Functions of the Complex Variable z

Transcription

1 Capter 2 Functions of te Complex Variable z Introduction We wis to examine te notion of a function of z were z is a complex variable. To be sure, a complex variable can be viewed as noting but a pair of real variables so tat in one sense a function of z is noting but a function of two real variables. Tis was tepointof viewwe took in te last section in discussing continuousfunctions. But someow tispoint of view is too general. Tere are some functions wic are direct functions of z = x + iy and not simply functions of te separate pieces x and y. Consider, for example, te function x 2 y 2 + 2i xy. Tis is adirect function of x + iy since x 2 y 2 + 2i xy = (x + iy) 2 ; it is te function squaring. On te oter and, te only sligtly different-looking function x 2 + y 2 2i xy is not expressible as a polynomial in x + iy. Tus we are led to distinguis a special class of functions, tose given by direct or explicit or analytic expressions in x + iy. Wen we finally do evolve a rigorous definition, tese functions will be called te analytic functions. For now we restrict our attention to polynomials. 2.1 Analytic Polynomials 2.1 Definition A polynomial P(x, y) will be called an analytic polynomial if tere exist (complex) constants α k suc tat P(x, y) = α 0 + α 1 (x + iy) + α 2 (x + iy) 2 + +α N (x + iy) N. We will ten say tat P isapolynomial in zand write it as P(z) = α 0 + α 1 z + α 2 z 2 + +α N z N. J. Bak, D. J. Newman, Complex Analysis, DOI / _2 Springer Science+Business Media, LLC

2 22 2 Functions of te Complex Variable z Indeed, x 2 y 2 + 2i xy is analytic. On te oter and, as we mentioned above, x 2 + y 2 2i xy is not analytic, and we now prove tis assertion. So suppose Setting y = 0, we obtain x 2 + y 2 2i xy N α k (x + iy) k. or N x 2 α k x k α 0 + α 1 x + (α 2 1)x 2 + +α N x N 0. Setting x = 0gives α 0 = 0; dividing out by x and again setting x = 0sowsα 1 = 0, etc. We conclude tat and so our assumption tat as led us to α 1 = α 3 = α 4 = =α N = 0 α 2 = 1, x 2 + y 2 2i xy N α k (x + iy) k x 2 + y 2 2i xy (x + iy) 2 = x 2 y 2 + 2i xy, wic issimply false! Abit of experimentation, using te metod described above (setting y = 0and comparing coefficients ) will sow ow rare te analytic polynomials are. A randomly cosen polynomial, P(x, y), will ardly ever be analytic. EXAMPLE x 2 + iv(x, y) is not analytic for any coice of te real polynomial v(x, y). For a polynomial in z can ave a real part of degree 2 in x only if it is of te form az 2 + bz + c wit a 0. In tat case, owever, te real part must containay 2 term as well. Anoter Way of Recognizing Analytic Polynomials We ave seen, in our metod of comparing coefficients, a perfectly adequate way of determining weter a given polynomial isoris not analytic. Tis metod, we point out, can be condensed to te statement: P(x, y) is analytic if and only if P(x, y) = P(x +iy, 0). Looking aead to

3 2.1 Analytic Polynomials 23 te time we will try to extend te notion of analytic beyond te class of polynomials, owever, we see tat we can expect trouble! Wat is sosimple for polynomials is totally intractable for more general functions. We can evaluate P(x +iy, 0) by simple aritmetic operations, but wat does it mean to speak of f (x + iy, 0)? For example, if f (x, y) = cos x + i sin y, we observe tat f (x, 0) = cos x. But wat sall we mean by cos(x + iy)? Wat is needed is anoter means of recognizing te analytic polynomials, and for tis we retreat to a familiar, real-variable situation. Suppose tat we ask of apolynomial P(x, y) weter it is a function of te single variable x + 2y. Again te answer can be given in tespirit of our previous one, namely: P(x, y) is a function of x + 2y if and only if P(x, y) = P(x + 2y, 0). But it can also be given in terms of partial derivatives! A function of x + 2y undergoes te same cange wen x canges by as wen y canges ɛ/2andtis means exactly tat its partial derivative wit te respect to y is twice its partial derivative wit respect to x. Tat is, P(x, y) is a function of x + 2y if and only if P y = 2P x. Of course, te 2 can be replaced by any real number, and we obtain te more general statement: P(x, y) is a function of x + λy if and only if P y = λp x. Indeed for polynomials, we can even ignore te limitation tat λ be real, wic yields te following proposition. 2.2 Definition Let f (x, y) = u(x, y)+iv(x, y) were u and v are real-valued functions. Te partial derivatives f x and f y are defined by u x + iv x and u y + iv y respectively, provided te latter exist. 2.3 Proposition A polynomial P(x, y) is analytic if and only if P y = i P x. Te necessity of te condition can be proven in a straigtforward manner. We leave te details as an exercise. To sow tat it isalsosufficient, note tat if P y = i P x, te condition must be met separately by te terms of any fixed degree. Suppose ten tat P as n-t degree terms of te form Q(x, y) = C 0 x n + C 1 x n 1 y + C 2 x n 2 y 2 + +C n y n. Since Q y = i Q x, C 1 x n 1 + 2C 2 x n 2 y + +nc n y n 1 = i[nc 0 x n 1 + (n 1)C 1 x n 2 y + +C n 1 y n 1 ].

4 24 2 Functions of te Complex Variable z Comparing coefficients, C 1 = inc 0 = i C 2 = i(n 1) 2 ( ) n C 1 0 ( ) 2 n(n 1) C 1 = i C 0 = i 2 n C 2 2 0, and in general C k = i k ( n k ) C 0 so tat Q(x, y) = n C k x n k y k = C 0 n ( n k ) x n k (iy) k = C 0 (x + iy) n. Tus P is analytic. Te condition f y = i f x is sometimes given in terms of te real and imaginary parts of f. Tat is, if f = u + iv,ten f x = u x + iv x f y = u y + iv y and te equation f y = i f x is equivalent to te twin equations u x = v y ; u y = v x. (1) Tese are usually called te Caucy-Riemann equations. EXAMPLES 1. A non-constant analytic polynomial cannot be real-valued, for ten bot P x and P y would be real and te Caucy-Riemann equations would not be satisfied. 2. Using te Caucy-Riemann equations, one can verify tat x 2 y 2 + 2i xy is analyticwile x 2 + y 2 2i xy is not. Finally, we note tat polynomials in z ave anoter property wic distinguises tem as functions of z: tey can be differentiated directly wit respect to z. We will make tismoreprecise below. 2.4 Definition A complex-valued function f, defined inaneigborood of z, issaidtobedifferentiable at z if lim 0 f (z + ) f (z) exists. In tat case, te limit is denoted f (z).

5 2.2 Power Series 25 It is important to note tat in Definition 2.4, is not necessarily real. Hence te limit must exist irrespectiveof te mannerin wic approaces0 in te complex plane. For example, f (z) = z is not differentiable at any point z since f (z + ) f (z) = wic equals +1 if is real and 1 if is purely imaginary. 2.5 Proposition If f and g are bot differentiable at z, ten so are 1 = f + g 2 = fg and, if g(z) 0, 3 = f g. In te respective cases, 1 (z) = f (z) + g (z) 2 (z) = f (z)g(z) + f (z)g (z) 3 (z) = [ f (z)g(z) f (z)g (z)]/g 2 (z). Exercise Proposition If P(z) = α 0 + α 1 z + + α N z N, ten P is differentiable at all points z and P (z) = α 1 + 2α 2 z + +Nα N z N 1. See Exercise Power Series We now considera wider class of direct functionsof z tosegiven by infinite polynomials or power series in z.

6 26 2 Functions of te Complex Variable z 2.7 Definition A power series in z isaninfinite series of te form C k z k. To study te convergence of a power series, we recall te notion of te lim ofa positive real-valued sequence. Tat is, ( ) lim a n = lim sup a k. n n k n Since sup k n a k is a non-increasing function of n, telimit alwaysexists or equals +. Te properties of te limwic will be of interest to us are te following. If lim n a n = L, i. for eac N and for eac ɛ>0, tere exists some k > N suc tat a k L ɛ; ii. for eac ɛ>0, tere is some N suc tat a k L + ɛ for all k > N. iii. lim ca n = cl for any nonnegative constant c. 2.8 Teorem Suppose lim C k 1/k = L. 1. If L = 0, C k z k converges for all z. 2. If L =, C k z k converges for z = 0 only. 3. If 0 < L <, setr= 1/L. Ten C k z k converges for z < R and diverges for z > R. (R is called te radius of convergence of te power series.) 1. L = 0. Since lim C k 1/k = 0, lim C k 1/k z =0forallz. Tus, for eac z,tereissome N suc tat k > N implies C k z k 1 2 k, so tat C k z k converges; terefore, by te Absolute Convergence Test, C k z k converges. 2. L =. For any z 0, C k 1/k 1 z for infinitely many values of k. Hence C k z k 1, te terms of te series do not approac zero, and te series diverges. (Te fact tat te series converges for z = 0 isobvious.)

7 2.2 Power Series < L <, R = 1/L. Assume first tat z < R and set z =R(1 2δ). Ten since lim C k 1/k z = (1 2δ), C k 1/k z < 1 δ for sufficiently large k and C k z k is absolutely convergent. On te order and, if z > R, lim C k 1/k z > 1, so tat for infinitely many values of k, C k z k as absolute value greater tan 1 and C k z k diverges. Note tat if C k z k as radius of convergence R, teseries converges uniformly in any smaller disc: z R δ. For ten C k z k C k (R δ) k, wic also converges. Hence a power series is continuous trougout its domainof convergence. (See Teorem 1.9.) All tree cases above can be combined by noting tat a power series always converges inside a disc of radius R = 1/lim C k 1/k. Here R = 0 means tat te series converges at z = 0 only and R = means tat te series converges for all z. In te cases were 0 < R <, wile te teorem assures us tat te series diverges for z > R, it says noting about te beavior of te power series on te circle of convergence z =R. As te following examples demonstrate, te series may converge for all or some or none of te points on te circle of convergence. EXAMPLES 1. Since n 1/n 1, n=1 nz n converges for z < 1anddiverges for z > 1. Te series also diverges for z =1forten nz n =n. (See Exercise 8.) 2. n=1 (z n /n 2 ) also as radius of convergence equal to 1. In tis case, owever, te series converges for all points z on te unitcircle since z n = 1 n 2 for z =1. n 2 3. n=1 (z n /n) as radius of convergence equal to 1. In tis case, te series converges at all points of te unitcircle except z = 1. (See Exercise 12.) 4. (z n /n!) converges for all z since 1 0. (n!) 1/n (See Exercise 13.) 5. [1 + ( 1) n ] n z n as radius of convergence 1 2 since lim[1 + ( 1)n ] = lim2= 2.

8 28 2 Functions of te Complex Variable z 6. z n2 = 1 + z + z 4 + z 9 + z 16 + as radius of convergence 1. In tis case lim C n 1/n = lim1= Any series of te form C n z n wit C n =±1foralln as radius of convergence equal to 1. It is easily seen tat te sum of two power series is convergent werever bot of te originaltwo powerseries are convergent. In fact, it follows directly fromte definition of infinite series tat (a n + b n )z n = a n z n + b n z n. Similarly if a n z n = A and b n z n = B, te Caucy product c n z n defined by c n = n a k b n k converges for appropriate values of z to te product AB. Te proof is te same as tat for real power series and is outlined inexercises 17 and Differentiability and Uniqueness of Power Series We now sow tat power series, like polynomials, are differentiable functions of z. Suppose ten tat C n z n converges in some disc D(0; R), R > 0. Ten te series nc n z n 1 obtained by differentiating C n z n term by term isconvergentin D(0; R), since lim nc n 1/(n 1) = lim( nc n 1/n ) n/(n 1) = lim C n 1/n. 2.9 Teorem Suppose f (z) = C n z n converges for z < R. Ten f (z) exists and equals nc n z n 1 trougout z < R. We will prove te teorem in two stages. First, we will assume tat R =,ten we will consider te more general situation. Of course, te second case contains te first, so te eager reader may skip te first proof. We include it since it contains te key ideas wit less cumbersome details. Case (1): Assume C n z n converges for all z. Ten f (z + ) f (z) = C n [(z + ) n z n ] and f (z + ) f (z) nc n z n 1 = C n b n n=2

9 2.3 Differentiability and Uniqueness of Power Series 29 were b n = (z + )n z n nz n 1 n ( ) n n ( ) = k 1 z n k n z n k = ( z +1) n k k k=2 for 1. Hence, for 1, f (z + ) f (z) nc n z n 1 C n ( z +1) n A since C n z n converges for all z. Letting 0, we conclude tat f (z) = nc n z n 1. Case (2): 0 < R <. Let z =R 2δ,δ > 0, and assume <δ. Ten z + < R and, as in te previous case. we can write f (z + ) f (z) nc n z n 1 = C n b n, n=2 were b n = n k=2 ( n k ) k 1 z n k. If z = 0, b n = n 1 and te proof follows easily. Oterwise, to obtain a useful estimate for b n we must be a little more careful. Note ten tat ( ) ( ) n n(n 1) (n k + 1) = n 2 n for k 2. k k! k 2 Hence, for z 0, b n n2 z 2 n2 z 2 n k=2 n j=0 = n2 z 2 ( z + )n n2 (R δ)n z 2 ( ) n k 2 z n (k 2) k 2 ( n j ) j z n j

10 30 2 Functions of te Complex Variable z and f (z + ) f (z) nc n z n 1 z 2 n 2 C n (R δ) n A, since z 0 is fixedandsince n 2 C n z n also converges for z < R. Again, letting 0, we conclude tat f (z) = nc n z n 1. EXAMPLE f (z) = (z n /n!) is convergent for all z and, according to Teorem 2.9, f (z) = nz n 1 n! = z n n! = f (z) Corollary Power series are infinitely differentiable witinteir domain of convergence. Applying te above results to f (z) = nc n z n 1 wic as te same radius of convergence as f, we see tat f is twice differentiable. By induction, f (n) is differentiable for all n Corollary If f (z) = C n z n as a nonzero radius of convergence, C n = f (n) (0) n! for alln. By definition f (0) = C 0. Differentiating te power series term-by-term gives f (z) = C 1 + 2C 2 z + 3C 3 z 2 + so tat f (0) = C 1. Similarly f (n) (z) = n!c n + (n + 1)!C n+1 z + and te result follows by setting z = 0. (n + 2)! C n+2 z 2 +, 2!

11 2.3 Differentiability and Uniqueness of Power Series 31 According to Corollary 2.11, if a power series is equal to zero trougout a neigboroodof te origin, it must be identically zero. For ten all its derivatives at te origin and ence all te coefficients of te power series would equal 0. By te same reasoning, ifapowerseries were equal to zero trougout an interval containing te origin, it would be identically zero. An even stronger result is proven below Uniqueness Teorem for Power Series Suppose C n z n iszeroatallpoints of a nonzero sequence {z k } wic converges to zero. Ten te power series is identically zero. [Note: If we set f (z) = C n z n, it follows from te continuity of power series tat f (0) = 0. We can sow by a similar argument tat f (0) = 0; owever, a sligtly different argument is needed to sow tat te iger coefficients are also 0.] Let f (z) = C 0 + C 1 z + C 2 z 2 +. By te continuity of f at te origin C 0 = f (0) = lim z 0 f (z) = lim k f (z k) = 0. But ten g(z) = f (z) = C 1 + C 2 z + C 3 z 2 + z is also continuous at te originand f (z) C 1 = g(0) = lim = lim z 0 z k Similarly, if C j = 0for0 j < n, ten f (z) C n = lim z 0 z n = lim k f (z k ) z n k f (z k ) z k = 0. = 0, so tat te power series is identically zero Corollary If a power series equals zero at all te pointsofasetwit an accumulation point at te origin, te power series is identically zero. Exercise 18. Te UniquenessTeoremderives its name from te following corollary.

12 32 2 Functions of te Complex Variable z 2.14 Corollary If a n z n and b n z n converge and agree on a set of points wit an accumulation point at te origin, ten a n = b n for all n. Apply 2.13 to te difference: (an b n )z n. Power SeriesExpansionaboutz = α All of te previous results on power series are easily adapted to power series of te form Cn (z α) n. By te simple substitution w = z α, we see, for example, tat series of te above form converge in adisc of radius R about z = α and are differentiable trougout z α < R were R = 1/lim C n 1/n. (See Exercises 22 and 23.) Exercises 1. Complete te proof of Proposition 2.3 bysowing tat for an analytic polynomial P, P y = i P x. [Hint: Prove it first for te monomials.] 2.* a. Suppose f (z) is real-valued and differentiable for all real z. Sow tat f (z) is also real-valued for real z. b. Suppose f (z) is real-valued and differentiable for all imaginary points z. Sow tat f (z) is imaginary for at all imaginary points z. 3. By comparing coefficients or by use of te Caucy-Riemann equations, determine wic of te following polynomials are analytic. a. P(x + iy) = x 3 3xy 2 x + i(3x 2 y y 3 y). b. P(x + iy) = x 2 + iy 2. c. P(x + iy) = 2xy + i(y 2 x 2 ). 4. Sow tat no nonconstant analytic polynomial can take imaginary values only. 5. Find te derivative P (z) of te analytic polynomials in (3). Sow tat in eac case P (z) = P x. Explain. 6. Prove Proposition 2.5 by arguments analogous to tose of real-variable calculus. 7. Prove Proposition 2.6. [Hint: Prove it for monomials and apply Proposition 2.5.] 8. Sow S n = n 1/n 1asn by considering log S n. 9. Find te radius of convergence of te following power series: a. z n!, b. (n + 2 n )z n. 10. Suppose c n z n as radius of convergence R. Find te radius of convergence of a. n p c n z n, b. cn z n, c. c 2 n z n.

13 Exercises Suppose a n z n and b n z n ave radii of convergence R 1 and R 2, respectively. Wat can be said about te radius of convergence of (a n + b n )z n? Sow, by example, tat te radius of convergence of te latter may be greater tan R 1 and R Sow tat n=1 (z n /n) converges at all points on te unitcircle except z = 1. [Hint: Let z = cis θ and analyze te real and imaginary parts of te series separately.] 13. a. Suppose {a n } is a sequence of positive real numbers and Sow ten tat lim n a 1/n n = L. b. Use te result above to prove a lim n+1 = L. n a n ( ) 1 1/n 0. n! 14. Use Exercise (13a) to find te radius of convergence of a. ( 1) n z n, b. z 2n+1 n! (2n + 1)!, c. n=1 n!z n n n, d. 2 n z n n! 15.* Find te radius of convergence of a. sin nz n, b. e n 2 z n, 16.* Find te radius of convergence of c n z n if c 2k = 2 k ; c 2k 1 = (1 + 1/k) k2, k = 1, 2, Suppose a k = A and b k = B. Suppose furter tat eac of te series is absolutely convergent. Sow tat if k c k = a j b k j j=0 ten c k = AB. Outline: Use te fact tat a k and b k converge to sow tat d k converges were. In particular, k d k = a j b k j. j=0 d n+1 + d n+2 + 0asn. Note ten tat if A n = a 0 + a 1 + +a n B n = b 0 + b 1 + +b n C n = c 0 + c 1 + +c n, A n B n = C n + R n,were R n d n+1 +d n+2 + +d 2n, and te result follows by letting n. 18. Suppose a n z n and b n z n ave radii of convergence R 1 and R 2 respectively. Sow tat te Caucy product c n z n converges for z < min(r 1, R 2 ).

14 34 2 Functions of te Complex Variable z 19. a. Using te identity sow tat (1 z)(1 + z + z 2 + +z N ) = 1 z N+1 z n = 1 for z < 1. 1 z b. By taking te Caucy product of z n wit itself, find a closed form for nz n. 20. Prove Corollary 2.13 by sowing tat if a set S as an accumulation point at 0, it contains a sequence of nonzero terms wic converge to Sow tat tere is nopowerseries f (z) = C n z n suc tat i. f (z) = 1forz = 1 2, 1 3, 1 4,..., ii. f (0) > Assume lim C n 1/n <. Sow tat if weset ten f (z) = C n (z α) n, C n = f (n) (α). n! 23. Find te domain of convergence of a. n(z 1) n, b. c. n 2 (2z 1) n. ( 1) n (z + 1) n, n!

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