Lecture 9. = 1+z + 2! + z3. 1 = 0, it follows that the radius of convergence of (1) is.

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1 The Exponential Function Lecture 9 The exponential function 1 plays a central role in analysis, more so in the case of complex analysis and is going to be our first example using the power series method. We define exp z := e z := n 0 z n n! z2 = 1+z + 2! + z3 +. (1) 3! n Since lim 1 = 0, it follows that the radius of convergence of (1) is. n n! Hence from the above theorem, we have, exp is differentiable throughout C and its derivative is given by exp (z) = n 1 n n! zn 1 = exp(z) (2) for all z. Also, exp(0) = 1. Thus we see that exp is a solution of the initial value problem: f (z) = f(z); f(0) = 1. (3) It can be easily seen that any analytic function which is a solution of (3) has to be equal to exp. We can verify that exp(a+b) = exp(a)exp(b) for all complex numbers a, b, directly by using the product formula for power series. (Use binomial expansion of (a+b) n.) This can also be proved by using (3) which we shall leave it you as an entertaining exercise. Thus, we have shown that exp defines a homomorphism from the additive group C to the multiplicative group C := C \ {0}. As a simple consequence of this rule we have, exp(nz) = exp(z) n for all integers n. In particular, let us denote exp(1) by e. Then we have, exp(n) = e n. This is the justification to have the notation e z for exp(z). It may be worth recalling some elementary facts about e that you know already. For instance, clearly 2 < e. By comparing with the geometric series n 2 n, it can be shown easily that e < 3. Also we have, ( e = lim n. (4) n n) 1 Euler was the first one to introuduce the complex exponential function and study it extensively.

2 Also observe that ē z = e z. (5) directly from the power series definition. This, in particular, implies that e ıy 2 = e ıy e ıy = e ıy e ıy = 1. Hence, e ıy = 1. (6) This is equivalent to the familiar trigonometric formula: cos 2 θ + sin 2 θ = 1. Trigonometric Functions. Recall that d d (sin x) = cos x, dx (cos x) = sin x. dx From this and from the Taylor s series for sin x and cos x, it follows that sin x = x x3 3! + x5 5! + ; cos x = 1 x2 2! + x4 4! +. These formulae are valid on the entire of R, since the radii of convergence of the two series are. Motivated by this, we can define the complex trigonometric functions by sin z = z z3 3! + z5 z2 + ; cos z = 1 5! 2! + z4 4! +. (7) Check that sin z = eız e ız 2ı ; cos z = eız + e ız. (8) 2 It turns out that these complex trigonometric functions also have differentiability properties similar to the real case. Also, from (8) additive properties of sin and cos can be derived. as usual. Other trigonometric functions are defined in terms of sin and cos For example, we have tan z = sin z/ cos z and its domain of definition is all points in C at which cos z 0. Indeed, it is possible to redo the entire theory of trigonometric functions starting with definitions

3 8 and not taking any of the properties of real trigonometric functions for granted. In what follows, we shall obtain other properties of the exponential function by the formula e ız = cos z + ı sin z. (9) In particular, e x+ıy = e x e ıy = e x (cos x + ı sin y). (10) It follows that e 2πı = 1. Indeed, we shall prove that e z = 1 iff z = 2nπı, for some integer n. Observe that e x 0 for all x R and that if x > 0 then e x > 1. Hence for all x < 0, we have, e x = 1/e x < 1. It follows that e x = 1 iff x = 0. Let now z = x + ıy and e z = 1. This means that e x cos y = 1 and e x sin y = 0. Since e x 0 for any x, we must have, sin y = 0. Hence, y = mπ, for some integer m. Therefore e x cos mπ = 1. Since e x 1 for all x R, it follows that cos mπ = 1 and e x = 1. Therefore x = 0 and m = 2n, as desired. Finally, let us prove: exp(c) = C. (11) Write 0 w = r(cos θ +ı sin θ), r 0. Since e x is a monotonically increasing function and has the property e x 0, as x and e x as x, it follows from Intermediate Value Theorem that there exist x such that e x = r. (Here x is nothing but ln r.) Now take z = x + ıθ and verify that e z = w. This is one place, where we are heavily depending on the intuitive properties of the angle and the corresponding properties of the real sin and cos functions. We remark that it is possible to avoid this by defining sin and cos by the above formulae in terms of exp and derive all these properties rigorously from the properties of exp alone. Remark 1 One of the most beautiful equations: Euler: e πı + 1 = 0 (12) which relates in a simple arithmetic way, five of the most fundamental numbers, made Euler 2 believe in the existence of God! 2 See E.T. Bell s book for some juicy stories

4 Example 1 Let us study the mapping properties of tan function. Since it follows that tan z = sin z cos z, it follows that tan is defined and holomorphic at all points where cos z 0. Also, tan(z + nπı) = tan z. In order to determine the range of this function, we have to take an arbitrary w C and solve for tan z = w. Putting e ız = X, temporarily, this equation reduces to X2 1 ı(x 2 + 1) = w. Hence X2 = 1 + ıw. This latter equation has 1 ıw solution in X iff w ı. Once we obtain the values for ±X, which are obviously not equal to 0, from the ontoness of exp : C C \ {0}, one concludes that e ız = ±X also can be solved. From this analysis, it also follows that tan z 1 = tan z 2 iff z 1 = z 2 + nπı. Likewise, the hyperbolic functions are defined by sinh z = ez e z ; cosh z = ez + e z. (13) 2 2 It is easy to see that these functions are holomorphic. Moreover, all the usual identities which hold in the real case amongst these functions also hold in the complex case and can be verified directly. One can study the mapping properties of these functions as well, which have wide range of applications. Remark 2 Before we proceed onto another example, we would like to draw your attention to some special properties of the exponential and trigonometric functions. You are familiar with the real limit lim exp(x) =. x However, such a result is not true when we replace the real x by a complex z. In fact, given any complex number w 0, we have seen that there exists z such that exp(z) = w. But then exp(z + 2nπı) = w for all n. Hence we can get z having arbitrarily large modulus such that exp(z ) = w. As a consequence, it follows that lim z exp(z) does not exist. Using the formula for sin and cos in terms of exp, it can be easily shown that sin and cos are both surjective mappings of C onto C. In particular, remember that they are not bounded unlike their real counter parts.

5 Logarithm: We would like to define the logarithm as the inverse of exponential. However, as we have observed, unlike in the real case, the complex exponential function e z is not one-one, and hence its inverse is going to be a multi-valued function, or rather a set valued function. Thus indeed, for any z 0 ( this is needed!) any w C satisfying the equation e w = z is defined as log z (or ln z). Putting w = u + ıv, z = re ıθ, we see that e u = r, and v = θ. Thus w = ln z := ln r + ıθ = ln z + ı arg z. Observe that the multi-valuedness of ln z is caused by that of arg z. We have the identity [which directly follows from e w 1+w 2 ln(z 1 z 2 ) = ln z 1 +ln z 2, (14) = e w 1 ew 2 ]. Here, we have to interpret this identity set-theoretically. The principal value of the logarithm is defined by Log z = ln z + ıarg z. Recall that π < Arg z π. The notation Ln z is also in use. We shall respect both of them. Caution: When z is a positive real number, ln z has two meanings! Unless mentioned otherwise one should stick to the older meaning, viz., ln z = Ln z in that case. The logarithmic function is all too important to be left as a mere set-valued function. If we restrict the domain suitably, then we see that the argument can be defined continuously. In fact for this to hold, we must be careful that in our domain, which necessarily excludes the origin, we are not able to go around the origin. Thus for instance, if we throw away an entire ray emerging from the origin, from the complex plane, then for each point of the remaining domain a continuous value of the argument can be chosen. This in turn, defines a continuous value of the logarithmic function also. We make a formal definition. Definition 1 Given a multi-valued function φ, on a domain Ω, by a branch of φ we mean a specific continuous function ψ : Ω C such that ψ(z) φ(z) for all z in a specified domain. For instance, if f is a function which is not one-one, then its inverse is a multi-valued function. Then any continuous function g such that

6 g f = Id over a suitable domain will be called a branch of f 1. In particular, branches of the inverse of the exponential function are called branches of the logarithmic function. Over domains such as C\L where L is an infinite half-ray from the origin, we easily see that ln has countably infinite number of branches. lemma: The idea of having such a definition is fully justified by the following Lemma 1 Let f : Ω 1 Ω 2 be a holomorphic function, g : Ω 2 Ω 1 be a continuous function such that f g(w) = w, w Ω 2. Suppose w 0 Ω 2 is such that f (z 0 ) 0, where z 0 = g(w 0 ). Then g is C differentiable at w 0, with g (w 0 ) = (f (z 0 )) 1. Proof : Since, it follows that, f(z 0 + h) f(z 0 ) lim h 0 h lim h 0 = f (z 0 ) 0, h f(z 0 + h) f(z 0 ) = 1 f (z 0 ). Therefore given ɛ > 0 there exists δ 1 > 0, such that, h f(z 0 + h) f(z 0 ) 1 f (z 0 ) < ɛ, 0 < h < δ 1. (15) Now, let δ 2 > 0 be sufficiently small so that B δ2 (w 0 ) Ω 2. For 0 < k < δ 2, put h := g(w 0 +k) g(w 0 ) 0. By the continuity of g, there exists δ 3 such that δ 2 > δ 3 > 0, and 0 < k < δ 3 = h = g(w 0 + k) g(w 0 ) < δ 1. (16) Since g(w 0 ) = z 0, we have, g(w 0 +k) = z 0 +h and hence, f(z 0 +h) = w 0 +k. Thus, k = f(z 0 +h) f(z 0 ). Therefore, whenever 0 < k < δ 2, from (15) and (16), we have, g(w 0 + k) g(w 0 ) k (f (g(w 0 ))) 1 = z 0 + h z 0 f(z 0 + h) f(z 0 ) 1 f (z 0 ) < ɛ. This completes the proof of the lemma. Remark 3

7 1. Observe that as a corollary, we have obtained holomorphic branches of the logarithmic function. For instance, l(z) := ln r + ıθ, 0 < θ < 2π, is one such branch defined over the entire of C minus the positive real axis. The question of the nature of domains on which ln has well defined branches will be discussed later on. 2. The hypothesis that f (z 0 ) 0 is indeed unnecessary in the above lemma. This stronger version of the above lemma will be proved in Ch.5. In contrast, in the real case, consider the function x x 3 which defines a continuous bijection of the real line onto itself. Its inverse is also continuous but not differentiable at 0 as can be seen easily in different ways. Example 2 Let us find out the derivative of a branch l(z) of the logarithm. We shall show that d dz (ln z) := l (z) = 1. Since, exp l = Id, it z follows from the chain rule that (exp) (l(z))l (z) = 1. Therefore, we have, zl (z) = 1 and hence, l (z) = 1/z, as claimed. Remark 4 Likewise, we could also discuss the inverse of trigonometric functions. They too are multi-valued. However, for continuous choice of a branch over a suitable open set, they will be holomorphic. Thus we now have a large class of holomorphic functions polynomials, exponential function, trigonometric functions. We can also take linear combinations of these and their quotients. We can even consider the well chosen branches of the inverses of these functions and so on. Together, all these constitute what is known as the class of elementary functions. In a latter section we shall again discuss mapping properties of some of them. Complex Exponents of Complex Numbers Recall that defining exponents was somewhat involved process, even with positive real numbers. Now, we want to deal with this concept with complex numbers. Here the idea is to use the logarithm function which converts multiplication into addition and hence the exponent] into multiplication. For any two complex numbers z, w C \ {0}, define z w := e w ln z. (17)

8 Observe that on the rhs the term lnz is a multi-valued function. Therefore, in general, this makes z w a set of complex numbers rather than a single number. For instance, 2 1/2 is a two element set viz., { 2, 2}. First, let us take the simplest case, viz., z 2 = n 1. Then irrespective of the value of z 1, (17) gives the single value which is equal toz 1 multiplied with itself n times. For negative integer exponents also, the story is the same. But as soon as z 2 is not an integer, we can no longer say that this single-valued. Does this definition follow the familiar laws of exponents: z w 1+w 2 = z1 w z2 w ; (z 1 z 2 ) w = z1 w z2 w? (18) Yes indeed. The only caution is that these formulae tell you that the two terms on either side of the equality sign are equal as sets. Just keep using (14). Finally we observe that if in some domain we have chosen a particular branch of lnz then in that domain (17) defines a (holomorphic) branch of the exponent.