If brute force isn t working, you re not using enough of it. -Tim Mauch

Transcription

1 Chapter 7 Functions of a Comple Variable If brute force isn t working, ou re not using enough of it. -Tim Mauch In this chapter we introduce the algebra of functions of a comple variable. We will cover the trigonometric and inverse trigonometric functions. The properties of trigonometric functions carr over directl from real-variable theor. However, because of multi-valuedness, the inverse trigonometric functions are significantl trickier than their real-variable counterparts. 7. Curves and Regions In this section we introduce curves and regions in the comple plane. This material is necessar for the stud of branch points in this chapter and later for contour integration. Curves. Consider two continuous functions (t) and (t) defined on the interval t [t..t ]. The set of points in the comple plane, {z(t) = (t) + ı(t) t [t...t ]}, 39

2 defines a continuous curve or simpl a curve. If the endpoints coincide ( z (t ) = z (t ) ) it is a closed curve. (We assume that t t.) If the curve does not intersect itself, then it is said to be a simple curve. If (t) and (t) have continuous derivatives and the derivatives do not both vanish at an point, then it is a smooth curve. This essentiall means that the curve does not have an corners or other nastiness. A continuous curve which is composed of a finite number of smooth curves is called a piecewise smooth curve. We will use the word contour as a snonm for a piecewise smooth curve. See Figure 7. for a smooth curve, a piecewise smooth curve, a simple closed curve and a non-simple closed curve. (a) (b) (c) (d) Figure 7.: (a) Smooth curve. (b) Piecewise smooth curve. (c) Simple closed curve. (d) Non-simple closed curve. Regions. A region R is connected if an two points in R can be connected b a curve which lies entirel in R. A region is simpl-connected if ever closed curve in R can be continuousl shrunk to a point without leaving R. A region which is not simpl-connected is said to be multipl-connected region. Another wa of defining simpl-connected is that a path connecting two points in R can be continuousl deformed into an other path that connects those points. Figure 7. shows a simpl-connected region with two paths which can be continuousl deformed into one another and two multipl-connected regions with paths which cannot be deformed into one another. Jordan curve theorem. A continuous, simple, closed curve is known as a Jordan curve. The Jordan Curve Theorem, which seems intuitivel obvious but is difficult to prove, states that a Jordan curve divides the plane into Wh is it necessar that the derivatives do not both vanish? 4

3 Figure 7.: A simpl-connected and two multipl-connected regions. a simpl-connected, bounded region and an unbounded region. These two regions are called the interior and eterior regions, respectivel. The two regions share the curve as a boundar. Points in the interior are said to be inside the curve; points in the eterior are said to be outside the curve. Traversal of a contour. Consider a Jordan curve. If ou traverse the curve in the positive direction, then the inside is to our left. If ou traverse the curve in the opposite direction, then the outside will be to our left and ou will go around the curve in the negative direction. For circles, the positive direction is the counter-clockwise direction. The positive direction is consistent with the wa angles are measured in a right-handed coordinate sstem, i.e. for a circle centered on the origin, the positive direction is the direction of increasing angle. For an oriented contour C, we denote the contour with opposite orientation as C. Boundar of a region. Consider a simpl-connected region. The boundar of the region is traversed in the positive direction if the region is to the left as ou walk along the contour. For multipl-connected regions, the boundar ma be a set of contours. In this case the boundar is traversed in the positive direction if each of the contours is traversed in the positive direction. When we refer to the boundar of a region we will assume it is given the positive orientation. In Figure 7.3 the boundaries of three regions are traversed in the positive direction. 4

4 Figure 7.3: Traversing the boundar in the positive direction. Two interpretations of a curve. Consider a simple closed curve as depicted in Figure 7.4a. B giving it an orientation, we can make a contour that either encloses the bounded domain Figure 7.4b or the unbounded domain Figure 7.4c. Thus a curve has two interpretations. It can be thought of as enclosing either the points which are inside or the points which are outside. 7. The Point at Infinit and the Stereographic Projection Comple infinit. In real variables, there are onl two was to get to infinit. We can either go up or down the number line. Thus signed infinit makes sense. B going up or down we respectivel approach + and. In the comple plane there are an infinite number of was to approach infinit. We stand at the origin, point ourselves in an direction and go straight. We could walk along the positive real ais and approach infinit via positive real numbers. We could walk along the positive imaginar ais and approach infinit via pure imaginar numbers. We could generalize the real variable notion of signed infinit to a comple variable notion of directional infinit, but this will not be useful A farmer wanted to know the most efficient wa to build a pen to enclose his sheep, so he consulted an engineer, a phsicist and a mathematician. The engineer suggested that he build a circular pen to get the maimum area for an given perimeter. The phsicist suggested that he build a fence at infinit and then shrink it to fit the sheep. The mathematician constructed a little fence around himself and then defined himself to be outside. 4

5 (a) (b) (c) Figure 7.4: Two interpretations of a curve. for our purposes. Instead, we introduce comple infinit or the point at infinit as the limit of going infinitel far along an direction in the comple plane. The comple plane together with the point at infinit form the etended comple plane. Stereographic projection. We can visualize the point at infinit with the stereographic projection. We place a unit sphere on top of the comple plane so that the south pole of the sphere is at the origin. Consider a line passing through the north pole and a point z = + ı in the comple plane. In the stereographic projection, the point point z is mapped to the point where the line intersects the sphere. (See Figure 7.5.) Each point z = + ı in the comple plane is mapped to a unique point (X, Y, Z) on the sphere. X = 4 z + 4, Y = 4 z + 4, Z = z z + 4 The origin is mapped to the south pole. The point at infinit, z =, is mapped to the north pole. In the stereographic projection, circles in the comple plane are mapped to circles on the unit sphere. Figure 7.6 shows circles along the real and imaginar aes under the mapping. Lines in the comple plane are also mapped to circles on the unit sphere. The right diagram in Figure 7.6 shows lines emanating from the origin under the mapping. 43

6 Figure 7.5: The stereographic projection. 44

7 Figure 7.6: The stereographic projection of circles and lines. 45

8 The stereographic projection helps us reason about the point at infinit. When we consider the comple plane b itself, the point at infinit is an abstract notion. We can t draw a picture of the point at infinit. It ma be hard to accept the notion of a jordan curve enclosing the point at infinit. However, in the stereographic projection, the point at infinit is just an ordinar point (namel the north pole of the sphere). 7.3 A Gentle Introduction to Branch Points In this section we will introduce the concepts of branches, branch points and branch cuts. These concepts (which are notoriousl difficult to understand for beginners) are tpicall defined in terms functions of a comple variable. Here we will develop these ideas as the relate to the arctangent function arctan(, ). Hopefull this simple eample will make the treatment in Section 7.9 more palateable. First we review some properties of the arctangent. It is a mapping from R to R. It measures the angle around the origin from the positive ais. Thus it is a multi-valued function. For a fied point in the domain, the function values differ b integer multiples of π. The arctangent is not defined at the origin nor at the point at infinit; it is singular at these two points. If we plot some of the values of the arctangent, it looks like a corkscrew with ais through the origin. A portion of this function is plotted in Figure 7.7. Most of the tools we have for analzing functions (continuit, differentiabilit, etc.) depend on the fact that the function is single-valued. In order to work with the arctangent we need to select a portion to obtain a single-valued function. Consider the domain (..) (..4). On this domain we select the value of the arctangent that is between and π. The domain and a plot of the selected values of the arctangent are shown in Figure 7.8. CONTINUE. 7.4 Cartesian and Modulus-Argument Form We can write a function of a comple variable z as a function of and or as a function of r and θ with the substitutions z = + ı and z = r e ıθ, respectivel. Then we can separate the real and imaginar components or write the function in modulus-argument form, f(z) = u(, ) + ıv(, ), or f(z) = u(r, θ) + ıv(r, θ), 46

9 Figure 7.7: Plots of R(log z) and a portion of I(log z) Figure 7.8: A domain and a selected value of the arctangent for the points in the domain. 47

10 f(z) = ρ(, ) e ıφ(,), or f(z) = ρ(r, θ) e ıφ(r,θ). Eample 7.4. Consider the functions f(z) = z, f(z) = z 3 and f(z) = and and separate them into their real and imaginar components. f(z) = z = + ı z. We write the functions in terms of f(z) = z 3 = ( + ı) 3 = 3 + ı ı 3 = ( 3 ) + ı ( 3) f(z) = z = ı = ı = + ı + ı ( ) + + ı ( ) + Eample 7.4. Consider the functions f(z) = z, f(z) = z 3 and f(z) = and θ and write them in modulus-argument form. f(z) = z = r e ıθ z. We write the functions in terms of r 48

11 f(z) = z = r e ıθ = r e ıθ r e ıθ r e ıθ = r e ıθ r e ıθ +r r cosθ + ır sin θ = r cosθ + r Note that the denominator is real and non-negative. = = = = f(z) = z 3 = ( r e ıθ) 3 = r 3 e ı3θ cos θ,r sin θ) r cosθ + ır sin θ eıarctan( r r cosθ + r ( r cosθ) r cosθ + r + r sin ıarctan( r cos θ,r sin θ) θ e r cosθ + r r cosθ + r cos θ + r sin ıarctan( r cos θ,r sin θ) θ e cos θ,r sin θ) eıarctan( r r cosθ + r 7.5 Graphing Functions of a Comple Variable We cannot directl graph functions of a comple variable as the are mappings from R to R. To do so would require four dimensions. However, we can can use a surface plot to graph the real part, the imaginar part, the modulus or the 49

12 argument of a function of a comple variable. Each of these are scalar fields, mappings from R to R. Eample 7.5. Consider the identit function, f(z) = z. In Cartesian coordinates and Cartesian form, the function is f(z) = + ı. The real and imaginar components are u(, ) = and v(, ) =. (See Figure 7.9.) In modulus argument form the function is Figure 7.9: The real and imaginar parts of f(z) = z = + ı. f(z) = z = r e ıθ = + e ıarctan(,). The modulus of f(z) is a single-valued function which is the distance from the origin. The argument of f(z) is a multivalued function. Recall that arctan(, ) has an infinite number of values each of which differ b an integer multiple of π. A few branches of arg(f(z)) are plotted in Figure 7.. The modulus and principal argument of f(z) = z are plotted in Figure 7.. Eample 7.5. Consider the function f(z) = z. In Cartesian coordinates and separated into its real and imaginar components the function is f(z) = z = ( + ı) = ( ) + ı. Figure 7. shows surface plots of the real and imaginar parts of z. The magnitude of z is z = z z = zz = ( + ı)( ı) = +. 5

13 Figure 7.: A few branches of arg(z) Figure 7.: Plots of z and Arg(z). 5

14 Figure 7.: Plots of R (z ) and I (z ). Note that In Figure 7.3 are plots of z and a branch of arg (z ). z = ( r e ıθ) = r e ıθ. 7.6 Trigonometric Functions The eponential function. Consider the eponential function e z. We can use Euler s formula to write e z = e +ı in terms of its real and imaginar parts. e z = e +ı = e e ı = e cos + ı e sin From this we see that the eponential function is ıπ periodic: e z+ıπ = e z, and ıπ odd periodic: e z+ıπ = e z. Figure 7.4 has surface plots of the real and imaginar parts of e z which show this periodicit. The modulus of e z is a function of alone. e z = e +ı = e 5

15 Figure 7.3: Plots of z and a branch of arg (z ) Figure 7.4: Plots of R (e z ) and I (e z ). 53

16 The argument of e z is a function of alone. In Figure 7.5 are plots of e z and a branch of arg (e z ). arg (e z ) = arg ( e +ı) = { + πn n Z} Figure 7.5: Plots of e z and a branch of arg (e z ). Eample 7.6. Show that the transformation w = e z maps the infinite strip, < <, < < π, onto the upper half-plane. Method. Consider the line z = + ıc, < <. Under the transformation, this is mapped to w = e +ıc = e ıc e, < <. This is a ra from the origin to infinit in the direction of e ıc. Thus we see that z = is mapped to the positive, real w ais, z = + ıπ is mapped to the negative, real ais, and z = + ıc, < c < π is mapped to a ra with angle c in the upper half-plane. Thus the strip is mapped to the upper half-plane. See Figure 7.6. Method. Consider the line z = c + ı, < < π. Under the transformation, this is mapped to w = e c+ı + e c e ı, < < π. 54

17 Figure 7.6: e z maps horizontal lines to ras. This is a semi-circle in the upper half-plane of radius e c. As c, the radius goes to zero. As c, the radius goes to infinit. Thus the strip is mapped to the upper half-plane. See Figure Figure 7.7: e z maps vertical lines to circular arcs. 55

18 The sine and cosine. We can write the sine and cosine in terms of the eponential function. e ız + e ız cos(z) + ı sin(z) + cos( z) + ı sin( z) = cos(z) + ı sin(z) + cos(z) ı sin(z) = = cosz e ız e ız ı cos(z) + ı sin(z) cos( z) ı sin( z) = cos(z) + ı sin(z) cos(z) + ı sin(z) = = sinz We separate the sine and cosine into their real and imaginar parts. cosz = coscosh ı sin sinh sin z = sincosh + ı cossinh For fied, the sine and cosine are oscillator in. The amplitude of the oscillations grows with increasing. See Figure 7.8 and Figure 7.9 for plots of the real and imaginar parts of the cosine and sine, respectivel. Figure 7. shows the modulus of the cosine and the sine. The hperbolic sine and cosine. The hperbolic sine and cosine have the familiar definitions in terms of the eponential function. Thus not surprisingl, we can write the sine in terms of the hperbolic sine and write the cosine in terms of the hperbolic cosine. Below is a collection of trigonometric identities. 56

19 Figure 7.8: Plots of R(cos(z)) and I(cos(z)) Figure 7.9: Plots of R(sin(z)) and I(sin(z)). 57

20 Figure 7.: Plots of cos(z) and sin(z). Result 7.6. e z = e (cos + ı sin ) cosz = eız + e ız sin z = eız e ız ı cosz = coscosh ı sin sinh sin z = sincosh + ı cossinh cosh z = ez + e z sinh z = ez e z cosh z = cosh cos + ı sinh sin sinh z = sinhcos + ı cosh sin sin(ız) = ı sinh z sinh(ız) = ı sin z cos(ız) = coshz cosh(ız) = cosz log z = ln z + ı arg(z) = ln z + ı Arg(z) + ıπn, n Z 58

21 7.7 Inverse Trigonometric Functions The logarithm. The logarithm, log(z), is defined as the inverse of the eponential function e z. The eponential function is man-to-one and thus has a multi-valued inverse. From what we know of man-to-one functions, we conclude that e log z = z, but log (e z ) z. This is because e log z is single-valued but log (e z ) is not. Because e z is ıπ periodic, the logarithm of a number is a set of numbers which differ b integer multiples of ıπ. For instance, e ıπn = so that log() = {ıπn : n Z}. The logarithmic function has an infinite number of branches. The value of the function on the branches differs b integer multiples of ıπ. It has singularities at zero and infinit. log(z) as either z or z. We will derive the formula for the comple variable logarithm. For now, let ln() denote the real variable logarithm that is defined for positive real numbers. Consider w = log z. This means that e w = z. We write w = u + ıv in Cartesian form and z = r e ıθ in polar form. e u+ıv = r e ıθ We equate the modulus and argument of this epression. e u = r u = ln r v = θ + πn v = θ + πn With log z = u + ıv, we have a formula for the logarithm. log z = ln z + ı arg(z) If we write out the multi-valuedness of the argument function we note that this has the form that we epected. log z = ln z + ı(arg(z) + πn), n Z We check that our formula is correct b showing that e log z = z e log z = e ln z +ıarg(z) = e ln r+ıθ+ıπn = r e ıθ = z 59

22 Note again that log (e z ) z. log (e z ) = ln e z + ı arg (e z ) = ln (e ) + ı arg ( e +ı) = + ı( + πn) = z + ınπ z The real part of the logarithm is the single-valued ln r; the imaginar part is the multi-valued arg(z). We define the principal branch of the logarithm Log z to be the branch that satisfies π < I(Log z) π. For positive, real numbers the principal branch, Log is real-valued. We can write Log z in terms of the principal argument, Arg z. Log z = ln z + ı Arg(z) See Figure 7. for plots of the real and imaginar part of Log z Figure 7.: Plots of R(Log z) and I(Log z). The form: a b. Consider a b where a and b are comple and a is nonzero. We define this epression in terms of the eponential and the logarithm as a b = e blog a. 6

23 Note that the multi-valuedness of the logarithm ma make a b multi-valued. First consider the case that the eponent is an integer. a m = e mlog a = e m(log a+ınπ) = e mlog a e ımnπ = e mlog a Thus we see that a m has a single value where m is an integer. Now consider the case that the eponent is a rational number. Let p/q be a rational number in reduced form. This epression has q distinct values as a p/q = e p q log a = e p q (Log a+ınπ) = e p q Log a e ınpπ/q. e ınpπ/q = e ımpπ/q if and onl if n = m mod q. Finall consider the case that the eponent b is an irrational number. a b = e blog a = e b(log a+ınπ) = e blog a e ıbnπ Note that e ıbnπ and e ıbmπ are equal if and onl if ıbnπ and ıbmπ differ b an integer multiple of ıπ, which means that bn and bm differ b an integer. This occurs onl when n = m. Thus e ıbnπ has a distinct value for each different integer n. We conclude that a b has an infinite number of values. You ma have noticed something a little fish. If b is not an integer and a is an non-zero comple number, then a b is multi-valued. Then wh have we been treating e b as single-valued, when it is merel the case a = e? The answer is that in the realm of functions of a comple variable, e z is an abuse of notation. We write e z when we mean ep(z), the single-valued eponential function. Thus when we write e z we do not mean the number e raised to the z power, we mean the eponential function of z. We denote the former scenario as (e) z, which is multi-valued. Logarithmic identities. Back in high school trigonometr when ou thought that the logarithm was onl defined for positive real numbers ou learned the identit log a = a log. This identit doesn t hold when the logarithm is defined for nonzero comple numbers. Consider the logarithm of z a. log z a = Log z a + ıπn 6

24 Note that Furthermore, since a log z = a(log z + ıπn) = a Log z + ıaπn Log z a = ln z a + ı Arg (z a ), log z a a log z and Arg (z a ) is not necessaril the same as a Arg(z) we see that Consider the logarithm of a product. Log z a a Log z. log(ab) = ln ab + ı arg(ab) a Log z = a ln z + ıa Arg(z) = ln a + ln b + ı arg(a) + ı arg(b) = log a + log b There is not an analogous identit for the principal branch of the logarithm since Arg(ab) is not in general the same as Arg(a) + Arg(b). Using log(ab) = log(a) + log(b) we can deduce that log (a n ) = n k= log a = n log a, where n is a positive integer. This result is simple, straightforward and wrong. I have led ou down the merr path to damnation. 3 In fact, log (a ) log a. Just write the multi-valuedness eplicitl, log ( a ) = Log ( a ) + ınπ, log a = (Log a + ınπ) = Log a + ı4nπ. You can verif that ( ) log = log a. a We can use this and the product identit to epand the logarithm of a quotient. ( a log = log a log b b) 3 Don t feel bad if ou fell for it. The logarithm is a trick bastard. 6

25 For general values of a, log z a a log z. However, for some values of a, equalit holds. We alread know that a = and a = work. To determine if equalit holds for other values of a, we eplicitl write the multi-valuedness. log z a = log ( e a log z) = a log z + ıπk, a log z = a ln z + ıa Arg z + ıaπm, k Z m Z We see that log z a = a log z if and onl if {am m Z} = {am + k k,m Z}. The sets are equal if and onl if a = /n, n Z ±. Thus we have the identit: log ( z /n) = log z, n Z± n 63

26 Result 7.7. Logarithmic Identities. a b = e b log a e log z = e Log z = z log(ab) = log a + log b log(/a) = log a log(a/b) = log a log b ) log (z /n = log z, n Z± n Logarithmic Inequalities. Log(uv) Log(u) + Log(v) log z a a log z Log z a a Log z log e z z Eample 7.7. Consider π. We appl the definition a b = e blog a. π = e π log() = e π(ln()+ınπ) = e ınπ Thus we see that π has an infinite number of values, all of which lie on the unit circle z = in the comple plane. However, the set π is not equal to the set z =. There are points in the latter which are not in the former. This is analogous to the fact that the rational numbers are dense in the real numbers, but are a subset of the real numbers. 64

27 Eample 7.7. We find the zeros of sin z. sin z = eız e ız = ı e ız = e ız e ız = z mod π = z = nπ, n Z Equivalentl, we could use the identit sin z = sincosh + ı cossinh =. This becomes the two equations (for the real and imaginar parts) sin cosh = and cossinh =. Since cosh is real-valued and positive for real argument, the first equation dictates that = nπ, n Z. Since cos(nπ) = ( ) n for n Z, the second equation implies that sinh =. For real argument, sinh is onl zero at =. Thus the zeros are z = nπ, n Z Eample Since we can epress sin z in terms of the eponential function, one would epect that we could epress 65

28 the sin z in terms of the logarithm. w = sin z z = sinw z = eıw e ıw ı e ıw ız e ıw = e ıw = ız ± z w = ı log (ız ± ) z Thus we see how the multi-valued sin is related to the logarithm. sin z = ı log (ız ± ) z Eample Consider the equation sin 3 z =. sin 3 z = sin z = /3 e ız e ız = /3 ı e ız ı() /3 e ız = e ız ı() /3 e ız = e ız = ı()/3 ± 4() /3 + 4 e ız = ı() /3 ± () /3 ( ) z = ı log ı() /3 ± /3 66

29 Note that there are three sources of multi-valuedness in the epression for z. The two values of the square root are shown eplicitl. There are three cube roots of unit. Finall, the logarithm has an infinite number of branches. To show this multi-valuedness eplicitl, we could write ( ) z = ı Log ı e ımπ/3 ± e ı4mπ/3 + πn, m =,,, n =...,,,,... Eample Consider the harmless looking equation, ı z =. Before we start with the algebra, note that the right side of the equation is a single number. ı z is single-valued onl when z is an integer. Thus we know that if there are solutions for z, the are integers. We now proceed to solve the equation. ı z = ( e ıπ/ ) z = Use the fact that z is an integer. ıπz/ = ınπ, e ıπz/ = z = 4n, for some n Z n Z Here is a different approach. We write down the multi-valued form of ı z. We solve the equation b requiring that all the values of ı z are. ı z = e z log ı = z log ı = ıπn, for some n Z z (ı π ) + ıπm = ıπn, m Z, for some n Z ı π z + ıπmz = ıπn, m Z, for some n Z 67

30 The onl solutions that satisf the above equation are z = 4k, k Z. Now let s consider a slightl different problem: ı z. For what values of z does ı z have as one of its values. ı z e z log ı {e z(ıπ/+ıπn) n Z} z(ıπ/ + ıπn) = ıπm, z = 4m + 4n, m, n Z m, n Z There are an infinite set of rational numbers for which ı z has as one of its values. For eample, ı 4/5 = /5 = {, e ıπ/5, e ı4π/5, e ı6π/5, e ı8π/5} 7.8 Riemann Surfaces Consider the mapping w = log(z). Each nonzero point in the z-plane is mapped to an infinite number of points in the w plane. w = {ln z + ı arg(z)} = {ln z + ı(arg(z) + πn) n Z} This multi-valuedness makes it hard to work with the logarithm. We would like to select one of the branches of the logarithm. One wa of doing this is to decompose the z-plane into an infinite number of sheets. The sheets lie above one another and are labeled with the integers, n Z. (See Figure 7..) We label the point z on the n th sheet as (z, n). Now each point (z, n) maps to a single point in the w-plane. For instance, we can make the zeroth sheet map to the principal branch of the logarithm. This would give us the following mapping. log(z, n) = Log z + ıπn 68

31 - - Figure 7.: The z-plane decomposed into flat sheets. This is a nice idea, but it has some problems. The mappings are not continuous. Consider the mapping on the zeroth sheet. As we approach the negative real ais from above z is mapped to ln z + ıπ as we approach from below it is mapped to ln z ıπ. (Recall Figure 7..) The mapping is not continuous across the negative real ais. Let s go back to the regular z-plane for a moment. We start at the point z = and selecting the branch of the logarithm that maps to zero. (log() = ıπn). We make the logarithm var continuousl as we walk around the origin once in the positive direction and return to the point z =. Since the argument of z has increased b π, the value of the logarithm has changed to ıπ. If we walk around the origin again we will have log() = ı4π. Our flat sheet decomposition of the z-plane does not reflect this propert. We need a decomposition with a geometr that makes the mapping continuous and connects the various branches of the logarithm. Drawing inspiration from the plot of arg(z), Figure 7., we decompose the z-plane into an infinite corkscrew with ais at the origin. (See Figure 7.3.) We define the mapping so that the logarithm varies continuousl on this surface. Consider a point z on one of the sheets. The value of the logarithm at that same point on the sheet directl above it is ıπ more than the original value. We call this surface, the Riemann surface for the logarithm. The mapping from the Riemann surface to the w-plane is continuous and one-to-one. 69

32 7.9 Branch Points Figure 7.3: The Riemann surface for the logarithm. Eample 7.9. Consider the function z /. For each value of z, there are two values of z /. We write z / in modulus-argument and Cartesian form. z / = z e ıarg(z)/ z / = z cos(arg(z)/) + ı z sin(arg(z)/) Figure 7.4 shows the real and imaginar parts of z / from three different viewpoints. The second and third views are looking down the ais and ais, respectivel. Consider R ( z /). This is a double laered sheet which intersects itself on the negative real ais. (I(z / ) has a similar structure, but intersects itself on the positive real ais.) Let s start at a point on the positive real ais on the lower sheet. If we walk around the origin once and return to the positive real ais, we will be on the upper sheet. If we do this again, we will return to the lower sheet. Suppose we are at a point in the comple plane. We pick one of the two values of z /. If the function varies continuousl as we walk around the origin and back to our starting point, the value of z / will have changed. We will 7

33 be on the other branch. Because walking around the point z = takes us to a different branch of the function, we refer to z = as a branch point. Now consider the modulus-argument form of z / : z / = z e ıarg(z)/. Figure 7.5 shows the modulus and the principal argument of z /. We see that each time we walk around the origin, the argument of z / changes b π. This means that the value of the function changes b the factor e ıπ =, i.e. the function changes sign. If we walk around the origin twice, the argument changes b π, so that the value of the function does not change, e ıπ =. z / is a continuous function ecept at z =. Suppose we start at z = = e ı and the function value (e ı ) / =. If we follow the first path in Figure 7.6, the argument of z varies from up to about π, down to about 4 π and back 4 to. The value of the function is still (e ı ) /. Now suppose we follow a circular path around the origin in the positive, counter-clockwise, direction. (See the second path in Figure 7.6.) The argument of z increases b π. The value of the function at half turns on the path is ( e ı ) / =, (e ıπ ) / = e ıπ/ = ı, ( e ıπ ) / = e ıπ = As we return to the point z =, the argument of the function has changed b π and the value of the function has changed from to. If we were to walk along the circular path again, the argument of z would increase b another π. The argument of the function would increase b another π and the value of the function would return to. ( e ı4π ) / = e ıπ = In general, an time we walk around the origin, the value of z / changes b the factor. We call z = a branch point. If we want a single-valued square root, we need something to prevent us from walking around the origin. We achieve this b introducing a branch cut. Suppose we have the comple plane drawn on an infinite sheet of paper. With a scissors we cut the paper from the origin to along the real ais. Then if we start at z = e ı, and draw a 7

34 Figure 7.4: Plots of R ( z /) (left) and I ( z /) (right) from three viewpoints. 7

35 Figure 7.5: Plots of z / and Arg ( z /). Im(z) Im(z) Re(z) Re(z) Figure 7.6: A path that does not encircle the origin and a path around the origin. 73

36 continuous line without leaving the paper, the argument of z will alwas be in the range π < arg z < π. This means that π < arg ( z /) < π. No matter what path we follow in this cut plane, z = has argument zero and ()/ =. B never crossing the negative real ais, we have constructed a single valued branch of the square root function. We call the cut along the negative real ais a branch cut. Eample 7.9. Consider the logarithmic function log z. For each value of z, there are an infinite number of values of log z. We write log z in Cartesian form. log z = ln z + ı arg z Figure 7.7 shows the real and imaginar parts of the logarithm. The real part is single-valued. The imaginar part is multi-valued and has an infinite number of branches. The values of the logarithm form an infinite-laered sheet. If we start on one of the sheets and walk around the origin once in the positive direction, then the value of the logarithm increases b ıπ and we move to the net branch. z = is a branch point of the logarithm Figure 7.7: Plots of R(log z) and a portion of I(log z). The logarithm is a continuous function ecept at z =. Suppose we start at z = = e ı and the function value log (e ı ) = ln() + ı =. If we follow the first path in Figure 7.6, the argument of z and thus the imaginar part of the logarithm varies from up to about π, down to about 4 π and back to. The value of the logarithm is still. 4 74

37 Now suppose we follow a circular path around the origin in the positive direction. (See the second path in Figure 7.6.) The argument of z increases b π. The value of the logarithm at half turns on the path is log ( e ı) =, log (e ıπ ) = ıπ, log ( e ıπ) = ıπ As we return to the point z =, the value of the logarithm has changed b ıπ. If we were to walk along the circular path again, the argument of z would increase b another π and the value of the logarithm would increase b another ıπ. Result 7.9. A point z is a branch point of a function f(z) if the function changes value when ou walk around the point on an path that encloses no singularities other than the one at z = z. Branch points at infinit : mapping infinit to the origin. Up to this point we have considered onl branch points in the finite plane. Now we consider the possibilit of a branch point at infinit. As a first method of approaching this problem we map the point at infinit to the origin with the transformation ζ = /z and eamine the point ζ =. Eample Again consider the function z /. Mapping the point at infinit to the origin, we have f(ζ) = (/ζ) / = ζ /. For each value of ζ, there are two values of ζ /. We write ζ / in modulus-argument form. ζ / = ζ e ıarg(ζ)/ Like z /, ζ / has a double-laered sheet of values. Figure 7.8 shows the modulus and the principal argument of ζ /. We see that each time we walk around the origin, the argument of ζ / changes b π. This means that the value of the function changes b the factor e ıπ =, i.e. the function changes sign. If we walk around the origin twice, the argument changes b π, so that the value of the function does not change, e ıπ =. Since ζ / has a branch point at zero, we conclude that z / has a branch point at infinit. 75

38 Figure 7.8: Plots of ζ / and Arg ( ζ /). Eample Again consider the logarithmic function log z. Mapping the point at infinit to the origin, we have f(ζ) = log(/ζ) = log(ζ). From Eample 7.9. we known that log(ζ) has a branch point at ζ =. Thus log z has a branch point at infinit. Branch points at infinit : paths around infinit. We can also check for a branch point at infinit b following a path that encloses the point at infinit and no other singularities. Just draw a simple closed curve that separates the comple plane into a bounded component that contains all the singularities of the function in the finite plane. Then, depending on orientation, the curve is a contour enclosing all the finite singularities, or the point at infinit and no other singularities. Eample Once again consider the function z /. We know that the function changes value on a curve that goes once around the origin. Such a curve can be considered to be either a path around the origin or a path around infinit. In either case the path encloses one singularit. There are branch points at the origin and at infinit. Now consider a curve that does not go around the origin. Such a curve can be considered to be either a path around neither of the branch points or both of them. Thus we see that z / does not change value when we follow a path that encloses neither or both of its branch points. 76

39 Eample Consider f(z) = (z ) /. We factor the function. f(z) = (z ) / (z + ) / There are branch points at z = ±. Now consider the point at infinit. f ( ζ ) = ( ζ ) / = ±ζ ( ζ ) / Since f (ζ ) does not have a branch point at ζ =, f(z) does not have a branch point at infinit. We could reach the same conclusion b considering a path around infinit. Consider a path that circles the branch points at z = ± once in the positive direction. Such a path circles the point at infinit once in the negative direction. In traversing this path, the value of f(z) is multiplied b the factor (e ıπ ) / (e ıπ ) / = e ıπ =. Thus the value of the function does not change. There is no branch point at infinit. Diagnosing branch points. We have the definition of a branch point, but we do not have a convenient criterion for determining if a particular function has a branch point. We have seen that log z and z α for non-integer α have branch points at zero and infinit. The inverse trigonometric functions like the arcsine also have branch points, but the can be written in terms of the logarithm and the square root. In fact all the elementar functions with branch points can be written in terms of the functions log z and z α. Furthermore, note that the multi-valuedness of z α comes from the logarithm, z α = e α log z. This gives us a wa of quickl determining if and where a function ma have branch points. Result 7.9. Let f(z) be a single-valued function. Then log(f(z)) and (f(z)) α ma have branch points onl where f(z) is zero or singular. Eample Consider the functions,. (z ) /. ( z /) 3. ( z /) 3 77

40 Are the multi-valued? Do the have branch points?. ( z ) / = ± z = ±z Because of the ( ) /, the function is multi-valued. The onl possible branch points are at zero and infinit. If ((e ı ) ) / =, then ( (e ıπ ) ) / = (e ı4π ) / = e ıπ =. Thus we see that the function does not change value when we walk around the origin. We can also consider this to be a path around infinit. This function is multi-valued, but has no branch points.. ( z / ) = ( ± z ) = z This function is single-valued. 3. ( z / ) 3 = ( ± z ) 3 = ± ( z ) 3 ( This function is multi-valued. We consider the possible branch point at z =. If (e ) /) 3 =, then ( (e ıπ ) /) 3 = (e ıπ ) 3 = e ı3π =. Since the function changes value when we walk around the origin, it has a branch point at z =. Since this is also a path around infinit, there is a branch point there. Eample Consider the function f(z) = log ( ) z. Since is onl zero at infinit and its onl singularit is at z z =, the onl possibilities for branch points are at z = and z =. Since ( ) log = log(z ) z and log w has branch points at zero and infinit, we see that f(z) has branch points at z = and z =. Eample Consider the functions, 78

41 . e log z. log e z. Are the multi-valued? Do the have branch points?.. This function is single-valued. e log z = ep(log z + ıπn) = e Log z e ıπn = z log e z = Log e z +ıπn = z + ıπm This function is multi-valued. It ma have branch points onl where e z is zero or infinite. This onl occurs at z =. Thus there are no branch points in the finite plane. The function does not change when traversing a simple closed path. Since this path can be considered to enclose infinit, there is no branch point at infinit. Consider (f(z)) α where f(z) is single-valued and f(z) has either a zero or a singularit at z = z. (f(z)) α ma have a branch point at z = z. If f(z) is not a power of z, then it ma be difficult to tell if (f(z)) α changes value when we walk around z. Factor f(z) into f(z) = g(z)h(z) where h(z) is nonzero and finite at z. Then g(z) captures the important behavior of f(z) at the z. g(z) tells us how fast f(z) vanishes or blows up. Since (f(z)) α = (g(z)) α (h(z)) α and (h(z)) α does not have a branch point at z, (f(z)) α has a branch point at z if and onl if (g(z)) α has a branch point there. Similarl, we can decompose log(f(z)) = log(g(z)h(z)) = log(g(z)) + log(h(z)) to see that log(f(z)) has a branch point at z if and onl if log(g(z)) has a branch point there. Result Consider a single-valued function f(z) that has either a zero or a singularit at z = z. Let f(z) = g(z)h(z) where h(z) is nonzero and finite. (f(z)) α has a branch point at z = z if and onl if (g(z)) α has a branch point there. log(f(z)) has a branch point at z = z if and onl if log(g(z)) has a branch point there. 79

42 Eample 7.9. Consider the functions,. sin z /. (sin z) / 3. z / sin z / 4. (sin z ) / Find the branch points and the number of branches.. sin z / = sin ( ± z ) = ± sin z sin z / is multi-valued. It has two branches. There ma be ( branch points at zero and infinit. Consider the unit circle which is a path around the origin or infinit. If sin (e ı ) /) ( = sin(), then sin (e ıπ ) /) = sin (e ıπ ) = sin( ) = sin(). There are branch points at the origin and infinit.. (sin z) / = ± sin z The function is multi-valued with two branches. The sine vanishes at z = nπ and is singular at infinit. There could be branch points at these locations. Consider the point z = nπ. We can write sin z = (z nπ) sin z z nπ Note that sin z z nπ is nonzero and has a removable singularit at z = nπ. lim z nπ sin z z nπ = lim cosz z nπ = ( ) n Since (z nπ) / has a branch point at z = nπ, (sin z) / has branch points at z = nπ. 8

43 Since the branch points at z = nπ go all the wa out to infinit. It is not possible to make a path that encloses infinit and no other singularities. The point at infinit is a non-isolated singularit. A point can be a branch point onl if it is an isolated singularit. 3. z / sin z / = ± z sin ( ± z ) = ± z ( ± sin z ) = z sin z The function is single-valued. Thus there could be no branch points. 4. ( sin z ) / = ± sin z This function is multi-valued. Since sin z = at z = (nπ) /, there ma be branch points there. First consider the point z =. We can write sin z = z sin z z where sin (z )/z is nonzero and has a removable singularit at z =. sin z lim z z = lim z z cosz z =. Since (z ) / does not have a branch point at z =, (sin z ) / does not have a branch point there either. Now consider the point z = nπ. sin z = ( z nπ ) sin z z nπ sin (z ) / (z nπ) in nonzero and has a removable singularit at z = nπ. lim z nπ sin z z nπ = lim z cosz z nπ = nπ( ) n 8

44 Since (z nπ) / has a branch point at z = nπ, (sin z ) / also has a branch point there. Thus we see that (sin z ) / has branch points at z = (nπ) / for n Z \ {}. This is the set of numbers: {± π, ± π,...,±ı π, ±ı π,...}. The point at infinit is a non-isolated singularit. Eample 7.9. Find the branch points of f(z) = ( z 3 z ) /3. Introduce branch cuts. If f() = 3 6 then what is f( )? We epand f(z). f(z) = z /3 (z ) /3 (z + ) /3. There are branch points at z =,,. We consider the point at infinit. f ( ) = ζ ( ) /3 ( ) /3 ( ) /3 ζ ζ ζ + = ζ ( ζ)/3 ( + ζ) /3 Since f(/ζ) does not have a branch point at ζ =, f(z) does not have a branch point at infinit. Consider the three possible branch cuts in Figure 7.9. The first and the third branch cuts will make the function single valued, the second will not. It is clear that the first set makes the function single valued since it is not possible to walk around an of the branch points. The second set of branch cuts would allow ou to walk around the branch points at z = ±. If ou walked around these two once in the positive direction, the value of the function would change b the factor e ı4π/3. The third set of branch cuts would allow ou to walk around all three branch points together. You can verif that if ou walk around the three branch points, the value of the function will not change (e ı6π/3 = e ıπ = ). Suppose we introduce the third set of branch cuts and are on the branch with f() = 3 6. f() = ( e ı) /3 ( e ı ) /3 ( 3 e ı ) /3 = 3 6 8

45 Figure 7.9: Three Possible Branch Cuts for f(z) = (z 3 z) /3. The value of f( ) is f( ) = ( e ıπ ) /3 (3 e ıπ ) /3 ( e ıπ ) /3 = 3 e ıπ/3 3 3 e ıπ/3 3 e ıπ/3 = 3 6 e ıπ = 3 6. Eample 7.9. Find the branch points and number of branches for f(z) = z z. z z = ep ( z log z ) There ma be branch points at the origin and infinit due to the logarithm. Consider walking around a circle of radius r centered at the origin in the positive direction. Since the logarithm changes b ıπ, the value of f(z) changes b the factor e ıπr. There are branch points at the origin and infinit. The function has an infinite number of branches. Eample Construct a branch of f(z) = ( z + ) /3 83

46 such that First we factor f(z). f() = ( + ı ) 3. f(z) = (z ı) /3 (z + ı) /3 There are branch points at z = ±ı. Figure 7.3 shows one wa to introduce branch cuts. φ ρ r θ Figure 7.3: Branch Cuts for f(z) = (z + ) /3. Since it is not possible to walk around an branch point, these cuts make the function single valued. We introduce the coordinates: z ı = ρ e ıφ, z + ı = r e ıθ. f(z) = ( ρ e ıφ) /3 ( r e ıθ ) /3 = 3 ρr e ı(φ+θ)/3 The condition f() = ( + ı ) 3 = e ı(π/3+πn) 84

47 can be stated 3 e ı(φ+θ)/3 = e ı(π/3+πn) φ + θ = π + 6πn The angles must be defined to satisf this relation. One choice is π < φ < 5π, π < θ < 3π. Principal branches. We construct the principal branch of the logarithm b putting a branch cut on the negative real ais choose z = r e ıθ, θ ( π, π). Thus the principal branch of the logarithm is Log z = ln r + ıθ, π < θ < π. Note that the if is a negative real number, (and thus lies on the branch cut), then Log is undefined. The principal branch of z α is z α = e α Log z. Note that there is a branch cut on the negative real ais. απ < arg ( e α Log z) < απ The principal branch of the z / is denoted z. The principal branch of z /n is denoted n z. Eample Construct z, the principal branch of ( z ) /. First note that since ( z ) / = ( z) / ( + z) / there are branch points at z = and z =. The principal branch of the square root has a branch cut on the negative real ais. z is a negative real number for z (... ) (... ). Thus we put branch cuts on (... ] and [... ). 85