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Unformatted text preview: Chapter 3 Examples of Functions Obvious is the most dangerous word in mathematics. E. T. Bell 3.1 Mbius Transformations o The first class of functions that we will discuss in some detail are built from linear polynomials. Definition 3.1. A linear fractional transformation is a function of the form f (z) = az + b , cz + d where a, b, c, d C. If ad - bc = 0 then f is called a Mbius1 transformation. o Exercise 11 of the previous chapter states that any polynomial (in z) is an entire function. From this fact we can conclude that a linear fractional transformation f (z) = az+b is holomorphic cz+d in C \ - d (unless c = 0, in which case f is entire). c One property of Mbius transformations, which is quite special for complex functions, is the o following. Lemma 3.2. Mbius transformations are bijections. In fact, if f (z) = az+b then the inverse o cz+d function of f is given by dz - b f -1 (z) = . -cz + a Remark. Notice that the inverse of a Mbius transformation is another Mbius transformation. o o Proof. Note that f : C \ {- d } C \ { a }. Suppose f (z1 ) = f (z2 ), that is, c c az1 + b az2 + b = . cz1 + d cz2 + d As the denominators are nonzero, this is equivalent to (az1 + b)(cz2 + d) = (az2 + b)(cz1 + d) , 1 Named after August Ferdinand Mbius (17901868). For more information about Mbius, see o o http://www-groups.dcs.st-and.ac.uk/history/Biographies/Mobius.html. 23 CHAPTER 3. EXAMPLES OF FUNCTIONS which can be rearranged to 24 Since ad - bc = 0 this implies that z1 = z2 , which means that f is one-to-one. The formula for f -1 : C \ { a } C \ {- d } can be checked easily. Just like f , f -1 is one-to-one, which implies that c c f is onto. Aside from being prime examples of one-to-one functions, Mbius transformations possess faso cinating geometric properties. En route to an example of such, we introduce some terminology. Special cases of Mbius transformations are translations f (z) = z + b, dilations f (z) = az, and ino 1 versions f (z) = z . The next result says that if we understand those three special transformations, we understand them all. Proposition 3.3. Suppose f (z) = az+b cz+d (ad - bc)(z1 - z2 ) = 0 . is a linear fractional transformation. If c = 0 then f (z) = a b z+ , d d + a . c if c = 0 then f (z) = bc - ad 1 c2 z + d c In particular, every linear fractional transformation is a composition of translations, dilations, and inversions. Proof. Simplify. With the last result at hand, we can tackle the promised theorem about the following geometric property of Mbius transformations. o Theorem 3.4. Mbius transformations map circles and lines into circles and lines. o Proof. Translations and dilations certainly map circles and lines into circles and lines, so by the 1 last proposition, we only have to prove the theorem for the inversion f (z) = z . Before going on we find a standard form for the equation of a straight line. Starting with ax + by = c (where z = x + iy), let = a + bi. Then z = ax + by + i(ay - bx) so z + z = z + z = 2 Re(z) = 2ax + 2by. Hence our standard equation for a line becomes z + z = 2c, or Re(z) = c. (3.1) Circle case: Given a circle centered at z0 with radius r, we can modify its defining equation |z - z0 | = r as follows: (z - z0 )(z - z0 ) = r2 |z - z0 |2 = r2 |z|2 - z0 z - zz0 + |z0 |2 - r2 = 0 . zz - z0 z - zz0 + z0 z0 = r2 CHAPTER 3. EXAMPLES OF FUNCTIONS 1 Now we want to transform this into an equation in terms of w, where w = z . If we solve w = 1 z we get z = w , so we make this substitution in our equation: 1 z 25 for (To get the second line we multiply by |w|2 = ww and simplify.) Now if r happens to be equal to |z0 |2 then this equation becomes 1 - z0 w - z0 w = 0, which is of the form (3.1) with = z0 , so we have a straight line in terms of w. Otherwise |z0 |2 - r2 is non-zero so we can divide our equation by it. We obtain z0 z0 1 |w|2 - w- w+ = 0. 2 2 2 2 2 |z0 | - r |z0 | - r |z0 | - r2 We define z0 w0 = , 2 |z0 | - r2 1 |z0 |2 |z0 |2 - r2 r2 s = |w0 | - = - = . |z0 |2 - r2 (|z0 |2 - r2 )2 (|z0 |2 - r2 )2 (|z0 |2 - r2 )2 2 2 2 1 - z0 1 - z0 1 + |z0 |2 - r2 = 0 w w w 1 - z0 w - z0 w + |w|2 |z0 |2 - r2 = 0 . Then we can rewrite our equation as |w|2 - w0 w - w0 w + |w0 |2 - s2 = 0 ww - w0 w - ww0 + w0 w0 = s2 (w - w0 )(w - w0 ) = s2 |w - w0 |2 = s2 . This is the equation of a circle in terms of w, with center w0 and radius s. Line case: We start with the equation of a line in the form (3.1) and rewrite it in terms of w, 1 as above, by substituting z = w and simplifying. We get z0 w + z0 w = 2cww . If c = 0, this describes a line in the form (3.1) in terms of w. Otherwise we can divide by 2c: z0 z0 ww - w - w = 0 2c 2c z0 z0 |z0 |2 w- w- - =0 2c 2c 4c2 2 2 w - z0 = |z0 | . 2c 4c2 z0 2c This is the equation of a circle with center and radius |z0 | 2|c| . There is one fact about Mbius transformations that is very helpful to understanding their o geometry. In fact, it is much more generally useful: CHAPTER 3. EXAMPLES OF FUNCTIONS 26 Lemma 3.5. Suppose f is holomorphic at a with f (a) = 0 and suppose 1 and 2 are two smooth curves which pass through a, making an angle of with each other. Then f transforms 1 and 2 into smooth curves which meet at f (a), and the transformed curves make an angle of with each other. In brief, an holomorphic function with non-zero derivative preserves angles. Functions which preserve angles in this way are also called conformal. Proof. For k = 1, 2 we write k parametrically, as zk (t) = xk (t) + iyk (t), so that zk (0) = a. The complex number zk (0), considered as a vector, is the tangent vector to k at the point a. Then f transforms the curve k to the curve f (k ), parameterized as f (zk (t)). If we differentiate f (zk (t)) at t = 0 and use the chain rule we see that the tangent vector to the transformed curve at the point f (a) is f (a)zk (0). Since f (a) = 0 the transformation from z1 (0) and z2 (0) to f (a)z1 (0) and (a)z (0) is a dilation. A dilation is the composition of a scale change and a rotation and both of f 2 these preserve the angles between vectors. 3.2 Infinity and the Cross Ratio Infinity is not a number--this is true whether we use the complex numbers or stay in the reals. However, for many purposes we can work with infinity in the complexes much more naturally and simply than in the reals. In the complex sense there is only one infinity, written . In the real sense there is also a "negative infinity", but - = in the complex sense. In order to deal correctly with infinity we have to realize that we are always talking about a limit, and complex numbers have infinite limits if they can become larger in magnitude than any preassigned limit. For completeness we repeat the usual definitions: Definition 3.6. Suppose G is a set of complex numbers and f is a function from G to C. (a) lim f (z) = means that for every M > 0 we can find > 0 so that, for all z G satisfying zz0 0 < |z - z0 | < , we have |f (z)| > M . z (b) lim f (z) = L means that for every > 0 we can find N > 0 so that, for all z G satisfying |z| > N , we have |f (z) - L| < . z (c) lim f (z) = means that for every M > 0 we can find N > 0 so that, for all z G satisfying |z| > N we have |f (z)| > M . In the first definition we require that z0 is an accumulation point of G while in the second and third we require that is an "extended accumulation point" of G, in the sense that for every B > 0 there is some z G with |z| > B. The usual rules for working with infinite limits are still valid in the complex numbers. In fact, it is a good idea to make infinity an honorary complex number so that we can more easily manipulate infinite limits. We then define algebraic rules for dealing with our new point, , based on the usual laws of limits. For example, if lim f (z) = and lim g(z) = a is finite then the usual "limit of zz0 zz0 CHAPTER 3. EXAMPLES OF FUNCTIONS 27 sum = sum of limits" rule gives lim (f (z) + g(z)) = . This leads us to want the rule + a = . ^ We do this by defining a new set, C: zz0 ^ Definition 3.7. The extended complex plane is the set C := C {}, together with the following algebraic properties: For any a C, (1) (2) (3) +a=a+= if a = 0 then if a = 0 then a = a = = a a = 0 and = 0 The extended complex plane is also called the Riemann sphere (or, in a more advanced course, the complex projective line, denoted CP1 ). If a calculation involving infinity is not covered by the rules above then we must investigate the limit more carefully. For example, it may seem strange that + is not defined, but if we take the limit of z + (-z) = 0 as z we will get 0, but the individual limits of z and -z are both . 1 Now we reconsider Mbius transformations with infinity in mind. For example, f (z) = z is o now defined for z = 0 and z = , with f (0) = and f () = 0, so the proper domain for ^ f (z) is actually C. Let's consider the other basic types of Mbius transformations. A translation o f (z) = z + b is now defined for z = , with f () = + b = , and a dilation f (z) = az (with a = 0) is also defined for z = , with f () = a = . Since every Mbius transformation o 1 can be expressed as a composition of translations, dilations and the inversion f (z) = z we see that ^ ^ every Mbius transformation may be interpreted as a transformation of C onto C. The general o case is summarized below: Lemma 3.8. Let f be the Mbius transformation o f (z) = az + b . cz + d ^ Then f is defined for all z C. If c = 0 then f () = , and, otherwise, a d f () = and f - = . c c 1 With this interpretation in mind we can add some insight to Theorem 3.4. Recall that f (z) = z transforms circles that pass through the origin to straight lines, but the point z = 0 must be excluded from the circle. However, now we can put it back, so f transforms circles that pass through the origin to straight lines plus . If we remember that corresponds to being arbitrarily far away from the origin we can visualize a line plus infinity as a circle passing through . If we make ^ this a definition then Theorem 3.4 can be expressed very simply: any Mbius transformation of C o transforms circles to circles. For example, the transformation f (z) = z+i z-i transforms -i to 0, i to , and 1 to i. The three points -i, i and 1 determine a circle--the unit circle |z| = 1--and the three image points 0, and i also determine a circle--the imaginary axis CHAPTER 3. EXAMPLES OF FUNCTIONS 28 plus the point at infinity. Hence f transforms the unit circle onto the imaginary axis plus the point at infinity. ^ This example relied on the idea that three distinct points in C determine uniquely a circle passing through them. If the three points are on a straight line or if one of the points is then ^ the circle is a straight line plus . Conversely, if we know where three distinct points in C are transformed by a Mbius transformation then we should be able to figure out everything about the o transformation. There is a computational device that makes this easier to see. ^ Definition 3.9. If z, z1 , z2 , and z3 are any four points in C with z1 , z2 , and z3 distinct, then their cross-ratio is defined by (z - z1 )(z2 - z3 ) [z, z1 , z2 , z3 ] = . (z - z3 )(z2 - z1 ) Here if z = z3 , the result is infinity, and if one of z, z1 , z2 , or z3 is infinity, then the two terms on the right containing it are canceled. Lemma 3.10. If f is defined by f (z) = [z, z1 , z2 , z3 ] then f is a Mbius transformation which o satisfies f (z1 ) = 0, f (z2 ) = 1, f (z3 ) = . Moreover, if g is any Mbius transformation which transforms z1 , z2 and z3 as above then g(z) = o f (z) for all z. Proof. Everything should be clear except the final uniqueness statement. By Lemma 3.2 the inverse f -1 is a Mbius transformation and, by Exercise 7 in this chapter, the composition h = g f -1 o is a Mbius transformation. Notice that h(0) = g(f -1 (0)) = g(z1 ) = 0. Similarly, h(1) = 1 and o h() = . If we write h(z) = az+b then cz+d b = b = 0 d a = h() = = c = 0 c a+b a+0 a 1 = h(1) = = = = a = d , c+d 0+d d 0 = h(0) = so h(z) = az+b = az+0 = a z = z. But since h(z) = z for all z we have h(f (z)) = f (z) and so cz+d 0+d d g(z) = g (f -1 f )(z) = (g f -1 ) f (z) = h(f (z)) = f (z). ^ So if we want to map three given points of C to 0, 1 and by a Mbius transformation then o the cross-ratio gives us the only way to do it. What if we have three points z1 , z2 and z3 and we want to map them to three other points, w1 , w2 and w3 ? ^ Theorem 3.11. Suppose z1 , z2 and z3 are distinct points in C and w1 , w2 and w3 are distinct ^ Then there is a unique Mbius transformation h satisfying h(z1 ) = w1 , h(z2 ) = w2 points in C. o and h(z3 ) = w3 . Proof. Let h = g -1 f where f (z) = [z, z1 , z2 , z3 ] and g(w) = [w, w1 , w2 , w3 ]. Uniqueness follows as in the proof of Lemma 3.10. This theorem gives an explicit way to determine h from the points zj and wj but, in practice, it is often easier to determine h directly from the conditions f (zk ) = wk (by solving for a, b, c and d). CHAPTER 3. EXAMPLES OF FUNCTIONS 29 3.3 Stereographic Projection The addition of to the complex plane C gives the plane a very useful structure. This structure is revealed by a famous function called stereographic projection. Stereographic projection also gives us a way of visualizing the extended complex plane that is, the point at infinity in R3 , as the unit sphere. It also provides a way of `seeing' that a line in the extended complex plane is really a circle, and of visualizing Mbius functions. o To begin, think of C as the xy-plane in R3 = {(x, y, z)}, C = {(x, y, 0) R3 }. To describe stereographic projection, we will be less concerned with actual complex numbers x + iy and more with their coordinates. Consider the unit sphere S2 := {(x, y, z) R3 |x2 + y 2 + z 2 = 1}. Then the sphere and the complex plane intersect in the set {(x, y, 0)|x2 + y 2 = 1}, corresponding to the equator on the sphere and the unit circle on the complex plane. Let N denote the North Pole (0, 0, 1) of S2 , and let S denote the South Pole (0, 0, -1). ^ ^ Definition 3.12. The stereographic projection of S2 to C from N is the map S2 C defined as 2 - {N }, as the z-coordinate of P is strictly less than 1, the line P follows. For any point P S N intersects C in exactly one point, Q. Define (P ) := Q. We also declare that (N ) = C. Proposition 3.13. The map is the bijection x y (x, y, z) = , ,0 , 1-z 1-z with inverse map -1 (p, q, 0) = 2p 2q p2 + q 2 - 1 , 2 , 2 p2 + q 2 + 1 p + q 2 + 1 p + q 2 + 1 , where we declare (0, 0, 1) = and -1 () = (0, 0, 1). Proof. That is a bijection follows from the existence of the inverse function, and is left as an exercise. For P = (x, y, z) S2 - {N }, the straight line N P through N and P is given by, for t , r(t) = N + t(P - N ) = (0, 0, 1) + t[(x, y, z) - (0, 0, 1)] = (tx, ty, 1 + t(z - 1)). 1 When r(t) hits C, the third coordinate is 0, so it must be that t = 1-z . Plugging this value of t into the formula for r yields as stated. To see the formula for the inverse map -1 , we begin with a point Q = (p, q, 0) C, and solve for a point P = (x, y, z) S2 so that (P ) = Q. The point P satisfies the equation x2 + y 2 + z 2 = 0. y x The equation (P ) = Q tells us that 1-z = p and 1-z = q. Thus, we solve 3 equations for 3 unknowns. The latter two equations yield p2 + q 2 = x2 + y 2 1 - z2 1+z = = . 2 2 (1 - z) (1 - z) 1-z Solving p2 + q 2 = 1+z for z, and then plugging this into the identities x = p(1 - z) and y = q(1 - z) 1-z proves the desired formula. It is easy to check that -1 and -1 are now both the identity; we leave these as exercises. This proves the proposition. CHAPTER 3. EXAMPLES OF FUNCTIONS We use the formulas above to prove the following. 30 Theorem 3.14. The stereographic projection takes the set of circles in S2 bijectively to the set ^ of circles in C, where for a circle S2 we have that () that is, () is a line in C if and only if N . Proof. A circle in S2 is the intersection of S2 with some plane P . If we have a normal vector (x0 , y0 , z0 ) to P , then there is a unique real number k so that the plane P is given by P = {(x, y, z) R3 |(x, y, z) (x0 , y0 , z0 ) = k} = {(x, y, z) R3 |xx0 + yy0 + zz0 = k}. Without loss of generality, we can assume that (x0 , y0 , z0 ) S2 by possibly changing k. We may also assume without loss of generality that 0 k 1, since if k < 0 we can replace (x0 , y0 , z0 ) with -(x0 , y0 , z0 ), and if k > 1 then P S2 = . Consider the circle of intersection P S2 . A point (p, q, 0) in the complex plane lies on the image of this circle under if and only if -1 (p, q, 0) satisfies the defining equation for P . Using the equations from Proposition 3.13 for -1 (p, q, 0), we see that (z0 - k)p2 + (2x0 )p + (z0 - k)q 2 + (2y0 )q = z0 + k. If z0 - k = 0, this is a straight line in the pq-plane. Moreover, every line in the pq-plane can be obtained in this way. Notice that z0 = k if and only if N P , which is if and only if the image under is a straight line. If z0 - k = 0, then completing the square yields 2 2 x0 y0 1 - k2 p+ + q+ = . z0 - k z0 - k (z0 - k)2 Depending on whether the right hand side of this equation is positive, 0, or negative, this is the equation of a circle, point, or the empty set in the pq-plane, respectively. These three cases happen when k < 1, k = 1, and k > 1, respectively. Only the first case corresponds to a circle in S2 . It is an exercise to verify that every circle in the pq-plane arises in this manner. We can now think of the extended complex plane as a sphere in R3 , called the Riemann sphere. It is particularly nice to think about the basic Mbius transformations via their effect on the o Riemann sphere. We will describe inversion. It is worth thinking about, though beyond the scope of these notes, how other basic Mbius functions behave. For instance, a rotation f (z) = ei z, o composed with -1 , can be seen to be a rotation of S2 . We encourage the reader to verify this to themselves, and consider the harder problems of visualizing a real dilation f (z) = rz or a translation, f (z) = z + b. We give the hint that a real dilation is in some sense `dual' to a rotation, in that each moves points `along' perpendicular sets of circles. Translations can also be visualized via how they move points `along' sets of circles. We now use stereographic projection to take another look at f (z) = 1/z. We want to know what this function does to the sphere S2 . We will take an (x, y, z) on S2 , project it to the plane by stereographic projection , apply f to the point that results, and then pull this point back to S2 by -1 . We know (x, y, z) = (x/(1 - z), y/(1 - z)) which we now regard as the complex number x y +i . 1-z 1-z CHAPTER 3. EXAMPLES OF FUNCTIONS We use 1 p - qi = 2 . p + qi p + q2 31 We know from a previous calculation that p2 + q 2 = (1 + z)/(1 - z). This gives x y x y 1-z x -y f +i = -i = +i . 1-z 1-z 1-z 1-z 1+z 1+z 1+z Rather than plug this result into the formulas for -1 , we can just ask what triple of numbers will go to this particular pair using the formulas (x, y, z) = (x/(1-z), y/(1-z)). The answer is clearly (x, -y, -z). Thus we have shown that the effect of f (z) = 1/z on S2 is to take (x, y, z) to (x, -y, -z). This is a rotation around the x-axis by 180 degrees. We now have a second argument that f (z) = 1/z takes circles and lines to circles and lines. A circle or line in C is taken to a circle on S2 by -1 . Then 1/z rotates the sphere which certainly takes circles to circles. Now takes circles back to circles and lines. We can also say that the circles that go to lines under f (z) = 1/z are the circles though 0. This is because 0 goes to (0, 0, -1) under -1 so a circle through 0 in C goes to a circle through the south pole in S2 . Now 180 rotation about the x-axis takes the south pole to the north pole, and our circle is now passing through N . But we know that will take this circle to a line in C. We end by mentioning that there is in fact a way of putting the complex metric on S2 . It is certainly not the (finite) distance function induced by R3 . Indeed, the origin in the complex plane corresponds to the South Pole S of S2 . We have to be able to get arbitrarily far away from the origin in C, so the complex distance function has to increase greatly with the z coordinate. The ^ closer points are to the North Pole N (corresponding to in C), the larger their distance to the origin, and to each other! In this light, a `line' in the Riemann sphere S2 corresponds to a circle in S2 through N . In the regular sphere, the circle has finite length, but as a line on the Riemann sphere with the complex metric, it has infinite length. 3.4 Exponential and Trigonometric Functions To define the complex exponential function, we once more borrow concepts from calculus, namely the real exponential function2 and the real sine and cosine, and--in addition--finally make sense of the notation eit = cos t + i sin t. Definition 3.15. The (complex) exponential function is defined for z = x + iy as exp(z) = ex (cos y + i sin y) = ex eiy . This definition seems a bit arbitrary, to say the least. Its first justification is that all exponential rules which we are used to from real numbers carry over to the complex case. They mainly follow from Lemma 1.2 and are collected in the following. 2 It is a nontrivial question how to define the real exponential function. Our preferred way to do this is through a P power series: ex = k0 xk /k!. In light of this definition, the reader might think we should have simply defined the complex exponential function through a complex power series. In fact, this is possible (and an elegant definition); however, one of the promises of these lecture notes is to introduce complex power series as late as possible. We agree with those readers who think that we are "cheating" at this point, as we borrow the concept of a (real) power series to define the real exponential function. CHAPTER 3. EXAMPLES OF FUNCTIONS Lemma 3.16. For all z, z1 , z2 C, (a) exp (z1 ) exp (z2 ) = exp (z1 + z2 ) (b) 1 exp(z) 32 = exp (-z) (c) exp (z + 2i) = exp (z) (d) |exp (z)| = exp (Re z) (e) exp(z) = 0 (f) d dz exp (z) = exp (z) . Remarks. 1. The third identity is a very special one and has no counterpart for the real exponential function. It says that the complex exponential function is periodic with period 2i. This has many interesting consequences; one that may not seem too pleasant at first sight is the fact that the complex exponential function is not one-to-one. 2. The last identity is not only remarkable, but we invite the reader to meditate on its proof. When proving this identity through the CauchyRiemann equations for the exponential function, one can get another strong reason why Definition 3.15 is reasonable. Finally, note that the last identity also says that exp is entire. We should make sure that the complex exponential function specializes to the real exponential function for real arguments: if z = x R then exp(x) = ex (cos 0 + i sin 0) = ex . 5 6 3 0 - 3 - 5 6 -1 0 1 2 exp Figure 3.1: Image properties of the exponential function. The trigonometric functions--sine, cosine, tangent, cotangent, etc.--have their complex analogues, however, they don't play the same prominent role as in the real case. In fact, we can define them as merely being special combinations of the exponential function. CHAPTER 3. EXAMPLES OF FUNCTIONS Definition 3.17. The (complex) sine and cosine are defined as sin z = 1 (exp(iz) - exp(-iz)) 2i and cos z = 1 (exp(iz) + exp(-iz)) , 2 33 respectively. The tangent and cotangent are defined as tan z = respectively. Note that to write tangent and cotangent in terms of the exponential function, we used the fact that exp(z) exp(-z) = exp(0) = 1. Because exp is entire, so are sin and cos. As with the exponential function, we should first make sure that we're not redefining the real sine and cosine: if z = x R then sin z = 1 1 (exp(ix) - exp(-ix)) = (cos x + i sin x - (cos(-x) + i sin(-x))) = sin x . 2i 2i sin z exp(2iz) - 1 = -i cos z exp(2iz) + 1 and cot z = cos z exp(2iz) + 1 =i , sin z exp(2iz) - 1 A similar calculation holds for the cosine. Not too surprisingly, the following properties follow mostly from Lemma 3.16. Lemma 3.18. For all z, z1 , z2 C, sin(z + 2) = sin z sin(-z) = - sin z cos(-z) = cos z cos(z + 2) = cos z cot(z + ) = cot z cos(z + /2) = - sin z cos (z1 + z2 ) = cos z1 cos z2 - sin z1 sin z2 cos z = - sin z . tan(z + ) = tan z sin(z + /2) = cos z sin (z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 cos2 z + sin2 z = 1 sin z = cos z cos2 z - sin2 z = cos(2z) Finally, one word of caution: unlike in the real case, the complex sine and cosine are not bounded--consider, for example, sin(iy) as y . We end this section with a remark on hyperbolic trig functions. The hyperbolic sine, cosine, tangent, and cotangent are defined as in the real case: Definition 3.19. 1 (exp(z) - exp(-z)) 2 sinh z exp(2z) - 1 tanh z = = cosh z exp(2z) + 1 sinh z = 1 (exp(z) + exp(-z)) 2 cosh z exp(2z) + 1 coth z = = . sinh z exp(2z) - 1 cosh z = As such, they are not only yet more special combinations of the exponential function, but they are also related with the trigonometric functions via sinh(iz) = i sin z and cosh(iz) = cos z . CHAPTER 3. EXAMPLES OF FUNCTIONS 34 3.5 The Logarithm and Complex Exponentials The complex logarithm is the first function we'll encounter that is of a somewhat tricky nature. It is motivated as being the inverse function to the exponential function, that is, we're looking for a function Log such that exp(Log z) = z = Log(exp z) . As we will see shortly, this is too much to hope for. Let's write, as usual, z = r ei , and suppose that Log z = u(z) + iv(z). Then for the first equation to hold, we need exp(Log z) = eu eiv = r ei = z , that is, eu = r = |z| u = ln |z| (where ln denotes the real natural logarithm; in particular we need to demand that z = 0), and eiv = ei v = +2k for some k Z. A reasonable definition of a logarithm function Log would hence be to set Log z = ln |z| + i Arg z where Arg z gives the argument for the complex number z according to some convention--for example, we could agree that the argument is always in (-, ], or in [0, 2), etc. The problem is that we need to stick to this convention. On the other hand, as we saw, we could just use a different argument convention and get another reasonable `logarithm.' Even worse, by defining the multi-valued map arg z = { : is a possible argument of z} and defining the multi-valued logarithm as log z = ln |z| + i arg z , we get something that's not a function, yet it satisfies exp(log z) = z . We invite the reader to check this thoroughly; in particular, one should note how the periodicity of the exponential function takes care of the multi-valuedness of our `logarithm' log. log is, of course, not a function, and hence we can't even consider it to be our sought-after inverse of the exponential function. Let's try to make things well defined. Definition 3.20. Any function Log : C \ {0} C which satisfies exp(Log z) = z is a branch of the logarithm. Let Arg z denote that argument of z which is in (-, ] (the principal argument of z). Then the principal logarithm is defined as Log z = ln |z| + i Arg z . The paragraph preceding this definition ensures that the principal logarithm is indeed a branch of the logarithm. Even better, the evaluation of any branch of the logarithm at z can only differ from Log z by a multiple of 2i; the reason for this is once more the periodicity of the exponential function. So what about the other equation Log(exp z) = z? Let's try the principal logarithm: Suppose z = x + iy, then Log(exp z) = Log ex eiy = ln ex eiy + i Arg ex eiy = ln ex + i Arg eiy = x + i Arg eiy . CHAPTER 3. EXAMPLES OF FUNCTIONS 35 The right-hand side is equal to z = x + iy only if y (-, ]. The same happens with any other branch of the logarithm Log--there will always be some (in fact, many) y-values for which Log(exp z) = z. To end our discussion of the logarithm on a happy note, we prove that any branch of the logarithm has the same derivative; one just has to be cautious about where each logarithm is holomorphic. Theorem 3.21. Suppose Log is a branch of the logarithm. Then Log is differentiable wherever it is continuous and 1 Log z = . z Proof. The idea is to apply Lemma 2.12 to exp and Log, but we need to be careful about the domains of these functions, so that we get actual inverse functions. Suppose Log maps C \ {0} to G (this is typically a half-open strip; you might want to think about what it looks like if Log = Log). We apply Lemma 2.12 with f : G C \ {0} , f (z) = exp(z) and g : C \ {0} G, g(z) = Log: if Log is continuous at z then Log z = 1 exp (Log z) = 1 1 = . exp(Log z) z We finish this section by defining complex exponentials. For two complex numbers a and b, the natural definition ab = exp(b log a) (which is a concept borrowed from calculus) would in general yield more than one value (Exercise 36), so it is not always useful. We turn instead to the principal logarithm and define the principal value of ab as ab = exp(b Log a) . A note about e. In calculus one proves the equivalence of the real exponential function (as given, for example, through a power series) and the function f (x) = ex where e is Euler's3 number and 1 n can be defined, for example, as e = limn 1 + n . With our definition of ab , we can now make a similar remark about the complex exponential function. Because e is a positive real number and hence Arg e = 0, we obtain ez = exp(z Log e) = exp (z (ln |e| + i Arg e)) = exp (z ln e) = exp (z) . A word of caution: this only works out this nicely because we carefully defined ab for complex numbers. Different definitions might lead to different outcomes of ez versus exp z! Exercises 1. Show that if f (z) = az+b cz+d is a Mbius transformation then f -1 (z) = o dz-b -cz+a . 2. Show that the derivative of a Mbius transformation is never zero. o 3. Prove that any Mbius transformation different from the identity map can have at most two o fixed points. (A fixed point of a function f is a number z such that f (z) = z.) 3 Named after Leonard Euler (17071783). For more information about Euler, see http://www-groups.dcs.st-and.ac.uk/history/Biographies/Euler.html. CHAPTER 3. EXAMPLES OF FUNCTIONS 4. Prove Proposition 3.3. 5. Show that the Mbius transformation f (z) = o onto the imaginary axis. 1+z 1-z 36 maps the unit circle (minus the point z = 1) 6. Suppose that f is holomorphic on the region G and f (G) is a subset of the unit circle. Show that f is constant. (Hint: Consider the function 1+f (z) and use Exercise 5 and a variation of 1-f (z) Exercise 13 in Chapter 2.) a b 7. Suppose A = is a 2 2 matrix of complex numbers whose determinant ad - bc is c d non-zero. Then we can define a corresponding Mbius transformation TA by TA (z) = az+b . o cz+d Show that TA TB = TAB . (Here denotes composition and denotes matrix multiplication.) 2z 8. Let f (z) = z+2 . Draw two graphs, one showing the following six sets in the z plane and the other showing their images in the w plane. Label the sets. (You should only need to calculate the images of 0, 2, and -1 - i; remember that Mbius transformations preserve angles.) o (b) The y-axis, plus . (a) The x-axis, plus . (c) The line x = y, plus . (d) The circle with radius 2 centered at 0. (e) The circle with radius 1 centered at 1. (f) The circle with radius 1 centered at -1. 9. Find Mbius transformations satisfying each of the following. Write your answers in standard o form, as az+b . cz+d (b) 1 0, 1 + i 1, 2 . (Use the cross-ratio.) (c) 0 i, 1 1, -i. 10. Let C be the circle with center 1 + i and radius 1. Using the cross-ratio, with different choices of zk , find two different Mbius transformations that transform C onto the real axis plus o infinity. In each case, find the image of the center of the circle. 11. Let C be the circle with center 0 and radius 1. Find a Mbius transformation which transforms o C onto C and transforms 0 to 1 . 2 12. Describe the image of the region under the transformation: (a) The disk |z| < 1 under w = iz-i z+1 . z-i z+i . (a) 1 0, 2 1, 3 . (Use the cross-ratio.) (b) The quadrant x > 0, y > 0 under w = (c) The strip 0 < x < 1 under w = z z-1 . CHAPTER 3. EXAMPLES OF FUNCTIONS 37 13. The Jacobian of a transformation u = u(x, y), v = v(x, y) is the determinant of the matrix u x v x u y v y . Show that if f = u + iv is holomorphic then the Jacobian equals |f (z)|2 . z 2 -1 2z+1 . ^ 14. Find the fixed points in C of f (z) = 15. Find the Mbius transformation f : o (b) f maps 1 1, -1 i, -i -1. (a) f maps 0 1, 1 , 0. ^ 16. Suppose z1 , z2 and z3 are distinct points in C. Show that z is on the circle passing through z1 , z2 and z3 if and only if [z, z1 , z2 , z3 ] is real or infinite. 17. Find the image under the stereographic projection of the following points: (0, 0, -1), (0, 0, 1), (1, 0, 0), (0, 1, 0), (1, 1, 0). 18. Prove that the stereographic projection of Proposition 3.13 is a bijection by verifying that that -1 and -1 are the identity. 19. Consider the plane P determined by x + y - z = 0. What is a unit normal vector to P ? Compute the image of P S 2 under the stereographic projection . 20. Prove that every circle in the extended complex plane is the image of some circle in S 2 under the stereographic projection . 21. Describe the effect of the basic Mbius transformations rotation, real dilation, and translation o on the Riemann sphere. hint: for the first two, consider all circles in S 2 centered on the NS axis, and all circles through both N and S. For translation, consider two families of circles through N, `orthogonal' to and `perpendicular' to the translation. 22. Describe the images of the following sets under the exponential function exp(z): (b) the line segment defined by z = 1 + iy, 0 y 2. 23. Prove Lemma 3.16. 24. Prove Lemma 3.18. 25. Let z = x + iy and show that (a) sin z = sin x cosh y + i cos x sinh y. (b) cos z = cos x cosh y - i sin x sinh y. 26. Let z = x + iy and show that (a) the line segment defined by z = iy, 0 y 2. (c) f maps x-axis to y = x, y-axis to y = -x, and the unit circle to itself. (c) the rectangle {z = x + iy C : 0 x 1, 0 y 2}. CHAPTER 3. EXAMPLES OF FUNCTIONS (a) |sin z|2 = sin2 x + sinh2 y = cosh2 y - cos2 x. (c) If cos x = 0 then |cot z|2 = cosh2 y-1 cosh2 y 38 (b) |cos z|2 = cos2 x + sinh2 y = cosh2 y - sin2 x. 1. (d) If |y| 1 then |cot z|2 27. Show that tan(iz) = i tanh z. 28. Find the principal values of (a) log i. (b) (-1)i . (c) log(1 + i). 29. Is arg(z) = - arg(z) true for the multiple-valued argument? What about Arg(z) = - Arg(z) for the principal branch? 30. Is there a difference between the set of all values of log z 2 and the set of all values of 2 log z? (Try some fixed numbers for z.) 31. For each of the following functions, determine all complex numbers for which the function is holomorphic. If you run into a logarithm, use the principal value (unless stated otherwise). (a) z 2 . (b) sin z . z 3 +1 sinh2 y+1 sinh2 y =1+ 1 sinh2 y 1+ 1 sinh2 1 2. (d) exp(z). (f) iz-3 . (c) Log(z - 2i + 1) where Log(z) = ln |z| + i Arg(z) with 0 Arg(z) < 2. (e) (z - 3)i . 32. Find all solutions to the following equations: (a) Log(z) = i. 2 (b) Log(z) = 3 2 i. (c) exp(z) = i. (d) sin z = cosh 4. (e) cos z = 0. (f) sinh z = 0. (g) exp(iz) = exp(iz). (h) z 1/2 = 1 + i. 33. Find the image of the annulus 1 < |z| < e under the principal value of the logarithm. 34. Show that |az | = aRe z if a is a positive real constant. CHAPTER 3. EXAMPLES OF FUNCTIONS 35. Fix c C \ {0}. Find the derivative of f (z) = z c . 39 36. Prove that exp(b log a) is single-valued if and only if b is an integer. (Note that this means that complex exponentials don't clash with monomials z n .) What can you say if b is rational? 37. Describe the image under exp of the line with equation y = x. To do this you should find an equation (at least parametrically) for the image (you can start with the parametric form x = t, y = t), plot it reasonably carefully, and explain what happens in the limits as t and t -. 38. For this problem, f (z) = z 2 . (a) Show that the image of a circle centered at the origin is a circle centered at the origin. (b) Show that the image of a ray starting at the origin is a ray starting at the origin. (c) Let T be the figure formed by the horizontal segment from 0 to 2, the circular arc from 2 to 2i, and then the vertical segment from 2i to 0. Draw T and f (T ). (d) Is the right angle at the origin in part (c) preserved? Is something wrong here? (Hint: Use polar coordinates.) 39. As in the previous problem, let f (z) = z 2 . Let Q be the square with vertices at 0, 2, 2 + 2i and 2i. Draw f (Q) and identify the types of image curves corresponding to the segments from 2 to 2 + 2i and from 2 + 2i to 2i. They are not parts of either straight lines or circles. (Hint: You can write the vertical segment parametrically as z(t) = 2 + it. Eliminate the parameter in u + iv = f (z(t)) to get a (u, v) equation for the image curve.) ...
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This note was uploaded on 02/27/2012 for the course MATH 417 taught by Professor Staff during the Fall '11 term at SUNY Albany.

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