Complex Derivative and Integral

Transcription

1 hapter 19 omplex Derivative and Integral So far we have concerned ourselves with the algebra of the complex numbers. The subject of complex analysis is extremely rich and important. The scope and the level of this book does not allow a comprehensive treatment of complex analysis. Therefore, we shall briefly review some of the more elementary topics and encourage the reader to refer to more advanced books for a more comprehensive treatment. We start here, as is done in real analysis, with the notion of a function omplex Functions Acomplexfunctionf(z) isarulethatassociatesonecomplexnumberto another. We write f(z) =w where both z and w are complex numbers. The function f can be geometrically thought of as acorrespondencebetweentwo complex planes, the z-plane and the w-plane. In the real case, this correspondence can be represented by a graph. It could also be represented by arrows from one real line (the x-axis) to another real line (the y-axis) joining a point of the first real line to the image point of the second real line. When the possibility of graph is available, the second representation of real functions appears prohibitively clumsy! For complex functions, no graph is available, because one cannot draw pictures in four dimensions! 1 Therefore, the second alternative is our only choice. The w-plane has a real axis and an imaginary axis, which we can call u and v, respectively.bothu and v are real functions of the coordinates of z, i.e.,x and y. Therefore,wemaywrite f(z) =u(x, y)+iv(x, y). (19.1) graph of a complex function is impossible to visualize because it lives in a four dimensional space. 1 The graph of a complex function would be a collection of pairs (z,f(z)) just as the graph of a real function is a collection of pairs (x, f(x)). While in the latter case the graph can be drawn in the (x, y) plane,theformerneedsfourdimensionsbecausebothz and f(z) have two components each.

2 498 omplex Derivative and Integral f z (x, y) w (u, v) Figure 19.1: A map from the z-plane to the w-plane. This equation gives a unique point (u, v) inthew-plane for each point (x, y) inthez-plane (see Figure 19.1). Under f, regionsofthez-plane are mapped onto regions of the w-plane. For instance, a curve in the z-plane may be mapped into a curve in the w-plane. Example Let us investigate the behavior of some elementary complex functions. In particular, we shall look at the way a line y = mx in the z-plane is mapped to lines and curves in the w-plane by the action of these functions. (a) Let us start with the simple function w = f(z) =z.wehave w =(x + iy) = x y +ixy with u(x, y) = x y and v(x, y) = xy. For y = mx, these equations yield u =(1 m )x and v =mx. Eliminating x in these equations, we find v = [m/(1 m )]u. This is a line passing through the origin of the w-plane [see Figure 19.(a)]. Note that the angle the line in the w-plane makes with its real axis is twice the angle the line in the z-plane makes with the x-axis. (b) Now let us consider w = f(z) =e z = e x+iy,whichgivesu(x, y) =e x cos y and v(x, y) =e x sin y. Substituting y = mx, weobtainu = e x cos mx and v = e x sin mx. Unlike part (a), we cannot eliminate x to find v as an explicit function of u. Nevertheless, the last pair of equations are the parametric equations of acurve(withx as the parameter) which we can plot in a uv-plane as shown in Figure 19.(b). y α x v α u y (a) v α x u (b) Figure 19.: (a) The map z takes a line with slope angle α and maps it onto a line with twice the angle in the w-plane. (b) The map e z takes the same line and maps it onto a spiral in the w-plane.

3 19.1 omplex Functions Derivatives of omplex Functions Limits of complex functions are defined in terms of absolute values. Thus, lim z a f(z) =w means that, given any real number ϵ >, we can find a corresponding real number δ > suchthat f(z) w < ϵ whenever z a < δ. Similarly, we say that a function f is continuous at z = a if lim z a f(z) = f(a), or if there exist ϵ > andδ > suchthat f(z) f(a) < ϵ whenever z a < δ. The derivative of a complex function is defined as usual: Definition Let f(z) be a complex function. The derivative of f at z is f(z + z) f(z ) dz = lim z z z provided the limit exists and is independent of z. In this definition independent of z meansindependentof x and y (the components of z) and,therefore,independentofthedirectionofap- proach to z.therestrictionsofthisdefinitionapplytotherealcaseaswell. For instance, the derivative of f(x) = x at x =doesnotexistbecauseit approaches +1 from the right and 1 fromtheleft. It can easily be shown that all the formal rules of differentiation that apply to the real case also apply to the complex case. For example, if f and g are differentiable, then f ± g, fg,and aslongasg is not zero f/g are also differentiable, and their derivatives are given by the usual rules of differentiation. Box Afunctionf(z) is called analytic at z if it is differentiable at z and at all other points in some neighborhood of z.apointatwhich f is analytic is called a regular point of f. A point at which f is not analytic is called a singular point or a singularity of f. Afunctionfor which all points in are regular is called an entire function. Example Let us examine the derivative of f(z) =x +iy at z =: f( z) f() x +i y dz = lim = lim z z= z x x + i y. y example illustrating path-dependence of derivative In general, along a line that goes through the origin, y = mx, thelimityields x +im x dz = lim x z= x + im x = 1+im 1+im. This indicates that we get infinitely many values for the derivative depending on the value we assign to m corresponding to different directions of approach to the origin. Thus, the derivative does not exist at z =.

4 5 omplex Derivative and Integral Aquestionarisesnaturallyatthispoint: Underwhatconditionsdoes the limit in the definition of derivative exist? We will find the necessary and sufficient conditions for the existence of that limit. It is clear from the definition that differentiability puts a severe restriction on f(z) becauseit requires the limit to be the same for all paths going through z,thepoint at which the derivative is being calculated. Another important point to keep in mind is that differentiability is a local property. To test whether or not a function f(z) isdifferentiable at z,wemoveawayfromz by a small amount z and check the existence of the limit in Definition For f(z) = u(x, y) + iv(x, y), Definition yields { u(x + x, y + y) u(x,y ) dz = lim z x x + i y y + i v(x } + x, y + y) v(x,y ). x + i y If this limit is to exist for all paths, it must exist for the two particular paths on which y =(paralleltothex-axis) and x =(paralleltothey-axis). For the first path we get dz z u(x + x, y ) u(x,y ) = lim x x v(x + x, y ) v(x,y ) + i lim x x For the second path ( x =),weobtain u(x,y + y) u(x,y ) dz = lim z y i y v(x,y + y) v(x,y ) + i lim y i y = u x = i u y (x,y ) (x,y ) + i v x + v y (x,y ) (x,y ) If f is to be differentiable at z,thederivativesalongthetwopathsmustbe equal. Equating the real and imaginary parts of both sides of this equation and ignoring the subscript z (x, y,orz is arbitrary), we obtain.. auchy Riemann conditions u x = v y and u y = v x. (19.) These two conditions, which are necessary for the differentiability of f, are called the auchy Riemann ( R) conditions. The arguments leading to Equation (19.) imply that the derivative, if it exists, can be expressed as dz = u x + i v x = v y i u y. (19.3) The R conditions assure us that these two equations are equivalent.

5 19.1 omplex Functions 51 Example Let us examine the differentiability of some complex functions. (a) We have already established that f(z) =x +iy is not differentiable at z =. We can now show that it is has no derivative at any point in the complex plane. This is easily seen by noting that u = x and v =y, andthat u/ x =1 v/ y =, and the first R condition is not satisfied. The second R condition is satisfied, but that is not enough. (b) Now consider f(z) =x y +ixy for which u = x y and v =xy. The R conditions become u/ x =x = v/ y and u/ y = y = v/ x. Thus,f(z) may be differentiable. Recall that R conditions are only necessary conditions; we do not know as yet if they are also sufficient. (c) Let u(x, y) =e x cos y and v(x, y) =e x sin y. Then u/ x = e x cos y = v/ y and u/ y = e x sin y = v/ x and the R conditions are satisfied. The requirement of differentiability is very restrictive: the derivative must exist along infinitely many paths. On the other hand, the R conditions seem deceptively mild: they are derived foronlytwopaths. Nevertheless, the two paths are, in fact, true representatives of all paths; that is, the R conditions are not only necessary, but also sufficient. This is the content of the auchy Riemann theorem which we state without proof: Theorem (auchy Riemann Theorem). The function f(z) = u(x, y)+iv(x, y) is differentiable in a region of the complex plane if and only if the auchy Riemann conditions u x = v y and u y = v x are satisfied and all first partial derivatives of u and v are continuous in that region. In that case dz = u x + i v x = v y i u y. The R conditions readily lead to u x + u y =, v x + v =, (19.4) y i.e., both real and imaginary parts of an analytic function satisfy the twodimensional Laplace equation [Equations (15.13) and (15.15)]. Such functions are called harmonic functions. harmonic functions defined Example Let us consider some examples of derivatives of complex functions. (a) f(z) =z. Here u = x and v = y; the R conditions are easily shown to hold, and for For a simple proof, see Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations, Springer-Verlag,1999,hapter9.

6 5 omplex Derivative and Integral any z, wehave /dz = u/ x + i v/ x =1. Therefore,thederivativeexistsatall points of the complex plane, i.e., f(z) = z is entire. (b) f(z) =z. Here u = x y and v =xy; the Rconditionshold,anorallpointsz of the complex plane, we have /dz = u/ x + i v/ x =x + iy =z. Therefore,f(z) is differentiable at all points. So, f(z) =z is also entire. (c) f(z) =z n for n 1. We can use mathematical induction and the fact that the product of two entire functions is an entire function to show that d dz (zn )=nz n 1. (d) f(z) =a + a 1z + + a n 1z n 1 + a nz n, where a i are arbitrary constants. That f(z) isentirefollowsdirectlyfrom(c)and the fact that the sum of two entire functions is entire. (e) f(z) =e z. Here u(x, y) =e x cos y and v(x, y) =e x sin y. Thus, u/ x = e x cos y = v/ y and u/ y = e x sin y = v/ x and the R conditions are satisfied at every point (x, y) ofthexy-plane. Furthermore, dz = u x + i v x = ex cos y + ie x sin y = e x (cos y + i sin y) =e x e iy = e x+iy = e z and e z is entire as well. (f) f(z) =1/z. The derivative can be found to be f (z) = 1/z which does not exist for z =. Thus, z =isasingularityoff(z). However, any other point is a regular point of f. (g) f(z) =1/ sin z. This gives /dz = cos z/ sin z. Thus, f has (infinitely many) singular points at z = ±nπ for n =, 1,,... complex trigonometric functions complex hyperbolic functions Example shows that any polynomial in z, aswellastheexponential function e z is entire. Therefore, any product and/or sum of polynomials and e z will also be entire. We can build other entire functions. For instance, e iz and e iz are entire functions; therefore, the complex trigonometric functions, definedby sin z = eiz e iz i and cos z = eiz + e iz (19.5) are also entire functions. Problem 19.7 shows that sin z and cos z have only real zeros. The complex hyperbolic functions can be defined similarly: sinh z = ez e z and cosh z = ez + e z. (19.6) Although the sum and product of entire functions are entire, the ratio is not. For instance, if f(z) andg(z) arepolynomialsofdegreesm and n, respectively, then for n>, the ratio f(z)/g(z) isnotentire,becauseatthe zeros of g(z) which always exist the derivative is not defined.

7 19.1 omplex Functions 53 The functions u(x, y)andv(x, y)ofananalyticfunction haveaninteresting property which the following example investigates. Example The family of curves u(x, y) = constantisperpendicularto the family of curves v(x, y) =constant ateachpointofthecomplexplanewhere f(z) = u + iv is analytic. This can easily be seen by looking at the normal to the curves. The normal to the curve u(x, y) = constant issimply u = u/ x, u/ y (see Theorem 1.3.). Similarly, the normal to the curve v(x, y) =constant is v = v/ x, v/ y. Taking the dot product of these two normals, we obtain ( u) ( v) = u v x x + u v y y = u ( u ) + u ( ) u = x y y x curves of constant u and v are perpendicular. by the R conditions. Historical Notes One can safely say that rigorous complex analysis was founded by a single man: auchy. Augustin-Louis auchy was one of the most influential French mathematicians of the nineteenth century. He began his career as a military engineer, but when his health broke down in 1813 he followed his natural inclination and devoted himself wholly to mathematics. In mathematical productivity auchy was surpassed only by Euler, andhiscol- lected works fill 7 fat volumes. He made substantial contributions to number theory and determinants; is considered to be the originator of the theory of finite groups; and did extensive work in astronomy, mechanics, optics, and the theory of elasticity. His greatest achievements, however, lay in the field of analysis. Together with his contemporaries Gauss and Abel, he was a pioneer in the rigorous treatment of limits, continuous functions, derivatives, integrals, and infinite series. Several of the basic tests for the convergence of series are associated with his name. He also provided the first existence proof for solutions of differential equations, gave the first proof of the convergence of a Taylor series, and was the first to feel the need for a careful study of the convergence behavior of Fourier series. However, his most important work was in the theory of functions of a complex variable, which in essence he created and which has continued to be one of the dominant branches of both pure and applied mathematics. In this field, auchy s integral theorem and auchy s integral formula are fundamental tools without which modern analysis could hardly exist. Unfortunately, his personality did not harmonize with the fruitful power of his mind. He was an arrogant royalist in politics and a self-righteous, preaching, pious believer in religion all this in an age of republican skepticism and most of his fellow scientists disliked him and considered him a smug hypocrite. It might be fairer to put first things first and describe him as a great mathematician who happened also to be a sincere but narrow-minded bigot. Augustin-Louis auchy Integration of omplex Functions We have thus far discussed the derivative of a complex function. The concept of integration is even more important because, as we shall see later, derivatives can be written in terms of integrals.

8 54 omplex Derivative and Integral complex integrals are path-dependent. The definite integral of a complex function is naively defined in analogy to that of a real function. However, a crucial difference exists: While in the real case, the limits of integration are real numbers and there is only one way to connect these two limits (along the real line), the limits of integration of acomplexfunctionarepointsinthecomplexplaneandthereareinfinitely many ways to connect these two points. Thus, we speak of a definite integral of a complex function along a path. It follows that complex integrals are, in general, path-dependent. α 1 α α 1 f(z) dz = lim N z i lim N z i i=1 N f(z i ) z i, (19.7) where z i is a small segment situated at z i of the curve that connects the complex number α 1 to the complex number α in the z-plane (see Figure 19.3). An immediate consequence of this equation is α f(z) dz = N N f(z i ) z i f(z i ) z i = lim N z i i=1 N f(z i ) z i = i=1 lim N z i α i=1 α 1 f(z) dz, (19.8) where we have used the triangle inequality as expressed in Equation (18.7). Since there are infinitely many ways of connecting α 1 to α,thereisno guarantee that Equation (19.7) has a unique value: It is possible to obtain different values for the integral of some functions for different paths. It may seem that we should avoid such functions and that they will have no use in physical applications. Quite to the contrary, most functions encountered, will not, in general, give the same result if we choose two completely arbitrary paths in the complex plane. In fact, it turns out that the only complex function that gives the same integral for any two arbitrary points connected by any two arbitrary paths is the constant function. Because of the importance of paths in complex integration, we need the following definition: z i α z i α 1 Figure 19.3: One of the infinitely many paths connecting two complex points α 1 and α.

9 19.1 omplex Functions 55 Box A contour is a collection of connected smooth arcs. When the beginning point of the first arc coincides with the end point of the last one, the contour is said to be a simple closed contour (or just closed contour). We encountered path-dependent integrals when we tried to evaluate the line integral of a vector field in hapter 14. The same argument for pathindependence can be used to prove (see Problem 19.1) Theorem (auchy Goursat Theorem). Letf(z) be analytic on a simple closed contour and at all points inside. Then f(z) dz = Equivalently, α α 1 f(z) dz is independent of the smooth path connecting α 1 and α as long as the path lies entirely in the region of analyticity of f(z). Example We consider a few examples of definite integrals. (a) Let us evaluate the integral I 1 = γ 1 zdz where γ 1 is the straight line drawn from the origin to the point (1, ) (see Figure 19.4). Along such a line y =x and thus γ 1(t) =t +it where t 1, and I 1 = zdz= (t +it)(dt+idt) = ( 3tdt +4itdt) = 3 +i. γ 1 For a different path γ,alongwhichy =x,wegetγ (t) =t+it where t 1, and 1 I 1 = zdz= (t +it )(dt +4itdt) = 3 +i. γ Therefore, I 1 = I 1. This is what is expected from the auchy Goursat theorem because the function f(z) =z is analytic on the two paths and in the region bounded by them. y (1, ) γ 1 γ γ 3 x Figure 19.4: The three different paths of integration corresponding to the integrals I 1, I 1, I,andI. 3 We are using the parameterization x = t, y =x =t for the curve.

10 56 omplex Derivative and Integral (b) Now let us consider I = γ 1 z dz with γ 1 as in part (a). Substituting for z in terms of t, weobtain 1 I = (t +it) (dt +idt) =(1+i) 3 t dt = 11 i. 3 3 γ 1 Next we compare I with I = γ 3 z dz where γ 3 is as shown in Figure This path can be described by { t for t 1, γ 3(t) = 1+i(t 1) for 1 t 3. Therefore, I = 1 t dt [1 + i(t 1)] (idt) = i = i, which is identical to I,onceagainbecausethefunctionisanalyticonγ 1 and γ 3 as well as in the region bounded by them. (c) An example of the case where equality for different paths is not attained is I 3 = γ 4 dz/z where γ 4 is the upper semicircle of unit radius, as shown in Figure A parametric equation for γ 4 can be given in terms of θ: γ 4(θ) =cosθ + i sin θ = e iθ dz = ie iθ dθ, θ π. Thus, we obtain On the other hand, I 3 = I 3 = γ 4 π 1 z dz = 1 e iθ ieiθ dθ = iπ. π π. 1 e iθ ieiθ dθ = iπ. Here the two integrals are not equal. From γ 4 and γ 4 we can construct a counterclockwise simple closed contour, along which the integral of f(z) = 1/z becomes dz/z = I3 I 3 =iπ. That the integral is not zero is a consequence of the fact that 1/z is not analytic at all points of the region bounded by the closed contour. γ 4 r = 1 θ γ 4 Figure 19.5: The two semicircular paths for calculating I 3 and I 3.

11 19.1 omplex Functions 57 Γ L (a) (b) Figure 19.6: A contour of integration can be deformed into another contour. The second contour is usually taken to be a circle because of the ease of its corresponding integration. (a) shows the original contour, and (b) shows the two contours as well as the (shaded) region between them in which the function is analytic. The auchy Goursat theorem applies to more complicated regions. When aregioncontainspointsatwhichf(z) isnotanalytic,thosepointscanbe avoided by redefining the region and the contour (Figure 19.6). Such a procedure requires a convention regarding the direction of motion along the contour. This convention is important enough to be stated separately. convention for positive sense of integration around aclosedcontour Box (onvention). When integrating along a closed contour, we agree to traverse the contour in such a way that the region enclosed by the contour lies to our left. An integration that follows this convention is called integration in the positive sense. Integration performed in the opposite direction acquires a minus sign. Suppose that we want to evaluate the integral f(z) dz where is some contour in the complex plane [see Figure 19.6(a)]. Let Γ be another usually simpler, say a circle contour which is either entirely inside or entirely outside. Figure 19.6 illustrates the case where Γ is entirely inside. We assume that Γ is such that f(z) doesnothaveanysingularityintheregion between and Γ. By connecting the two contours with a line as shown in Figure 19.6(b), we construct a composite closed contour consisting of, Γ, and twice the line segment L, onceinthepositivedirectionsandoncein the negative. Within this composite contour, the function f(z) isanalytic. Therefore, by the auchy Goursat theorem, we have f(z) dz + f(z) dz + f(z) dz f(z) dz =. Γ The negative sign for is due to the convention above. It follows from this equation that the integral along is the same as that along the circle Γ. This result can be interpreted by saying that L L

12 58 omplex Derivative and Integral Box We can always deform the contour of an integral in the complex plane into a simpler contour, as long as in the process of deformation we encounter no singularity of the function auchy Integral Formula One extremely important consequence of the auchy Goursat theorem, the centerpiece of complex analysis, is the auchy integral formula which we state without proof. 4 Theorem Let f(z) be analytic on and within a simple closed contour integrated in the positive sense. Let z be any interior point of. Then f(z )= 1 f(z) dz. πi z z This is called the auchy integral formula (IF). Example We can use the IF to evaluate the following integrals: z dz I 1 = (z +3) (z i), I = (z 1) dz (z 1 )(z 4), 3 1 I 3 = 3 e z/ dz (z iπ)(z ) 4, where 1,,and 3 are circles centered at the origin with radii r 1 = 3, r =1, and r 3 =4,respectively. For I 1 we note that f(z) =z /(z +3) is analytic within and on 1,andz = i lies in the interior of 1.Thus, I 1 = f(z)dz 1 z i =πif(i) =πi (i +3) = i π. i why analytic functions remote sense their values at distant points Similarly, f(z) =(z 1)/(z 4) 3 for I is analytic on and within,andz = 1 is an interior point of.thus,theifgives 1 f(z)dz I = =πif( 1 z )=πi 1 4 3π = 1 ( 1 4)3 115 i. 4 For the last integral, f(z) =e z/ /(z ) 4,andtheinteriorpointisz = iπ: f(z)dz e iπ/ I 3 = =πif(iπ) =πi z iπ ( π ) = π 4 (π +). 4 3 The IF gives the value of an analytic function at every point inside a simple closed contour when it is given the value of the function only at points on the contour. It seems as though analytic functions have no freedom within acontour: Theyarenotfreetochangeinsidearegiononcetheirvalueis fixed on the contour enclosing that region. There is an analogous situation in 4 For a proof, see Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations, Springer-Verlag,1999,hapter9.

13 19.1 omplex Functions 59 certain areas of physics, for example, electrostatics: The specification of the potential at the boundaries, such as conductors, automatically determines it at any other point in the region of space bounded by the conductors. This is the content of the uniqueness theorem (to be discussed later in this book) used in electrostatic boundary-value problems. However, the electrostatic potential Φ is bound by another condition, Laplace s equation; and the combination of Laplace s equation and the boundary conditions furnishes the uniqueness of Φ. It seems, on the other hand, as though the merespecificationofananalytic function on a contour, without any other condition, is sufficient to determine the function s value at all points enclosed within that contour. This is not the case. An analytic function, by its very definition, satisfies another restrictive condition: Its real and imaginary parts separately satisfy Laplace s equation in two dimensions! [see Equation (19.4)]. Thus, it should come as no surprise that the value of an analytic function at a boundary (contour) determines the function at all points inside the boundary Derivatives as Integrals The IF is a very powerful tool for working with analytic functions. One of the applications of this formula is in evaluating the derivatives of such functions. It is convenient to change the dummy integration variable to ξ and write the IF as f(z) = 1 f(ξ) dξ πi ξ z, (19.9) where is a simple closed contour in the ξ-plane and z is a point within. By carrying the derivative inside the integral, we get dz = 1 πi d dz f(ξ) dξ ξ z = 1 πi d dz [ ] f(ξ) dξ = 1 ξ z πi f(ξ) dξ (ξ z). By repeated differentiation, we can generalize this formula to the nth derivative, and obtain Theorem The derivatives of all orders of an analytic function f(z) exist in the domain of analyticity of the function and are themselves analytic in that domain. The nth derivative of f(z) is given by f (n) (z) = dn f dz n = n! f(ξ) dξ. (19.1) πi (ξ z) n+1 Example Let us apply Equation (19.1) directly to some simple functions. In all cases, we will assume that the contour is a circle of radius r centered at z. (a) Let f(z) =K, aconstant.then,forn =1wehave dz = 1 Kdξ πi (ξ z).

14 51 omplex Derivative and Integral Since ξ is always on the circle centered at z, ξ z = re iθ and dξ = rie iθ dθ. So we have dz = 1 π Kire iθ dθ = K π e iθ dθ =. πi (re iθ ) πr That is, the derivative of a constant is zero. (b) Given f(z) =z, itsfirstderivativewillbe dz = 1 ξ dξ πi (ξ z) = 1 π (z + re iθ )ire iθ dθ πi (re iθ ) = 1 ( z π π ) e iθ dθ + dθ = 1 ( + π) =1. π r π = 1 π (c) Given f(z) =z,forthefirstderivativeequation(19.1)yields dz = 1 ξ dξ πi (ξ z) = 1 π (z + re iθ ) ire iθ dθ πi (re iθ ) π [ z +(re iθ ) +zre iθ] (re iθ ) 1 dθ = 1 π ( z r π e iθ dθ + r π e iθ dθ +z π ) dθ =z. It can be shown that, in general, (d/dz)z m = mz m 1.TheproofisleftasProblem The IF is a central formula in complex analysis. However, due to space limitations, we cannot explore its full capability here. Nevertheless, one of its applications is worth discussing at this point. Suppose that f is a bounded entire function and consider dz = 1 πi f(ξ) dξ (ξ z). Since f is analytic everywhere in the complex plane, the closed contour can be chosen to be a very large circle of radius R with center at z. Taking the absolute value of both sides yields dz = 1 π 1 π π π f(z + Re iθ ) (Re iθ ) ire iθ dθ f(z + Re iθ ) dθ 1 R π π M R dθ = M R, where we used Equation (19.8) and e iθ =1. M is the maximum of the function in the complex plane. 5 Now as R,thederivativegoestozero. The only function whose derivative is zero is the constant function. Thus Box Aboundedentirefunctionisnecessarilyaconstant. 5 M exists because f is assumed to be bounded.

15 19. Problems 511 There are many interesting and nontrivial real functions that are bounded and have derivatives (of all orders) on the entire real line. For instance, e x is such afunction. Nosuchfreedomexistsforcomplex analytic functions according to Box ! Any nontrivial analytic function is either not bounded (goes to infinity somewhere on the complex plane) or not entire [it is not analytic at some point(s) of the complex plane]. AconsequenceofProposition19.1.5isthefundamental theorem of algebra which states that any polynomial of degree n 1hasn roots (some of which may be repeated). In other words, the polynomial p(x) =a + a 1 x + + a n x n for n 1 any nontrivial function is either unbounded or not entire. fundamental theorem of algebra proved can be factored completely as p(x) =c(x z 1 )(x z )...(x z n )wherec is aconstantandthez i are, in general, complex numbers. To see how Proposition implies the fundamental theorem of algebra, we let f(z) =1/p(z) andassumethecontrary,i.e.,thatp(z) isneverzerofor any (finite) z. Then f(z) isboundedandanalyticforallz, andproposition says that f(z) isaconstant. Thisisobviouslywrong. Thus,theremust be at least one z, sayz = z 1,forwhichp(z) iszero. So,wecanfactorout (z z 1 )fromp(z) andwritep(z) =(z z 1 )q(z) whereq(z) isofdegreen 1. Applying the above argument to q(z), we have p(z) =(z z 1 )(z z )r(z) where r(z) isofdegreen. ontinuing in this way, we can factor p(z) into linear factors. The last polynomial will be a constant (a polynomial of degree zero) which we have denoted as c. 19. Problems Show that f(z) = z maps a line that makes an angle α with the real axis of the z-plane onto a line in the w-plane which makes an angle α with the real axis of the w-plane. Hint: Use the trigonometric identity tan α =tanα/(1 tan α) Show that the function w =1/z maps the straight line y = 1 z-plane onto a circle in the w-plane. in the (a) Using the chain rule, find f/ z and f/ z in terms of partial derivatives with respect to x and y. (b) Evaluate f/ z and f/ z assuming that the R conditions hold (a) Show that, when z is represented by polar coordinates, the R conditions on a function f(z) are U r = 1 V r θ, U θ = r V r, where U and V are the real and imaginary parts of f(z) writteninpolar coordinates.

16 51 omplex Derivative and Integral (b) Show that the derivative of f can be written as ( ) U dz = e iθ r + i V. r Hint: Start with the R conditions in artesian coordinates and apply the chain rule to them using x = r cos θ and y = r sin θ Prove the following identities for differentiation by finding the real and imaginary parts of the function u(x, y) andv(x, y) and differentiating them: (a) d (f + g) = dz dz + dg dz. (c) d ( ) f dz g (b) d (fg)= dz dz g + f dg dz. = f (z)g(z) g (z)f(z) [g(z)], where g(z) Show that d/dz(ln z) = 1/z. Hint: Find u(x, y) and v(x, y) for lnz using the exponential representation of z, then differentiate them Show that sin z and cos z have only real roots. Hint: Use definition of sine and cosine in terms of exponentials Use mathematical induction and the product rule for differentiation to show that d dz (zn )=nz n Use Equations (19.5) and (19.6), to establish the following identities: (a) Re(sin z) =sinx cosh y, Im(sin z) =cosx sinh y. (b) Re(cos z) =cosx cosh y, Im(cos z) = sin x sinh y. (c) Re(sinh z) =sinhx cos y, Im(sinh z) =coshx sin y. (d) Re(cosh z) =coshx cos y, Im(cosh z) =sinhx sin y. (e) sin z =sin x +sinh y, cos z =cos x +sinh y. (f) sinh z =sinh x +sin y, cosh z =sinh x +cos y Find all the zeros of sinh z and cosh z Verify the following trigonometric identities: (a) cos z +sin z =1. (b) cos(z 1 + z )=cosz 1 cos z sin z 1 sin z. (c) sin(z 1 + z )=sinz 1 cos z +cosz 1 sin z. ( π ) ( π ) (d) sin z =cosz, cos z =sinz. (e) cos z =cos z sin z, sin z =sinz cos z. (f) tan(z 1 + z )= tan z 1 +tanz 1 tan z 1 tan z.

17 19. Problems Verify the following hyperbolic identities: (a) cosh z sinh z =1. (b) cosh(z 1 + z )=coshz 1 cosh z +sinhz 1 sinh z. (c) sinh(z 1 + z )=sinz 1 cosh z +coshz 1 sinh z. (d) cosh z =cosh z +sinh z, sinh z =sinhz cosh z. (e) tanh(z 1 + z )= tanh z 1 +tanhz 1+tanhz 1 tanh z Show that ( z ) (a) tanh = Prove the following identities: sinh x + i sin y ( z ) cosh x +cosy. (b) coth sinh x i sin y = cosh x cos y. (a) cos 1 z = i ln(z ± z 1). (b) sin 1 z = i ln[iz ± 1 z )]. (c) tan 1 z = 1 ( ) i z i ln. (d) cosh 1 z =ln(z ± z i + z 1). (e) sinh 1 z =ln(z ± z +1). (f) tanh 1 z = 1 ln ( 1+z 1 z Prove that exp(z )isnotanalyticanywhere Show that e iz =cosz + i sin z for any z Show that both the real and imaginary parts of an analytic function are harmonic Show that each of the following functions call each one u(x, y) is harmonic, and find the function s harmonic partner, v(x, y), such that u(x, y)+iv(x, y) isanalytic.hint:use Rconditions. (a) x 3 3xy. (b) e x cos y. (c) (d) e y cos x. (e) e y x cos xy. ). x x + y where x + y. (f) e x (x cos y y sin y)+sinhy sin x + x 3 3xy + y Describe the curve defined by each of the following equations: (a) z =1 it, t. (b) z = t + it, <t<. π (c) z = a(cos t + i sin t) t 3π. (d) z = t + i <t<. t 19.. Let f(z) =w = u + iv. Supposethat Φ x + Φ y =. Showthatiff is analytic, then Φ u + Φ v =. Thatis,analyticfunctionsmapharmonic functions in the z-plane to harmonic functions in the w-plane.

18 514 omplex Derivative and Integral (a) Show that f(z) dz can be written as A dr + i B dr, where A = u, v,, B = v, u,, anddr = dx, dy,. (b) Show that both A and B have vanishing curls when f is analytic. (c) Now use the Stokes theorem to prove the auchy Goursat theorem Find the value of the integral [(z +)/z] dz, where is: (a) the semicircle z =e iθ,for θ π; (b)thesemicirclez =e iθ,forπ θ π; and (c) the circle z =e iθ,for π θ π Evaluate the integral dz/(z 1 i) whereγ is: (a) the line joining γ z 1 =i and z =3;and(b)thepathfromz 1 to the origin and from there to z Use Equation (19.1) to show that d dz (zm )=mz m 1.Hint:Usethe binomial theorem Let be the boundary of a square whose sides lie along the lines x = ±3 andy = ±3. For the positive sense of integration, evaluate each of the following integrals by using IF or the derivative formula (19.1): e z (a) z iπ/ dz. (b) sinh z (d) z 4 dz. (e) cos z (g) dz. (h) (z iπ/) e z (j) z 5z +4 dz. (k) z (m) (z )(z 1) dz. e z z(z dz. (c) +1) cosh z z 4 dz. (f) e z dz. (i) (z iπ) sinh z dz. (l) (z iπ/) cos z (z π 4 )(z 1) dz. cos z dz. z 3 cos z z + iπ dz. cosh z (z π/) dz.