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 .
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James Brown, Ruel Churchill Complex Variables and Applications.pdf
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Current bid:. Increase your maximum bid:. Upon wriling lhis as I:. Our final example here illustrnlcs lhc JXm1cr of geomclric reasoning in complex analysis when slraighlforward computation can be somcwhm ledious. If a point I This important inequality is geometrically evident in Fig. Th is is ill ustratcd in the following example.
Inequality 2 tells us. Because I. An immediate consequence of the triangle inequality is the facl that 2 To derive inequality 2. IR 11 Geometrically. I The triangle inequality I can be generalized by means of mathematical induction lo sums involving any finite number or terms: This important property of fXllynomials will be used lacer on in Sec.
We shall shm. To give details or the induction proof here. Reduce this inequality to Ix I. Locale lhc numhers:: Ill I:. Statement 6 follmvs immediately from this. Verify inequalities 0. In each case. Verify that This tel Is us th at or laol la1I la2I la.
The \\falter Rudin Student Series in Ad\'anced l\ilathematics
I so la. I" whenever 1: Show that for R sutticiently large. Note that. C sing the fact that I. C sc the tlnal result in Exercise 8 and mathematical induction to show that I. Thal is. Observe that there is a positive number R such that the modulus of each quotient in inequality GATES So lhe conjugale of che sum is che sum of lhe conjugaces: As an illuscracion Hence if: Prove that a: Verify property 9 of moduli in Sec. Verify properties 3 and 4 of conjugates in Sec.. Sketch the set of points determined by the con di ti on al Re: I and recalling that a modulus is never negative.
Brown-Churchill Complex Variables
C sc results in Sec.. Show thac when I: By factoring: Give an alternative prcxlf based on Che corresponding result for real numhers and using idencity 8. C se property 4 of conjugates in Sec. Propcny 8 tells us that 1: I denote real numbers.
As in calculus Each value of H is called an argument of:. Show that the equation I:: With the aid of the results in Exercise The pri11cipal value of arg:. In complex analysis. The real number 8 rcprcsellls the an- gle. The choice of the symhol. According to equation 2. Jr of the principal argument -. The number -I.. Its use in Sec. The complex number. Nocc how expression 4 v. Values of J 11 arc. This can be seen vecto1ially Fig. It is. To he specific. In view of expression l for the product of two nonzero complex numbers in exponential form.
Another important result that can be obtained fonnally hy applying rules for real numbers to:. Expression 4 is now established for all integral powers. It is easily verified for positive values of 11 hy mathematical induction. The following example uses a special case of ic. Sec also Exercises I 0 and I I. In order lo pul. I Equalion 2 is lo be illlcrprclcd as saying lhal if values of lv.
Verification when values of arg:: I and:. Jr Jr 3: Since equation 2 is evidently satisfied when the value arg:: We stan the vcri fication of statemcn l 2 by Jelling - 1 and f-J2 denote any val ucs of arg:. Sec Fig. In order co i lluscracc slmcmerH Scaccmcrll 4 is.
Hence Find the principal argument Arg: Slatcmerll 3 is. C se mathematical induction to show that t! To derive the second identity. Prove that two nonzero complex numbers: As for the first identity. By writing the individual factors on the left in exponential form.. Show that if Re: I0 that two In particular.
M nERS 25 We arc able to sec immediately from this exponential fonn of the rools that they all lie on 1he circle I denote these dislinct rools and write.
This observation. The method starts v.. The number co here can. Observe that if we write expression I for the roots or:: The examples in lhc nexl section serve lo illuslrale this method for hnding roots of complex numbers. Thus when:: I form compare with Example 2 in Sec When JI In order lO determine the 11lh roots of unity. Without any further calculations. They lie at the vertices of a square.
Let a denote any positive real number. Because ci: S from expression 6 and 7. Find the four zeros of the polynomial: According to Sec.. Then use those zeros to factor: Show that if c is any 11th root of unity other than unity itself.
Cse the tirst identity in Exercise 9. Find The lolality of all boundary poinls is called the boundary of S. Some secs arc. For a set S to be not open there must be a boundary poim that is contained in the set. The circle When the value of F: Our basic tool is the concept of an:: Then define: Note that rhc lirsl of sets 3 is open and chm the second is its closure. A set is open if il docs not contain any of ils boundary points. A set is closed if it cofllains all of its boundary points. A boundary poifll is.
I -S l is neither open nor closed. An open set S is co1l11ected if each pair of points z. Note that any neighborhood is a domain. First of all The set of all complex numbers is. Both of the sets 3 arc bounded regions. The open set l.. A domain together with some.
A nonempty open set that is connected is called a domai By completing the square.. Sketch the following sets and determine which arc domains: Note lhal lhc origin is lhc only accumulation poi Ill of lhe scl:: Which sets in Exercise I arc bounded'!
I So inequalily 4 represents the region imcrior lo lhc circle Fig. It is lcfl as an exercise lo show lhat the converse is. Which sets in Exercise I arc neither open nor closed? For if an accumulalion point:: Thus a scl is closed if and only if il conlains all of ils accumulalion poillls.
II A11s. Prove that a linite set of points:: Prove that if a set contains each of its accumulation points. Show that any poi Ill: Determine the accumulation points of each of the following sets: Show that a set Sis open if and only if each point in Sis an interior point.
Let S be the open set consisting of all points:: When chc domain ordclinicion is not mentioned. If f is defined on the set The main goal or che chapter is to introduce analytic funccions.
The SC sis called che domtlill of defi11itio11 or. A real-valued function that is used to illustrate some important concepts lalcr in lhis chapter is In thal case.
If the function v in equation I always has value zcm. Note that the sum here has a finite number of tcnns and that the domain of definition is the entire If 11 is a positive integer and if a 0. These multiple-valueclfunctio11s occur. Polynomials and rational functions constitute elementary. Hence 1 Quotients P A generalization of the concepl of function is a rule that assigns more lhan one value to a point In such cases.
Since zero is the only square roo1 of zero. Propc11ics of a real-valued funciion of a real variable arc oflcn cxhibi1cd by 1hc graph of!
The image of a poi One can.. The i11verse image of a point is the set of all poinls The last case occurs.. We know from Sec. The funciion. When a funciion f is thought of in this way. I 0 that:. Terms such as lra11sll1tion. The image of! To do 1his. The inverse image of a point may con1ain just one poinl. In the next section. It is easy to show.
This form of lhe mapping is especially useful in hnding chc images of cenain hyperbolas. Inasmuch as lhe image of lhe lower branch has paramcuic representation c. To verify chis.. Suppose chal. Hence as 0. This mapping of lhc firsl quadranl onto the upper half plane can also be vc1ified using lhc rays indicated by dashes in Fig.
In view of equalions I. Details of lhc vcrilication arc lefl co Exercise 7. Observe chal points:. Our next example illustrates how polar coordinates can be used in analyzing cercain mappings. The point:. The mapping u: Suppose thal f: Use the expressions sec Sec. For each oft he functions helow.
In part b. Such a lransfonnation maps the entire:. Indicate graphically the vector fields represented hy - ll u: The function assigns a vu: Find and sketch. Sketch the region onto which the sector r By referring to the discussion in Sec.
J sc rays indicated hy dashed half lines in Fig.. Appendix 2.
Complex Variables and Applications
To prove lhis.. Let us show lhal if f: Such a deleted neighborhood. I I2 e'f.. Thus comlilion 2 is salisf1cd by points in the region The next example emphasizes this Ohscrvc lhal when:: To prove che cheorem. Since limits of the latter cypc arc scudicd in calculus. Sttppose thllt. Since a limit is unique. Theorem 1. Bue lu. Suppose that Let us now scart v. Theorem 2. This escahlishes Ii mies I.. Hence it follmvs from inequalities 5 and 6 lhal Corre- sponding vcrificalions of properties 8 and I 0 can be given.
So the real and imaginary componems of the producl. Property 9 is thus cslablishcd. Wo This importanl lhcorcm can be proved dircclly by using lhc dcfinilion of lhc limil of a funclion of a complex variable..
To verify properly To each point:: In the definition of limit in Sec. FIGURE 25 Observe thm che exterior or lhc unil circle centered al the origin in the complex plane crnTcsponds co chc upper hemisphere v. A meaning is now readily given to chc statement lim. By lclling the poinl N or chc sphere correspond co lhc poinc al infinity.: In like manner.
We chus call the scl The poinc P is che poim where the line through:: The sphere is known as che Riem sphere. Lee us agree lhal in referring co a poilll The proof of che following theorem illuscraces how chis is done. The complex plane together with this point is called lhe exte11ded complex pla11e.
To visualize the poim al infinily. Observe that. This means that for each positive number f Suppose now that the second of limits 2 holds. The continuity of the composition g[f: These observations arc di rcct consequences of Theorem 2.
I 3 ofthat ncighborhtxxl under f.: But the continuity off at: Observe that statement 3 actually contains statements I and 2. A function of a complex variable is said to be continuous in a region R if it is continuous at each point in R. A precise statement of this theorem is contained in the proof to follov.
Suppose now that f is continuous at: Our proofs depend on definition 4 of continuity.. If two functions arc continuous at a point. In view of the continuity of g at f: Statement 3 says. We turn now co cwo expected properties of continuous funccions whose verifica- tions arc not so immediate. So if lhere is a poilll:. The proof follows immediately from Theorem I in Sec.
Before staling lhe lheorem. The conlinuity of a function Uf is contins at:: Theorem 3. This lells us that there is a positive number r5 such that I. C sc mathematical induction and property 9. Then use definition 2. Lise detlnition 2. Cse Theorem 2 in Sec. Thus show that the limit off: Show that the function has the value I al all nonzero points on the real and imaginary axes.
Prove statement 8 in Theorem 2 of Sec. Use the theorem in Sec. With the aid of the theorem in Sec. State why limits involving the point at infinity arc unique. Show that a set Sis unbounded Sec. Write il:: Show that Jim f Observe how in1. 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. 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:Let the. Listed in category: Observe that log. Thcpri11cipal value of Chere arc cxccpcions '''hen certain numbers arc involved..
Thus 3 Re:.
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