K, then f(z)=0f(z)=0 for every z∈S.z∈S.

Proof

It follows from (Reference) that there exists an r>0r>0 such that f(z)=0f(z)=0 for all z∈Br(c).z∈Br(c). Now let ww be another point in S0,S0, and let us show that f(w)f(w) must equal 0.0. Using part (f) of (Reference), let φ:[a^,b^]→Cφ:[a^,b^]→C be a piecewise smooth curve, joining cc to w,w, that lies entirely in S0.S0. Let AA be the set of all t∈[a^,b^]t∈[a^,b^] such that f(φ(s))=0f(φ(s))=0 for all s∈[a^,t).s∈[a^,t). We claim first that AA is nonempty. Indeed, because φφ is continuous, there exists an ϵ>0ϵ>0 such that |φ(s)-c|=|φ(s)-φ(a^)|<r|φ(s)-c|=|φ(s)-φ(a^)|<r if |s-a^|<ϵ.|s-a^|<ϵ. Therefore f(φ(s))=0f(φ(s))=0 for all s∈[a^,a^+ϵ),s∈[a^,a^+ϵ), whence, a^+ϵ∈A.a^+ϵ∈A. Obviously, AA is bounded above by b^,b^, and we write t0t0 for the supremum of A.A. We wish to show that t0=b^,t0=b^, whence, since φφ is continuous at B^,B^,f(w)=f(φ(b^))=f(φ(t0))=0.f(w)=f(φ(b^))=f(φ(t0))=0. Suppose, by way of contradiction, that t0<b^,t0<b^, and write z0=φ(t0).z0=φ(t0). Now z0∈S0,z0∈S0, and z0=limφ(t0-1/k)z0=limφ(t0-1/k) because φφ is continuous at t0.t0. But f(φ(t0-1/k))=0f(φ(t0-1/k))=0 for all k.k. So, again using (Reference), we know that there exists an r'>0r'>0 such that f(z)=0f(z)=0 for all z∈Br'(z0).z∈Br'(z0). As before, because φφ is continuous at t0,t0, there exists a δ>0δ>0 such that t0+δ<b^t0+δ<b^ and |φ(s)-φ(t0)|<r'|φ(s)-φ(t0)|<r' if |s-t0|<δ.|s-t0|<δ. Hence, f(φ(s))=0f(φ(s))=0 for all s∈(t0-δ,t0+δ),s∈(t0-δ,t0+δ), which implies that t0+δt0+δ belongs to A.A. But then t0t0 could not be the supremum of A,A, and therefore we have arrived at a contradiction. Consequently, t0=b^,t0=b^, and therefore f(w)=0f(w)=0 for all w∈S0.w∈S0. Of course, since every point in SS is a limit of points from S0,S0, and since ff is continuous on S,S, we see that f(z)=0f(z)=0 for all z∈S,z∈S, and the theorem is proved.

The next exercise gives some consequences of the Identity Theorem. Part (b) may appear to be a contrived example, but it will be useful later on.

Exercise 3

  1. Suppose ff and gg are two functions, both continuous on a piecewise smooth geometric set SS and both differentiable on its interior. Suppose {zk}{zk} is a sequence of elements of S0S0 that converges to a point c∈S0,c∈S0, and assume that f(zk)=g(zk)f(zk)=g(zk) for all k.k. Prove that f(z)=g(z)f(z)=g(z) for all z∈S.z∈S.
  2. Suppose ff is a nonconstant differentiable function defined on the interior of a piecewise smooth geometric set S.S. If c∈S0c∈S0 and Bϵ(c)⊆S0,Bϵ(c)⊆S0, show that there must exist an 0<r<ϵ0<r<ϵ for which f(c)≠f(z)f(c)≠f(z) for all zz on the boundary of the disk Br(c).Br(c).

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