# Complex variables and applications churchill pdf

Complex variables and applications / James Ward Brown, Ruel V. Churchill. coauthor with Dr. Churchill of Fourier Series and Boundary Value Problems, now. theory and application of functions of a complex variable. This edition the late Ruel V. Churchill alone. V, Churchill's "Operational Mathematics," where. COMPLEX VARIABLES AND APPLICATIONS Eighth Edition James Ward Brown Professor of Mathematics The University of Michigan–Dearborn Ruel V.

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Brown-Churchill-Complex Variables and Application 8th edition. Pages· · Complex Variables And Applications Sol 7ed JW Brown RV Churchill. Complex variables and applications (4th edition), by Ruel V. Churchill and James Ward Brown. Pp ISBN COMPLEX VARIABLES AND APPLICATIONS. Bro\\'D and Churchill Series Complex \laria/Jle.\' aml 1\pplications. 9 Edition Fourier Series and Boundary .

Table of Tr. Thal edilion has served. This new edilion preserves lhe basic conlenl and slyle or lhe earlier ones. Churchill alone. The hook has always had lwo main ohjeclives. The applications of residues include their use in evaluating real improper integrals. Considerable anention is paid to! The firsl nine chapters of this hook have for many years formed the basis of a lhree- hourcourse given each Lenn al The University of Michigan.

The firsl nine chapters of this hook have for many years formed the basis of a lhree- hourcourse given each Lenn al The University of Michigan. The final three chapters have fewer changes and are mostly intended for self-study and reference. The classes using the hook have consisled mainly of seniors concentrating in mathemalics. Before taking! Ir mapping by elementary functions is desired earlier in lhe course.

We mention here a sample of the changes in this edition. For example. Another example is the moving of the derivation of an important inequality needed in proving the fundamental theorem of algebra Chap. The introduction lo the corn. This was suggested by some users of the last edition. This edition contains many new examples. Boldface leuers have been used to make definitions more easily identified.

The book has lirteen new figures. The derivative off al:: Because f is defined throughout a neighborhood of:: Jim cl I sufficiemly small Fig. Al each nonzero poinl Ir the limit of t MPLE I if ii exists. Hence the limit must be. Herc I Since limits arc unique Sec In that case. Hence if the limil of. Procccdi ng as in Example 2. Example 3 illustrates the following three facts. To sec this. Hence if the limit of b. We defer the development of such interprclations until Chap.

Hence chc basic differentiation rules given below can he derived from lhc dcfinicion in Sec. Thus D. Herc we have used chc fact that g is continuous at the point F Let us derive rule 4. To do this. If the derivatives of two functions f and g exist at a point It is easy to shmv that d d d In scaling such rules.

To find the derivative of I. There is. Then d. With that substitution. U'o Note that in view of the definition of derivative. L'se results in Sec I is differentiable everywhere Prove that expression Derive expression 2.

With the aid of the binomial formula D in Sec. We use the convention 1ha1 the derivative of order i'. I at each nonzero point ilx. Show that if: I 0 of thc author. Starting with the assumption chat f'.

Conclude from these observations that f' 0 does not exist. Compare with Exercise Horiw11tlll llpproaclz In particular. To obtain those condi- tions. This lasl equation enables us lO write.

I 1I Ill. In Exercise I. Thus Cauchy To verify that the Cauchy-Riemann equations arc satisfied everywhere. Yo and urLto. We summarize the above results as follows..

Riemann The lhcorcm in lhc ncxl scclion will. In Example 2. If lhc Cauchy-Riemann cqualions arc lo hold at a poirll. If lhc function f UR Dll: This is illustrated in our next example. L where b. Let the. Thus I b. To prove the theorem.

Yo and b.

Th ell. The assumption! It Xo. Because the Cauchy-Riemann equations arc assumed to be sat is lied at x Bue l. The expression for f':. Consider the function. Herc It also follows from our 1hcorcm that! Since Condition 7 1hus! We saw in Example Suppose tha1 the firsi-order panial derivatives of 11 and l' with respect to.

When using 1hc 1hcorcm in 1his sccrion to find a derivative at a point:. Sec also Example 3. The first-order partial derivatives of 11 and v with respect 10 r and H also have those propenics. In vicv. Depending on whcrhcr we write In view of equations 6 and the expression for I': Then I':. Ec1umions 6 arc. If I i2.

## Brown-Churchill Complex Variables

Inasmuch as Jr The theorem also tells us that I R I.. Leta function f: Thus complete the verification that equations 6. Csc the theorem in Sec. Solve equations 2. Lo dcnotc analytil'ity. If we should speak of a function that is analytic in a set S that is not open. Sec the statement in italics near the end of Sec. Sec Example 3. It is tuwlytic at a poi11t:: Satisfaction of the Cauchy-Riemann equations is also necessary.

In panicular. An e11tire function is a function that is analytic al each point in the entire plane. Thus derive the complex form af. The den vati ves of the sum and product or two functions exist wherever the functions themselves have derivatives. Sufficicm conditions for analyticity in D arc provided by the theorems in Secs.

Other useful sufficient conditions arc obtained from the rules for c.. We start the proof by writing f Because u. I is analytic in D.. The following properly of analytic functions is especially useful. We Jct s denote the distance along L from the point P and let U denote the unit vectoralong Lin the direction of increasing s sec Fig. Assuming that f':. Then the composition gf f Singular poinls will play an imponanl role in our developmenl of complex analysis in chapters lo fol low.

When a function is given in terms of its component functions 11 and v. If a funccion f fails to be analycic al a poinl:. Thal is.. In face. The quotienl:. The analyticity is due to che existence of familiar differentiation rules. Because It is slmightforn: We may conclude. As in Example 3.

For then U.. According to expression 8 in Sec. In view of relations 2. By adding corresponding sides of the first of equations J and This result is needed to obtain an important result later on in Chap. Assuming further that the modulus if: JI is constant throughout D.

Apply the theorem in Sec. According to Example State why a composition of two entire functions is entire. The main result in Example 3 just above thus ensures that So c- f:. It is easy co verify chat chc funccion T x.: Observe that Jm:. Prove that if f It also assumes the values on chc edges of lhc strip chat arc indicaccd in Fig. Observe that Re Let a function f be analytic everywhere in a domain D. Harmonic functions play an imporlalll role in applied mathematics. Jim T x. Then show that the composite function G:.

Cse results in Sec. I 0 and in pans of chaplers follmving iL" Thal lhco11 is based on lhe lheorem below. To shov. Assuming lhal.. Fourier Scric. Diffcrcnliating bolh sides of lhese c4ua1ions wilh resp1xl lox. Hence ils real component which is lhe lcmperaturc funclion T. Prove that these families arc orthogonal. Observe that the curves Since the funclion f:: Let the function f. Show that the same is true of the function t 1 r. The funccion.

Further discussion of harmonic functions related to the theory of functions of a complex variable appears in Chaps. The reader may pass directly co Chap. We let d be the shortest distance from points on L co the boundary or D.

To prove chis lenuna While these seccions arc of considerable theorecical inccrcst. Do Exercise 4 using polar coordinates. Since D is a con11ected open sec Sec. Bul lhc poirll This complclcs lhc proof of lhc lemma. A more general rcsuh. Theorem 3 in Sec. I in Appcnc According co the lemma.. Nole lhal lhcsc neighborhoods arc all comaincd in D and dun lhc cenlcr. If h is the analytic continuation of fi from a domain D 1 into a domain D2.

The theorem here. Exercise 2. We note. Suppose that a. Once we show that the function 2 F. That is To cstahlish the analyt. The theorem tells us. To prove lhc converse in the theorem. Starting with the function. Then show that the function. Because of definition 2 of the funccion F: We al so noted that: According to the theorem in Sec. Cse the theorem in Sec. Hence f. Suppose that f.. Just prior to the statement of the theorem. Stale why the function. Then verify this directly.. Point out how i l follows from the rellcclion principle Sec.

We know from Example I. We sec from this definition that e. Ji when:. This is an exception to the convention Sec. To be specific. Nole chat the diffcrcmiability of e. The cxccnsion 1 5 to complex analysis is easy to vc1ify. Accord- ing to Example l in Sec. In addition co properly 4. According co definition l. Observe hmv property 5 enables us to wrice e.

Show in two ways that the function f: State why the function f:: We recall Sec.. Show that. L'se the Cauchy-Riemann equations and the theorem in Sec. There arc. In facl. This is shown in lhc ncxl scclion. In order lO find numbers Write Re t! Compare with Exercise 4. Show that exp i I in terms of x and y. Prove that jcxp Show that jexp Then show that.

Then use the result in part ll.. Why is this function harmonic in every domain that does not contain the origin'! Describe the behavior of e: Thus 6.

As anticipated. From expression 2 in Sec. Juccs to the usual logarithm in calculus when Observe that log. It rec. Special care must be taken in anticipating that familiar properties of ln.

While slalcmcnl I might he cxpcclcd. It is shown in Exercise The rcquiremelll of analyticity. For if Poinls on the branch cut for F arc singular points Sec. The funclion 6 Log. The origin is cvidcnlly a branch poim for branches of the mulliplc-valucd logarithmic funclion. A hrallch of a mulliple-valucd funclion f is any single-valued function F lhal is analy1ic in some domain al each poinl. Observe lhal for each lixcd ex.. The following example docs show.

In Sec. A reader who wishes lo pass to Sec. Show that log i 2 2 log i when the branch.. Compare this with the example in Sec. I log i: Find al I roots of the equation log ICompare with equation 5. Show that a branch Sec.: Conclude that I og t Suppose that the point: Given that the branch log: The vcrificalion of slatcmcnl I can he based on stalcmcnt 2 in lhe following way.

Show that I. Show that a the function fC To illuslrntc slalcmclll I. I and recall from Examples 2 and 3 in Sec. Why must this function satisfy Laplace's equation when Thm is. In our next example. Noting that Thus where Jr and.. We include here two 01hcr properties of log IL on the other hand. Compare with Example 2 in Sec. Equation 5 is readily veri lied by writing.. Show that all of the values of log: Thal right-hand side is. Do this by writing: Verify expression By choosing specitic nonzero values of: To prove this.

This establishes property 6. Show that for any two nonzero complex numbers: II Thal is. Show that property 6. Thus 7 I exp. Equation I provides a consistelll dcfi nition of:. The de1ivativc of such a bra11clt of:. The otherpropc11y is a differentiation rule for:. Thus show that log: Because of the logarithm. When a specific branch Sec 33 log:. IThe result in Exercise 5. I S11ggeJfio We mention here two other expected properties of lhe power function:.

That yields the result d exp c log:. This wi 11 be illustrntcd in the next section Dcfinilion I is. I log When a value of log c is specified. According lo definition I.

Considcnhc pov. Thcpri11cipal value of This is because the principal value of loge is unity. Py - Equal. In fact. Consider chc nonzero complex numbers: When principal values of lhc powers arc cakcn.

The principal hrnnch of:: On chc ocher hand. Chere arc cxccpcions '''hen certain numbers arc involved.. Prove th. Let c. Show that the result in Exercise 3 could have been obt. I" is to be taken.

Assuming that f': Hence eix For i nstancc sec Exercises 2 and 3. Euler"s formula Sec.. From these. This is.. Knowing the derivatives d i Inasmuch as sinh y tends to infinity as y tends to infinity.

To obtain expressions 13 and Sec the definition of a bounded function at the end of Sec. Observe that once expres- sion 13 is obtained. When y is any real number. I 0 sin:. II is possible chat a function of a real variable can have more zeros when the domain of definition is enlarged. Hence the zeros or sin:: The function f. Since sin:: The other four trigonomcuic functions arc defined in tcnns of the sine and cosine functions hy the expected relations: I tan:: To show that there are no other:: In order to prove this theorem.

One might ask if there arc other zeros in the elllirc plane. But the function f:: As for the cosine function. Give details in the derivation of expressions 2. Then use relations 0.

The periodicity of each of the Lrigonomclric functions defined by equations l and 2 follov. By diffcrenliacing chc righc-hand sides of cquacions l and A reader who wishes al this cimc co learn some of those propc11ies is sufficiently prepared lo read Secs Point out how it follows from expressions 15 and 16 in Sec Csc the rellection principle Sec.

Establish differentiation formulas 3 and 4 in Sec.. Show that a cos i Verify identity 9 in Sec. In Sec I has zero values along the.. With the aid of expression Nole how il follows readily from rclalions 4 and lhc pcriodicily or sin. Then express them in the form:: Because of lhc way in which lhe cxponcnlial function appears in dclinilions I and in lhc definitions Sec 37 e'.

Some of lhc mosl frcqucmly used idcn1i1ics involving hyperbolic sine and cosine runclions arc 5 sinh -. Expressions 14 and 15 now combine lo yield rclmion We prcscnl lhe resulls as a lhcorem in order lo emphasize their imporlancc in laler chapler. We lum now to lhc zeros of sinh: Using rclalions 3 lo replace sin: To illuslrmc lhc melhod of proof jusl suggeslcd.

In fa cl. I COS. Now we alreadv know from rclalion 16 in Sec. Let us verify expression 12 using lhe second of rclalions 4. I 0 cosh: Verify that the derivatives of sinh:: Show that jsinh. Show that a sinh:: Write sinh:.

The functions colh It is slmightforward lo vcri ry the following diffcrcmiation fonnulas. I 6 lanh:: Give detai Is showing that the zeros of sinh:: Show that tanh:: Derive expression I I in Sec. Prove that sin h Derive differentiation formulas Show how identities 6 and 8 in Sec. In order lo define che inverse sine funclion sin By accepting that the stated identity is valid when:. Jlnd all roots of the equation I ti sinh:.

Write its real component as a function of x and y. By using one of the identities 9 and IO in Sec. Compare this exercise with Exercise The derivacives of the firsl two depend on the values chosen for che square roo1s: Since I In. The derivaci ves of chesc chrce fun cl ions arc readily obcai ned from cheir logarichmic expressions.

The derivacivc of chc last one. Solve the equation sin: Derive expression 9. Derive expression 4. It turns out that sinh 1. Derive expression 7. Solve the equation cos: Derive expression 5. The 1heorems are generally concise and pmverful. Various rules learned in calculus..

The derivacive I d u. The cheory of inicgration. Yeriticalions can of1en be based on corresponding rules in calculus. To verify this. F- Then d -lu: Suppose that w I is continuous on an interval a The following example illustrates this. To he precise. Another expected rule li.. I is a complex-valued function of a real variahle t and is written I u: As for the quotient on the right in equation 5.

COS I o' ]: Of course. Thus j''. Hing sums of such functions. Those rules. When bolh 11 and F arc piecewise conlinuous. Such a function is continuous everywhere in lhc stated interval cxccpl possibly for a linilc number of points where.. Sec Exercise 2 d. For an illustration of dclinilion 2. RALll or: Thus special care musl continue lO be used in applying rules from calculus.

To be spcci fie. Lel w I be a corllinuous complex-valued function oft defined on an interval a S I S b. I We recall from Example 3 in Sec.

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Our final example here shows thm the mean value theorem for integral. Since sec Example 2 in Sec. I when a: According 10 definition 2. Evaluate the following integrals: In pan a.. The arc C is a simple arc. Such a curve isposilivel ' orie11ted when it is in the counterclockwise direction. The geometric nature of a particular arc often suggests different notation for the parameter t in equation 2.

C is simple if:: Classes of curves that arc adequate for the study of such integrals arc intnxluccd in this section. This definition establishes a continuous mapping of the interval a: S b into the xy. In each part of this exercise. A set of points:: So is the circle jl 6: The arc ill The arc here differs.: The parametric representation used for any given arc C is.

The points on the arc. The polygonal line Sec.

## Complex Variables and Applications

The set of points is the same. The unit circle ill 5: The same set of points can make up different arcs. The value of L is inva1ianl under cercain changes in the representation for C l. Representation 2 is then lransformcd by equation 9 into I S bin rcprcscntalion 2.

## Complex Variables And Applications 8th Edition Textbook Solutions | billpercompzulbe.ga

This is illustrated in Exercise 3. The arc is then called a differe11tillhle arc. The polygonal line 4 is. I arc che same. I of lhc: I is continuous. Such an arc is said to be smooth. Thus the same length of C would be obtained if representation JO were to be used. The length of a comour or a simple closed contour is chc sum of che lengchs of che smooth arcs chat make up the concour.

The points on any simple closed curve or simple closed contour C arc boundary poi ms of two distinct domains.

If equation 2 represellls a differentiable arc and if:. The ocher. I is inter- preted as a radius vector. Hence if equation 2 represents a concour. These identities can be obtained by noting that they are valid for relll- rnlued functions of r. Examples are che circles 5 and 6. This expression for T is the one learned in calculus when:. In referring to a smooth arc Show that if I represents an arc C that intersects the real axis at the points: Suppose that a function f: Let C denote the right-hand half of the circle I: Write f: Derive the equation of the line through the points a.

Excepc in special cases. We assume that. Suppose chac the equacion I:. Dclinice integrals in calculus can be interpreted as c.

Such an integral is defined in cenns of the values f:. This can be seen by following chc same general procedure chm was used in Sec. J Note chm since C is a contour. The flrsc propeny is 4. We then define the line i nlcgral. Observe char if C has the representation I. By using rcprescnlation 3 and referring to Exercise I b C 1 is represented by S '''here.

There is a value c oft. In order lO cval uale the in leg ml jr. I sec Fig.! We begin here by lclling C denote an arbitrary smooth arc Scc.. According lO ucl1nition 2. To evaluate this imcgral. Then 2. Let us evaluate the contour integral where C 1 is the lop half ill:: Expression 4 is also valid when C is a concour cha! Observe hov. I on this segmcn t. This means thal f:: To sec lhal lhis is so.

The next two examples illustrate this. Jo Ji 0. Using chc principal branch f:. Let C denote the positively oriented unit circle I: R'' 1s where the pos1t1ve. I Log: I and C is the arc from:: HGURE 46 Let C denote the semicircular palh shown in Fig Wilh lhe aid of lhe result in Exercise 3.