Students need encouragement. So if a student gets an answer right, tell them it was a lucky guess. That way, they develop a good, lucky feeling.

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1 Chapter 8 Analytic Functions Stuents nee encouragement. So if a stuent gets an answer right, tell them it was a lucky guess. That way, they evelop a goo, lucky feeling Complex Derivatives -Jack Haney Functions of a Real Variable. The erivative of a function of a real variable is f(x + x) f(x) f(x). x x 0 x If the limit exists then the function is ifferentiable at the point x. Note that x can approach zero from above or below. The limit cannot epen on the irection in which x vanishes. Consier f(x) = x. The function is not ifferentiable at x = 0 since 1 Quote slightly moifie. 0 + x 0 lim x 0 + x = 1 360

2 an 0 + x 0 lim x 0 x = 1. Analyticity. The complex erivative, (or simply erivative if the context is clear), is efine, f(z) z z 0 f(z + z) f(z). z The complex erivative exists if this limit exists. This means that the value of the limit is inepenent of the manner in which z 0. If the complex erivative exists at a point, then we say that the function is complex ifferentiable there. A function of a complex variable is analytic at a point z 0 if the complex erivative exists in a neighborhoo about that point. The function is analytic in an open set if it has a complex erivative at each point in that set. Note that complex ifferentiable has a ifferent meaning than analytic. Analyticity refers to the behavior of a function on an open set. A function can be complex ifferentiable at isolate points, but the function woul not be analytic at those points. Analytic functions are also calle regular or holomorphic. If a function is analytic everywhere in the finite complex plane, it is calle entire. Example Consier z n, n Z +, Is the function ifferentiable? Is it analytic? What is the value of the erivative? We etermine ifferentiability by trying to ifferentiate the function. We use the limit efinition of ifferentiation. We will use Newton s binomial formula to expan (z + z) n. (z + z) n z n z zn z 0 ( z ) z n + nz n 1 z + n(n 1) z n 2 z z n z n 2 z 0 z 0 = nz n 1 ( nz n 1 + z ) n(n 1) z n 2 z + + z n

3 The erivative exists everywhere. The function is analytic in the whole complex plane so it is entire. The value of the erivative is z = nzn 1. Example We will show that f(z) = z is not ifferentiable. Consier its erivative. f(z) z z 0 f(z + z) f(z). z z z z + z z z 0 z z 0 z z First we take z = x an evaluate the limit. Then we take z = ı y. x lim x 0 x = 1 ı y lim y 0 ı y = 1 Since the limit epens on the way that z 0, the function is nowhere ifferentiable. Thus the function is not analytic. Complex Derivatives in Terms of Plane Coorinates. Let z = ζ(ξ, ψ) be a system of coorinates in the complex plane. (For example, we coul have Cartesian coorinates z = ζ(x, y) = x + ıy or polar coorinates z = ζ(r, θ) = r e ıθ ). Let f(z) = φ(ξ, ψ) be a complex-value function. (For example we might have a function in the form φ(x, y) = u(x, y) + ıv(x, y) or φ(r, θ) = R(r, θ) e ıθ(r,θ).) If f(z) = φ(ξ, ψ) is analytic, its complex erivative is 362

4 equal to the erivative in any irection. In particular, it is equal to the erivatives in the coorinate irections. ( ) 1 f z f(z + z) f(z) φ(ξ + ξ, ψ) φ(ξ, ψ) ζ φ ξ 0, ψ=0 z ξ 0 ζ ξ = ξ ξ ξ f z f(z + z) f(z) ξ=0, ψ 0 z φ(ξ, ψ + ψ) φ(ξ, ψ) ψ 0 ζ ψ = ψ ( ζ ψ ) 1 φ ψ Example Consier the Cartesian coorinates z = x + ıy. We write the complex erivative as erivatives in the coorinate irections for f(z) = φ(x, y). ( ) 1 f (x + ıy) z = φ x x = φ x ( ) 1 f (x + ıy) z = φ y y = ı φ y We write this in operator notation. z = x = ı y. Example In Example we showe that z n, n Z +, is an entire function an that z zn = nz n 1. Now we corroborate this by calculating the complex erivative in the Cartesian coorinate irections. z zn = (x + ıy)n x = n(x + ıy) n 1 = nz n 1 z zn = ı (x + ıy)n y = ıın(x + ıy) n 1 = nz n 1 363

5 Complex Derivatives are Not the Same as Partial Derivatives Recall from calculus that f(x, y) = g(s, t) f x = g s s x + g t t x Do not make the mistake of using a similar formula for functions of a complex variable. If f(z) = φ(x, y) then f z φ x x z + φ y y z. This is because the operator means The erivative in any irection in the complex plane. Since f(z) is analytic, z f (z) is the same no matter in which irection we take the erivative. Rules of Differentiation. For an analytic function efine in terms of z we can calculate the complex erivative using all the usual rules of ifferentiation that we know from calculus like the prouct rule, z f(z)g(z) = f (z)g(z) + f(z)g (z), or the chain rule, z f(g(z)) = f (g(z))g (z). This is because the complex erivative erives its properties from properties of limits, just like its real variable counterpart. 364

6 Result The complex erivative is, f(z) z z 0 f(z + z) f(z). z The complex erivative is efine if the limit exists an is inepenent of the manner in which z 0. A function is analytic at a point if the complex erivative exists in a neighborhoo of that point. Let z = ζ(ξ, ψ) efine coorinates in the complex plane. The complex erivative in the coorinate irections is z = ( ) 1 ( ) 1 ζ ζ ξ ξ = ψ ψ. In Cartesian coorinates, this is z = x = ı y. In polar coorinates, this is z = e ıθ r = ı r e ıθ θ Since the complex erivative is efine with the same limit formula as real erivatives, all the rules from the calculus of functions of a real variable may be use to ifferentiate functions of a complex variable. Example We have shown that z n, n Z +, is an entire function. Now we corroborate that z zn = nz n 1 by 365

7 calculating the complex erivative in the polar coorinate irections. z zn = e ıθ r rn e ınθ = e ıθ nr n 1 e ınθ = nr n 1 e ı(n 1)θ = nz n 1 z zn = ı r e ıθ θ rn e ınθ = ı r e ıθ r n ın e ınθ = nr n 1 e ı(n 1)θ = nz n 1 Analytic Functions can be Written in Terms of z. Consier an analytic function expresse in terms of x an y, φ(x, y). We can write φ as a function of z = x + ıy an z = x ıy. ( z + z f (z, z) = φ 2, z z ) ı2 We treat z an z as inepenent variables. We fin the partial erivatives with respect to these variables. z = x z x + y z y = 1 ( 2 x ı ) y z = x z x + y z y = 1 ( 2 x + ı ) y Since φ is analytic, the complex erivatives in the x an y irections are equal. φ x = ı φ y 366

8 The partial erivative of f (z, z) with respect to z is zero. f z = 1 2 ( φ x + ı φ y ) = 0 Thus f (z, z) has no functional epenence on z, it can be written as a function of z alone. If we were consiering an analytic function expresse in polar coorinates φ(r, θ), then we coul write it in Cartesian coorinates with the substitutions: r = x 2 + y 2, θ = arctan(x, y). Thus we coul write φ(r, θ) as a function of z alone. Result Any analytic function φ(x, y) or φ(r, θ) can be written as a function of z alone. 8.2 Cauchy-Riemann Equations If we know that a function is analytic, then we have a convenient way of etermining its complex erivative. We just express the complex erivative in terms of the erivative in a coorinate irection. However, we on t have a nice way of etermining if a function is analytic. The efinition of complex erivative in terms of a limit is cumbersome to work with. In this section we remey this problem. A necessary conition for analyticity. Consier a function f(z) = φ(x, y). If f(z) is analytic, the complex erivative is equal to the erivatives in the coorinate irections. We equate the erivatives in the x an y irections to obtain the Cauchy-Riemann equations in Cartesian coorinates. φ x = ıφ y (8.1) This equation is a necessary conition for the analyticity of f(z). Let φ(x, y) = u(x, y) + ıv(x, y) where u an v are real-value functions. We equate the real an imaginary parts of Equation 8.1 to obtain another form for the Cauchy-Riemann equations in Cartesian coorinates. u x = v y, u y = v x. 367

9 Note that this is a necessary an not a sufficient conition for analyticity of f(z). That is, u an v may satisfy the Cauchy-Riemann equations but f(z) may not be analytic. At this point, Cauchy-Riemann equations give us an easy test for etermining if a function is not analytic. Example In Example we showe that z is not analytic using the efinition of complex ifferentiation. Now we obtain the same result using the Cauchy-Riemann equations. u x = 1, z = x ıy v y = 1 We see that the first Cauchy-Riemann equation is not satisfie; the function is not analytic at any point. A sufficient conition for analyticity. A sufficient conition for f(z) = φ(x, y) to be analytic at a point z 0 = (x 0, y 0 ) is that the partial erivatives of φ(x, y) exist an are continuous in some neighborhoo of z 0 an satisfy the Cauchy-Riemann equations there. If the partial erivatives of φ exist an are continuous then φ(x + x, y + y) = φ(x, y) + xφ x (x, y) + yφ y (x, y) + o( x) + o( y). Here the notation o( x) means terms smaller than x. We calculate the erivative of f(z). f f(z + z) f(z) (z) z 0 z x, y 0 x, y 0 x, y 0 φ(x + x, y + y) φ(x, y) x + ı y φ(x, y) + xφ x (x, y) + yφ y (x, y) + o( x) + o( y) φ(x, y) x + ı y xφ x (x, y) + yφ y (x, y) + o( x) + o( y) x + ı y 368

10 Here we use the Cauchy-Riemann equations. ( x + ı y)φ x (x, y) o( x) + o( y) + lim x, y 0 x + ı y x, y 0 x + ı y = φ x (x, y) Thus we see that the erivative is well efine. Cauchy-Riemann Equations in General Coorinates Let z = ζ(ξ, ψ) be a system of coorinates in the complex plane. Let φ(ξ, ψ) be a function which we write in terms of these coorinates, A necessary conition for analyticity of φ(ξ, ψ) is that the complex erivatives in the coorinate irections exist an are equal. Equating the erivatives in the ξ an ψ irections gives us the Cauchy-Riemann equations. ( ) 1 ( ) 1 ζ φ ζ ξ ξ = φ ψ ψ We coul separate this into two equations by equating the real an imaginary parts or the moulus an argument. 369

11 Result A necessary conition for analyticity of φ(ξ, ψ), where z = ζ(ξ, ψ), at z = z 0 is that the Cauchy-Riemann equations are satisfie in a neighborhoo of z = z 0. ( ) 1 ( ) 1 ζ φ ζ ξ ξ = φ ψ ψ. (We coul equate the real an imaginary parts or the moulus an argument of this to obtain two equations.) A sufficient conition for analyticity of f(z) is that the Cauchy-Riemann equations hol an the first partial erivatives of φ exist an are continuous in a neighborhoo of z = z 0. Below are the Cauchy-Riemann equations for various forms of f(z). f(z) = φ(x, y), φ x = ıφ y f(z) = u(x, y) + ıv(x, y), u x = v y, u y = v x f(z) = φ(r, θ), φ r = ı r φ θ f(z) = u(r, θ) + ıv(r, θ), u r = 1 r v θ, u θ = rv r f(z) = R(r, θ) e ıθ(r,θ), R r = R r Θ θ, 1 r R θ = RΘ r f(z) = R(x, y) e ıθ(x,y), R x = RΘ y, R y = RΘ x Example Consier the Cauchy-Riemann equations for f(z) = u(r, θ) + ıv(r, θ). From Exercise 8.3 we know that the complex erivative in the polar coorinate irections is z = e ıθ r = ı r e ıθ θ. 370

12 From Result we have the equation, e ıθ r [u + ıv] = ı r e ıθ [u + ıv]. θ We multiply by e ıθ an equate the real an imaginary components to obtain the Cauchy-Riemann equations. u r = 1 r v θ, u θ = rv r Example Consier the exponential function. e z = φ(x, y) = e x (cosy + ı sin(y)) We use the Cauchy-Riemann equations to show that the function is entire. φ x = ıφ y e x (cosy + ı sin(y)) = ı e x ( sin y + ı cos(y)) e x (cosy + ı sin(y)) = e x (cosy + ı sin(y)) Since the function satisfies the Cauchy-Riemann equations an the first partial erivatives are continuous everywhere in the finite complex plane, the exponential function is entire. Now we fin the value of the complex erivative. z ez = φ x = ex (cosy + ı sin(y)) = e z The ifferentiability of the exponential function implies the ifferentiability of the trigonometric functions, as they can be written in terms of the exponential. In Exercise 8.13 you can show that the logarithm log z is ifferentiable for z 0. This implies the ifferentiability of z α an the inverse trigonometric functions as they can be written in terms of the logarithm. 371

13 Example We compute the erivative of z z. z (zz ) = log z ez z = (1 + log z) e z log z = (1 + log z)z z = z z + z z log z 8.3 Harmonic Functions A function u is harmonic if its secon partial erivatives exist, are continuous an satisfy Laplace s equation u = 0. 2 (In Cartesian coorinates the Laplacian is u u xx + u yy.) If f(z) = u + ıv is an analytic function then u an v are harmonic functions. To see why this is so, we start with the Cauchy-Riemann equations. u x = v y, u y = v x We ifferentiate the first equation with respect to x an the secon with respect to y. (We assume that u an v are twice continuously ifferentiable. We will see later that they are infinitely ifferentiable.) u xx = v xy, u yy = v yx Thus we see that u is harmonic. One can use the same metho to show that v = 0. u u xx + u yy = v xy v yx = 0 If u is harmonic on some simply-connecte omain, then there exists a harmonic function v such that f(z) = u + ıv is analytic in the omain. v is calle the harmonic conjugate of u. The harmonic conjugate is unique up to an aitive 2 The capital Greek letter is use to enote the Laplacian, like u(x, y), an ifferentials, like x. 372

14 constant. To emonstrate this, let w be another harmonic conjugate of u. Both the pair u an v an the pair u an w satisfy the Cauchy-Riemann equations. We take the ifference of these equations. u x = v y, u y = v x, u x = w y, u y = w x v x w x = 0, v y w y = 0 On a simply connecte omain, the ifference between v an w is thus a constant. To prove the existence of the harmonic conjugate, we first write v as an integral. v(x, y) = v (x 0, y 0 ) + (x,y) (x 0,y 0 ) v x x + v y y On a simply connecte omain, the integral is path inepenent an efines a unique v in terms of v x an v y. We use the Cauchy-Riemann equations to write v in terms of u x an u y. v(x, y) = v (x 0, y 0 ) + (x,y) (x 0,y 0 ) u y x + u x y Changing the starting point (x 0, y 0 ) changes v by an aitive constant. The harmonic conjugate of u to within an aitive constant is v(x, y) = u y x + u x y. This proves the existence 3 of the harmonic conjugate. This is not the formula one woul use to construct the harmonic conjugate of a u. One accomplishes this by solving the Cauchy-Riemann equations. 3 A mathematician returns to his office to fin that a cigarette tosse in the trash has starte a small fire. Being calm an a quick thinker he notes that there is a fire extinguisher by the winow. He then closes the oor an walks away because the solution exists. 373

15 Result If f(z) = u + ıv is an analytic function then u an v are harmonic functions. That is, the Laplacians of u an v vanish u = v = 0. The Laplacian in Cartesian an polar coorinates is = 2 x y 2, = 1 r ( r ) + 1r 2 r r 2 θ 2. Given a harmonic function u in a simply connecte omain, there exists a harmonic function v, (unique up to an aitive constant), such that f(z) = u + ıv is analytic in the omain. One can construct v by solving the Cauchy-Riemann equations. Example Is x 2 the real part of an analytic function? The Laplacian of x 2 is [x 2 ] = x 2 is not harmonic an thus is not the real part of an analytic function. Example Show that u = e x (x sin y y cosy) is harmonic. u x = e x sin y e x (x sin y y cosy) = e x sin y x e x sin y + y e x cosy 2 u x 2 = e x sin y e x sin y + x e x sin y y e x cosy = 2 e x sin y + x e x sin y y e x cosy u y = e x (x cosy cosy + y sin y) 374

16 Thus we see that 2 u x u y 2 = 0 an u is harmonic. 2 u y 2 = e x ( x sin y + siny + y cosy + siny) = x e x sin y + 2 e x sin y + y e x cosy Example Consier u = cos x cosh y. This function is harmonic. u xx + u yy = cosxcosh y + cosxcosh y = 0 Thus it is the real part of an analytic function, f(z). We fin the harmonic conjugate, v, with the Cauchy-Riemann equations. We integrate the first Cauchy-Riemann equation. v y = u x = sin x cosh y v = sin x sinh y + a(x) Here a(x) is a constant of integration. We substitute this into the secon Cauchy-Riemann equation to etermine a(x). Here c is a real constant. Thus the harmonic conjugate is v x = u y cosxsinh y + a (x) = cosxsinh y a (x) = 0 a(x) = c v = sin x sinh y + c. The analytic function is We recognize this as f(z) = cosxcosh y ı sin x sinh y + ıc f(z) = cosz + ıc. 375

17 Example Here we consier an example that emonstrates the nee for a simply connecte omain. Consier u = Log r in the multiply connecte omain, r > 0. u is harmonic. Log r = 1 (r r ) r r Log r + 1r 2 2 θ Log r = 0 2 We solve the Cauchy-Riemann equations to try to fin the harmonic conjugate. u r = 1 r v θ, u θ = rv r v r = 0, v θ = 1 v = θ + c We are able to solve for v, but it is multi-value. Any single-value branch of θ that we choose will not be continuous on the omain. Thus there is no harmonic conjugate of u = Log r for the omain r > 0. If we ha instea consiere the simply-connecte omain r > 0, arg(z) < π then the harmonic conjugate woul be v = Arg(z) + c. The corresponing analytic function is f(z) = Log z + ıc. Example Consier u = x 3 3xy 2 + x. This function is harmonic. u xx + u yy = 6x 6x = 0 Thus it is the real part of an analytic function, f(z). We fin the harmonic conjugate, v, with the Cauchy-Riemann equations. We integrate the first Cauchy-Riemann equation. v y = u x = 3x 2 3y v = 3x 2 y y 3 + y + a(x) Here a(x) is a constant of integration. We substitute this into the secon Cauchy-Riemann equation to etermine a(x). v x = u y 6xy + a (x) = 6xy a (x) = 0 a(x) = c 376

18 Here c is a real constant. The harmonic conjugate is v = 3x 2 y y 3 + y + c. The analytic function is f(z) = x 3 3xy 2 + x + ı ( 3x 2 y y 3 + y ) + ıc f(z) = x 3 + ı3x 2 y 3xy 2 ıy 2 + x + ıy + ıc f(z) = z 3 + z + ıc 8.4 Singularities Any point at which a function is not analytic is calle a singularity. In this section we will classify the ifferent flavors of singularities. Result Singularities. If a function is not analytic at a point, then that point is a singular point or a singularity of the function Categorization of Singularities Branch Points. If f(z) has a branch point at z 0, then we cannot efine a branch of f(z) that is continuous in a neighborhoo of z 0. Continuity is necessary for analyticity. Thus all branch points are singularities. Since function are iscontinuous across branch cuts, all points on a branch cut are singularities. Example Consier f(z) = z 3/2. The origin an infinity are branch points an are thus singularities of f(z). We choose the branch g(z) = z 3. All the points on the negative real axis, incluing the origin, are singularities of g(z). Removable Singularities. 377

19 Example Consier f(z) = sin z z. This function is unefine at z = 0 because f(0) is the ineterminate form 0/0. f(z) is analytic everywhere in the finite complex plane except z = 0. Note that the limit as z 0 of f(z) exists. sin z lim z 0 z z 0 cosz 1 If we were to fill in the hole in the efinition of f(z), we coul make it ifferentiable at z = 0. Consier the function g(z) = = 1 { sin z z z 0, 1 z = 0. We calculate the erivative at z = 0 to verify that g(z) is analytic there. f (0) z 0 f(0) f(z) z z 0 1 sin(z)/z z z 0 z sin(z) z 2 1 cos(z) z 0 2z sin(z) z 0 2 = 0 We call the point at z = 0 a removable singularity of sin(z)/z because we can remove the singularity by efining the value of the function to be its limiting value there. 378

20 Consier a function f(z) that is analytic in a elete neighborhoo of z = z 0. If f(z) is not analytic at z 0, but lim z z0 f(z) exists, then the function has a removable singularity at z 0. The function { f(z) z z 0 g(z) = lim z z0 f(z) z = z 0 is analytic in a neighborhoo of z = z 0. We show this by calculating g (z 0 ). g g (z 0 ) g(z) (z 0 ) z z0 z 0 z g (z) z z0 1 f (z) z z0 This limit exists because f(z) is analytic in a elete neighborhoo of z = z 0. Poles. If a function f(z) behaves like c/ (z z 0 ) n near z = z 0 then the function has an n th orer pole at that point. More mathematically we say lim z z 0 (z z 0 ) n f(z) = c 0. We require the constant c to be nonzero so we know that it is not a pole of lower orer. We can enote a removable singularity as a pole of orer zero. Another way to say that a function has an n th orer pole is that f(z) is not analytic at z = z 0, but (z z 0 ) n f(z) is either analytic or has a removable singularity at that point. Example / sin (z 2 ) has a secon orer pole at z = 0 an first orer poles at z = (nπ) 1/2, n Z ±. lim z 0 z 2 sin (z 2 ) z 0 2z 2z cos (z 2 ) z cos (z 2 ) 4z 2 sin (z 2 ) = 1 379

21 z (nπ) 1/2 lim z (nπ) 1/2 sin (z 2 ) 1 z (nπ) 1/2 2z cos (z 2 ) 1 = 2(nπ) 1/2 ( 1) n Example e 1/z is singular at z = 0. The function is not analytic as lim z 0 e 1/z oes not exist. We check if the function has a pole of orer n at z = 0. lim z 0 zn e 1/z e ζ ζ ζ n e ζ ζ n! Since the limit oes not exist for any value of n, the singularity is not a pole. We coul say that e 1/z is more singular than any power of 1/z. Essential Singularities. If a function f(z) is singular at z = z 0, but the singularity is not a branch point, or a pole, the the point is an essential singularity of the function. The point at infinity. We can consier the point at infinity z by making the change of variables z = 1/ζ an consiering ζ 0. If f(1/ζ) is analytic at ζ = 0 then f(z) is analytic at infinity. We have encountere branch points at infinity before (Section 7.9). Assume that f(z) is not analytic at infinity. If lim z f(z) exists then f(z) has a removable singularity at infinity. If lim z f(z)/z n = c 0 then f(z) has an n th orer pole at infinity. 380

22 Result Categorization of Singularities. Consier a function f(z) that has a singularity at the point z = z 0. Singularities come in four flavors: Branch Points. Branch points of multi-value functions are singularities. Removable Singularities. If lim z z0 f(z) exists, then z 0 is a removable singularity. It is thus name because the singularity coul be remove an thus the function mae analytic at z 0 by reefining the value of f (z 0 ). Poles. If lim z z0 (z z 0 ) n f(z) = const 0 then f(z) has an n th orer pole at z 0. Essential Singularities. Instea of efining what an essential singularity is, we say what it is not. If z 0 neither a branch point, a removable singularity nor a pole, it is an essential singularity. A pole may be calle a non-essential singularity. This is because multiplying the function by an integral power of z z 0 will make the function analytic. Then an essential singularity is a point z 0 such that there oes not exist an n such that (z z 0 ) n f(z) is analytic there Isolate an Non-Isolate Singularities Result Isolate an Non-Isolate Singularities. Suppose f(z) has a singularity at z 0. If there exists a elete neighborhoo of z 0 containing no singularities then the point is an isolate singularity. Otherwise it is a non-isolate singularity. 381

23 If you on t like the abstract notion of a elete neighborhoo, you can work with a elete circular neighborhoo. However, this will require the introuction of more math symbols an a Greek letter. z = z 0 is an isolate singularity if there exists a δ > 0 such that there are no singularities in 0 < z z 0 < δ. Example We classify the singularities of f(z) = z/ sin z. z has a simple zero at z = 0. sin z has simple zeros at z = nπ. Thus f(z) has a removable singularity at z = 0 an has first orer poles at z = nπ for n Z ±. We can corroborate this by taking limits. lim f(z) z 0 z 0 z sin z z 0 1 cosz = 1 (z nπ)z lim (z nπ)f(z) z nπ z nπ sin z z nπ = nπ ( 1) n 0 2z nπ cosz Now to examine the behavior at infinity. There is no neighborhoo of infinity that oes not contain first orer poles of f(z). (Another way of saying this is that there oes not exist an R such that there are no singularities in R < z <.) Thus z = is a non-isolate singularity. We coul also etermine this by setting ζ = 1/z an examining the point ζ = 0. f(1/ζ) has first orer poles at ζ = 1/(nπ) for n Z \ {0}. These first orer poles come arbitrarily close to the point ζ = 0 There is no elete neighborhoo of ζ = 0 which oes not contain singularities. Thus ζ = 0, an hence z = is a non-isolate singularity. The point at infinity is an essential singularity. It is certainly not a branch point or a removable singularity. It is not a pole, because there is no n such that lim z z n f(z) = const 0. z n f(z) has first orer poles in any neighborhoo of infinity, so this limit oes not exist. 382

24 8.5 Application: Potential Flow Example We consier 2 imensional uniform flow in a given irection. The flow correspons to the complex potential Φ(z) = v 0 e ıθ 0 z, where v 0 is the flui spee an θ 0 is the irection. We fin the velocity potential φ an stream function ψ. φ = v 0 (cos(θ 0 )x + sin(θ 0 )y), These are plotte in Figure 8.1 for θ 0 = π/6. Φ(z) = φ + ıψ ψ = v 0 ( sin(θ 0 )x + cos(θ 0 )y) Figure 8.1: The velocity potential φ an stream function ψ for Φ(z) = v 0 e ıθ 0 z. Next we fin the stream lines, ψ = c. v 0 ( sin(θ 0 )x + cos(θ 0 )y) = c c y = v 0 cos(θ 0 ) + tan(θ 0)x 383

25 Figure 8.2: Streamlines for ψ = v 0 ( sin(θ 0 )x + cos(θ 0 )y). Figure 8.2 shows how the streamlines go straight along the θ 0 irection. Next we fin the velocity fiel. v = φ v = φ xˆx + φ y ŷ v = v 0 cos(θ 0 )ˆx + v 0 sin(θ 0 )ŷ The velocity fiel is shown in Figure 8.3. Example Steay, incompressible, invisci, irrotational flow is governe by the Laplace equation. We consier flow aroun an infinite cyliner of raius a. Because the flow oes not vary along the axis of the cyliner, this is a two-imensional problem. The flow correspons to the complex potential ( ) Φ(z) = v 0 z + a2. z 384

26 Figure 8.3: Velocity fiel an velocity irection fiel for φ = v 0 (cos(θ 0 )x + sin(θ 0 )y). We fin the velocity potential φ an stream function ψ. φ = v 0 ( r + a2 r Φ(z) = φ + ıψ ) cosθ, ψ = v 0 (r a2 r ) sin θ These are plotte in Figure

27 ( ) Figure 8.4: The velocity potential φ an stream function ψ for Φ(z) = v 0 z + a2. z Next we fin the stream lines, ψ = c. ( ) v 0 r a2 sin θ = c r r = c ± c 2 + 4v 0 sin 2 θ 2v 0 sin θ Figure 8.5 shows how the streamlines go aroun the cyliner. Next we fin the velocity fiel. 386

28 ( ) Figure 8.5: Streamlines for ψ = v 0 r a2 sin θ. r The velocity fiel is shown in Figure 8.6. v = φ v = φ rˆr + φ θ r ˆθ ( ) ) v = v 0 1 a2 cosθˆr v r 2 0 (1 + a2 sin θˆθ r 2 387