Chapter2 - Chapter 2 Differentiation Mathematical study and research are very suggestive of mountaineering Whymper made several efforts before he

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Unformatted text preview: Chapter 2 Differentiation Mathematical study and research are very suggestive of mountaineering. Whymper made several efforts before he climbed the Matterhorn in the 1860's and even then it cost the life of four of his party. Now, however, any tourist can be hauled up for a small cost, and perhaps does not appreciate the difficulty of the original ascent. So in mathematics, it may be found hard to realise the great initial difficulty of making a little step which now seems so natural and obvious, and it may not be surprising if such a step has been found and lost again. Louis Joel Mordell (18881972) 2.1 First Steps A (complex) function f is a mapping from a subset G C to C (in this situation we will write f : G C and call G the domain of f ). This means that each element z G gets mapped to exactly one complex number, called the image of z and usually denoted by f (z). So far there is nothing that makes complex functions any more special than, say, functions from Rm to Rn . In fact, we can construct many familiar looking functions from the standard calculus repertoire, such 1 as f (z) = z (the identity map), f (z) = 2z + i, f (z) = z 3 , or f (z) = z . The former three could be defined on all of C, whereas for the latter we have to exclude the origin z = 0. On the other hand, we could construct some functions which make use of a certain representation of z, for example, f (x, y) = x - 2iy, f (x, y) = y 2 - ix, or f (r, ) = 2rei(+) . Maybe the fundamental principle of analysis is that of a limit. The philosophy of the following definition is not restricted to complex functions, but for sake of simplicity we only state it for those functions. Definition 2.1. Suppose f is a complex function with domain G and z0 is an accumulation point of G. Suppose there is a complex number w0 such that for every > 0, we can find > 0 so that for all z G satisfying 0 < |z - z0 | < we have |f (z) - w0 | < . Then w0 is the limit of f as z approaches z0 , in short lim f (z) = w0 . zz0 This definition is the same as is found in most calculus texts. The reason we require that z0 is an accumulation point of the domain is just that we need to be sure that there are points z of the domain which are arbitrarily close to z0 . Just as in the real case, the definition does not require 13 CHAPTER 2. DIFFERENTIATION 14 that z0 is in the domain of f and, if z0 is in the domain of f , the definition explicitly ignores the value of f (z0 ). That is why we require 0 < |z - z0 |. Just as in the real case the limit w0 is unique if it exists. It is often useful to investigate limits by restricting the way the point z "approaches" z0 . The following is a easy consequence of the definition. Lemma 2.2. Suppose limzz0 f (z) exists and has the value w0 , as above. Suppose G0 G, and suppose z0 is an accumulation point of G0 . If f0 is the restriction of f to G0 then limzz0 f0 (z) exists and has the value w0 . The definition of limit in the complex domain has to be treated with a little more care than its real companion; this is illustrated by the following example. z Example 2.3. lim does not exist. z0 z To see this, we try to compute this "limit" as z 0 on the real and on the imaginary axis. In the first case, we can write z = x R, and hence z x x = lim = lim = 1 . z0 z x0 x x0 x lim In the second case, we write z = iy where y R, and then z iy -iy = lim = lim = -1 . z0 z y0 iy y0 iy lim So we get a different "limit" depending on the direction from which we approach 0. Lemma 2.2 then implies that limz0 z does not exist. z On the other hand, the following "usual" limit rules are valid for complex functions; the proofs of these rules are everything but trivial and make for nice exercises. Lemma 2.4. Let f and g be complex functions and c, z0 C. If limzz0 f (z) and limzz0 g(z) exist, then: (a) lim f (z) + c lim g(z) = lim (f (z) + c g(z)) zz0 zz0 zz0 zz0 zz0 (b) lim f (z) lim g(z) = lim (f (z) g(z)) zz0 zz0 (c) lim f (z)/ lim g(z) = lim (f (z)/g(z)) ; zz0 zz0 In the last identity we also require that limzz0 g(z) = 0. Because the definition of the limit is somewhat elaborate, the following fundamental definition looks almost trivial. Definition 2.5. Suppose f is a complex function. If z0 is in the domain of the function and either z0 is an isolated point of the domain or zz0 lim f (z) = f (z0 ) then f is continuous at z0 . More generally, f is continuous on G C if f is continuous at every z G. CHAPTER 2. DIFFERENTIATION Just as in the real case, we can "take the limit inside" a continuous function: 15 Lemma 2.6. If f is continuous at w0 and limzz0 g(z) = w0 then limzz0 f (g(z)) = f (w0 ). In other words, lim f (g(z)) = f lim g(z) . zz0 zz0 2.2 Differentiability and Holomorphicity The fact that limits such as limz0 z do not exist points to something special about complex z numbers which has no parallel in the reals--we can express a function in a very compact way in one variable, yet it shows some peculiar behavior "in the limit." We will repeatedly notice this kind of behavior; one reason is that when trying to compute a limit of a function as, say, z 0, we have to allow z to approach the point 0 in any way. On the real line there are only two directions to approach 0--from the left or from the right (or some combination of those two). In the complex plane, we have an additional dimension to play with. This means that the statement "A complex function has a limit..." is in many senses stronger than the statement "A real function has a limit..." This difference becomes apparent most baldly when studying derivatives. Definition 2.7. Suppose f : G C is a complex function and z0 is an interior point of G. The derivative of f at z0 is defined as f (z0 ) = lim zz0 f (z) - f (z0 ) , z - z0 provided this limit exists. In this case, f is called differentiable at z0 . If f is differentiable for all points in an open disk centered at z0 then f is called holomorphic at z0 . The function f is holomorphic on the open set G C if it is differentiable (and hence holomorphic) at every point in G. Functions which are differentiable (and hence holomorphic) in the whole complex plane C are called entire. The difference quotient limit which defines f (z0 ) can be rewritten as f (z0 ) = lim f (z0 + h) - f (z0 ) . h0 h This equivalent definition is sometimes easier to handle. Note that h is not a real number but can rather approach zero from anywhere in the complex plane. The fact that the notions of differentiability and holomorphicity are actually different is seen in the following examples. Example 2.8. The function f (z) = z 3 is entire, that is, holomorphic in C: For any z0 C, zz0 lim 3 f (z) - f (z0 ) z 3 - z0 = lim zz0 z - z0 z - z0 2 (z 2 + zz0 + z0 )(z - z0 ) zz0 z - z0 2 2 2 = lim z + zz0 + z0 = 3z0 . = lim zz0 CHAPTER 2. DIFFERENTIATION 16 Example 2.9. The function f (z) = z 2 is differentiable at 0 and nowhere else (in particular, f is not holomorphic at 0): Let's write z = z0 + rei . Then z2 - z0 = z - z0 2 z0 + rei z0 + rei - z0 2 - z0 2 2 z0 + re-i - z0 2 = rei 2 + 2z re-i + r 2 e-2i - z 2 z0 0 0 = rei 2z0 re-i + r2 e-2i = = 2z0 e-2i + re-3i . rei If z0 = 0 then the limit of the right-hand side as z z0 does not exist since r 0 and we get different answers for horizontal approach ( = 0) and for vertical approach ( = /2). (A more entertaining way to see this is to use, for example, z(t) = z0 + 1 eit , which approaches z0 as t .) t On the other hand, if z0 = 0 then the right-hand side equals re-3i = |z|e-3i . Hence 2 z lim = lim |z|e-3i = lim |z| = 0 , z z0 z0 z0 which implies that z2 = 0. z0 z lim Example 2.10. The function f (z) = z is nowhere differentiable: lim z - z0 z - z0 z = lim = lim z - z0 zz0 z - z0 z0 z zz0 does not exist, as discussed earlier. The basic properties for derivatives are similar to those we know from real calculus. In fact, one should convince oneself that the following rules follow mostly from properties of the limit. (The `chain rule' needs a little care to be worked out.) Lemma 2.11. Suppose f and g are differentiable at z C, and that c C, n Z, and h is differentiable at g(z). (a) f (z) + c g(z) = f (z) + c g (z) (b) f (z) g(z) = f (z)g(z) + f (z)g (z) f (z)g(z) - f (z)g (z) (c) f (z)/g(z) = g(z)2 (d) z n = nz n-1 (e) h(g(z)) = h (g(z))g (z) . In the third identity we have to be aware of division by zero. CHAPTER 2. DIFFERENTIATION 17 We end this section with yet another differentiation rule, that for inverse functions. As in the real case, this rule is only defined for functions which are bijections. A function f : G H is one-to-one if for every image w H there is a unique z G such that f (z) = w. The function is onto if every w H has a preimage z G (that is, there exists a z G such that f (z) = w). A bijection is a function which is both one-to-one and onto. If f : G H is a bijection then g is the inverse of f if for all z H, f (g(z)) = z. Lemma 2.12. Suppose G and H are open sets in C, f : G H is a bijection, g : H G is the inverse function of f , and z0 H. If f is differentiable at g(z0 ), f (g(z0 )) = 0, and g is continuous at z0 then g is differentiable at z0 with g (z0 ) = Proof. We have: g (z0 ) = lim zz0 1 . f (g(z0 )) g(z) - g(z0 ) g(z) - g(z0 ) 1 = lim = lim . zz0 f (g(z)) - f (g(z0 )) zz0 f (g(z)) - f (g(z0 )) z - z0 g(z) - g(z0 ) 1 . f (g(z)) - f (g(z0 )) g(z) - g(z0 ) Because g(z) g(z0 ) as z z0 , we obtain: g (z0 ) = g(z)g(z0 ) lim Finally, as the denominator of this last term is continuous at z0 , by Lemma 2.6 we have: g (z0 ) = 1 1 = . f (g(z)) - f (g(z0 )) f (g(z0 ) lim g(z) - g(z0 ) g(z)g(z0 ) 2.3 The CauchyRiemann Equations When considering real-valued functions f (x, y) : R2 R of two variables, there is no notion of `the' derivative of a function. For such functions, we instead only have partial derivatives fx (x, y) and fy (x, y) (and also directional derivatives) which depend on the way in which we approach a point (x, y) R2 . For a complex-valued function f (z) = f (x, y) : C R, we now have a new concept of derivative, f (z), which by definition cannot depend on the way in which we approach a point (x, y) C. It is logical, then, that there should be a relationship between the complex derivative f (z) and the partial derivatives f (z) and f (z) (defined exactly as in the real-valued x y case). The relationship between the complex derivative and partial derivatives is very strong and is a powerful computational tool. It is described by the CauchyRiemann Equations, named after Augustin Louis Cauchy (17891857)1 and Georg Friedrich Bernhard Riemann (18261866)2 .: For more information about Cauchy, see http://www-groups.dcs.st-and.ac.uk/history/Biographies/Cauchy.html. 2 For more information about Riemann, see http://www-groups.dcs.st-and.ac.uk/history/Biographies/Riemann.html. 1 CHAPTER 2. DIFFERENTIATION 18 Theorem 2.13. (a) Suppose f is differentiable at z0 = x0 + iy0 . Then the partial derivatives of f satisfy f f (z0 ) = -i (z0 ) . (2.1) x y (b) Suppose f is a complex function such that the partial derivatives fx and fy exist in an open disk centered at z0 and are continuous at z0 . If these partial derivatives satisfy (2.1) then f is differentiable at z0 . In both cases (a) and (b), f is given by f (z0 ) = f (z0 ) . x Remarks. 1. It is traditional, and often convenient, to write the function f in terms of its real and imaginary parts. That is, we write f (z) = f (x, y) = u(x, y) + iv(x, y) where u is the real part of f and v is the imaginary part. Then fx = ux + ivx and -ify = -i(uy + ivy ) = vy - iuy . Using this terminology we can rewrite the equation (2.1) equivalently as the following pair of equations: ux (x0 , y0 ) = vy (x0 , y0 ) uy (x0 , y0 ) = -vx (x0 , y0 ) . (2.2) 2. As stated, (a) and (b) are not quite converse statements. However, we will later show that if f is holomorphic at z0 = x0 + iy0 then u and v have continuous partials (of any order) at z0 . That is, later we will prove that f = u + iv is holomorphic in an open set G if and only if u and v have continuous partials that satisfy (2.2) in G. 3. If u and v satisfy (2.2) and their second partials are also continuous then we obtain uxx (x0 , y0 ) = vyx (x0 , y0 ) = vxy (x0 , y0 ) = -uyy (x0 , y0 ) , that is, uxx (x0 , y0 ) + uyy (x0 , y0 ) = 0 and an analogous identity for v. Functions with continuous second partials satisfying this partial differential equation are called harmonic; we will study such functions in Chapter 6. Again, as we will see later, if f is holomorphic in an open set G then the partials of any order of u and v exist; hence we will show that the real and imaginary part of a function which is holomorphic on an open set are harmonic on that set. Proof of Theorem 2.13. (a) If f is differentiable at z0 = (x0 , y0 ) then f (z0 ) = lim f (z0 + z) - f (z0 ) . z0 z As we saw in the last section we must get the same result if we restrict z to be on the real axis and if we restrict it to be on the imaginary axis. In the first case we have z = x and f (z0 ) = lim f (z0 + x) - f (z0 ) f (x0 + x, y0 ) - f (x0 , y0 ) f = lim = (x0 , y0 ). x0 x0 x x x CHAPTER 2. DIFFERENTIATION In the second case we have z = iy and f (z0 ) = lim f (z0 + iy) - f (z0 ) 1 f (x0 , y0 + y) - f (x0 , y0 ) f = lim = -i (x0 , y0 ) iy0 y0 i iy y y 19 (using 1 = -i). Thus we have shown that f (z0 ) = fx (z0 ) = -ify (z0 ). i (b) To prove the statement in (b), "all we need to do" is prove that f (z0 ) = fx (z0 ), assuming the CauchyRiemann equations and continuity of the partials. We first rearrange a difference quotient for f (z0 ), writing z = x + iy: f (z0 + z) - f (z0 ) f (z0 + z) - f (z0 + x) + f (z0 + x) - f (z0 ) = z z f (z0 + x + iy) - f (z0 + x) f (z0 + x) - f (z0 ) = + z z y f (z0 + x + iy) - f (z0 + x) x f (z0 + x) - f (z0 ) = + . z y z x Now we rearrange fx (z0 ): fx (z0 ) = z iy + x y x fx (z0 ) = fx (z0 ) = ifx (z0 ) + fx (z0 ) z z z z y x = fy (z0 ) + fx (z0 ) , z z where we used equation (2.1) in the last step to convert ifx to i(-ify ) = fy . Now we subtract our two rearrangements and take a limit: f (z0 + z) - f (z0 ) - fx (z0 ) z0 z y f (z0 + x + iy) - f (z0 + x) = lim - fy (z0 ) z0 z y x f (z0 + x) - f (z0 ) + lim - fx (z0 ) . z0 z x lim (2.3) We need to show that these limits are both 0. The fractions x/z and y/z are bounded by 1 in modulus so we just need to see that the limits of the expressions in parentheses are 0. The second term in (2.3) has a limit of 0 since, by definition, fx (z0 ) = lim f (z0 + x) - f (z0 ) x0 x and taking the limit as z 0 is the same as taking the limit as x 0. We can't do this for the first expression since both x and y are involved, and both change as z 0. For the first term in (2.3) we apply Theorem 1.16, the real mean-value theorem, to the real and imaginary parts of f . This gives us real numbers a and b, with 0 < a, b < 1, so that u(x0 + x, y0 + y) - u(x0 + x, y0 ) = uy (x0 + x, y0 + ay) y v(x0 + x, y0 + y) - v(x0 + x, y0 ) = vy (x0 + x, y0 + by) . y CHAPTER 2. DIFFERENTIATION Using these expressions, we have f (z0 + x + iy) - f (z0 + x) - fy (z0 ) y = uy (x0 + x, y0 + ay) + ivy (x0 + x, y0 + by) - (uy (x0 , y0 ) + ivy (x0 , y0 )) 20 = (uy (x0 + x, y0 + ay) - uy (x0 , y0 )) + i (vy (x0 + x, y0 + ay) - vy (x0 , y0 )) . Finally, the two differences in parentheses have zero limit as z 0 because uy and vy are continuous at (x0 , y0 ). 2.4 Constant Functions As an example application of the Cauchy-Riemann Equations, we consider functions which have a derivative of 0. One of the first applications of the Mean-Value Theorem for real-valued functions, Theorem 1.16, is to show that if a function has zero derivative everywhere on an interval then it must be constant. Lemma 2.14. If f : I R is a real-valued function with f (x) defined and equal to 0 for all x I, then there is a constant c R such that f (x) = c for all x I. Proof. The proof is easy: The Mean-Value Theorem says that for any x, y I, f (y) - f (x) = f (x + a(y - x))(y - x) for some 0 < a < 1. If we know that f is always zero then we know that f (x + a(y - x)) = 0, so the above equation yields f (y) = f (x). Since this is true for any x, y I, f must be constant. There is a complex version of the Mean-Value Theorem, but we defer its statement to another course. Instead, we will use a different argument to prove that complex functions with derivative that are always 0 must be constant. Lemma 2.14 required two key features of the function f , both of which are somewhat obviously necessary. The first is that f be differentiable everywhere in its domain. In fact, if f is not differentiable everywhere, we can construct functions which have zero derivative `almost' everywhere but which have infinitely many values in their range. The second key feature is that the interval I is connected. It is certainly important for the domain to be connected in both the real and complex cases. For instance, if we define 1 if Re z > 0, f (z) = -1 if Re z < 0, then f (z) = 0 for all z in the domain of f but f is not constant. This may seem like a silly example, but it illustrates a pitfall to proving a function is constant that we must be careful of. Recall that a region of C is an open connected subset. Theorem 2.15. If the domain of f is a region G C and f (z) = 0 for all z in G then f is a constant. CHAPTER 2. DIFFERENTIATION 21 Proof. We will show that f is constant along horizontal segments and along vertical segments in G. Then, if x and y are two points in G which can be connected by horizontal and vertical segments, we have that f (x) = f (y). But any two points of a region may be connected by finitely many such segments by Theorem 1.14, so f has the same value at any two points of G, proving the theorem. To see that f is constant along horizontal segments, suppose that H is a horizontal line segment in G. Since H is a horizontal segment, there is some value y0 R so that the imaginary part of any z H is Im(z) = y0 . Consider the real part u(z) of the function f . Since Im(z) is constant on H, we can consider u(z) to be just a function of x, the real part of z = x + iy0 . By assumption, f (z) = 0, so for z H we have ux (z) = Re(f (z)) = 0. Thus, by Lemma 2.14, u(z) is constant on H. We can argue the same way to see that the imaginary part v(z) of f (z) is constant on H, since vx (z) = Im(f (z)) = 0. Since both the real and imaginary parts of f are constant on H, f itself is constant on H. This same argument works for vertical segments, interchanging the roles of the real and imaginary parts, so we're done. There are a number of surprising applications of this basic theorem; see Exercises 13 and 14 for a start. Exercises 1. Use the definition of limit to show that limzz0 (az + b) = az0 + b. 2. Evaluate the following limits or explain why they don't exist. (a) lim iz -1 . z+i zi 3 (b) z1-i lim x + i(2x + y). 3. Prove Lemma 2.4. 4. Prove Lemma 2.4 by using the formula for f given in Theorem 2.13. 1 5. Apply the definition of the derivative to give a direct proof that f (z) = - z12 when f (z) = z . 6. Show that if f is differentiable at z then f is continuous at z. 7. Prove Lemma 2.6. 8. Prove Lemma 2.11. 9. If u(x, y) and v(x, y) are continuous (respectively differentiable) does it follow that f (z) = u(x, y) + iv(x, y) is continuous (resp. differentiable)? If not, provide a counterexample. 10. Where are the following functions differentiable? Where are they holomorphic? Determine their derivatives at points where they are differentiable. (a) f (z) = e-x e-iy . (b) f (z) = 2x + ixy 2 . CHAPTER 2. DIFFERENTIATION (c) f (z) = x2 + iy 2 . (d) f (z) = ex e-iy . (e) f (z) = cos x cosh y - i sin x sinh y. (f) f (z) = Im z. (g) f (z) = |z|2 = x2 + y 2 . (i) f (z) = ix+1 y . 22 (h) f (z) = z Im z. (k) f (z) = 2xy - i(x + y)2 . (l) f (z) = z 2 - z 2 . (j) f (z) = 4(Re z)(Im z) - i(z)2 . 11. Prove that if f (z) is given by a polynomial in z then f is entire. What can you say if f (z) is given by a polynomial in x = Re z and y = Im z? 12. Consider the function xy(x + iy) f (z) = x2 + y 2 0 if z = 0, if z = 0. (As always, z = x + iy.) Show that f satisfies the CauchyRiemann equations at the origin z = 0, yet f is not differentiable at the origin. Why doesn't this contradict Theorem 2.13 (b)? 13. Prove: If f is holomorphic in the region G C and always real valued, then f is constant in G. (Hint: Use the CauchyRiemann equations to show that f = 0.) 14. Prove: If f (z) and f (z) are both holomorphic in the region G C then f (z) is constant in G. 15. Suppose f (z) is entire, with real and imaginary parts u(z) and v(z) satisfying u(z)v(z) = 3 for all z. Show that f is constant. 16. Is x x2 +y 2 harmonic? What about x2 ? x2 +y 2 17. The general real homogeneous quadratic function of (x, y) is u(x, y) = ax2 + bxy + cy 2 , where a, b and c are real constants. (b) If u is harmonic then show that it is the real part of a function of the form f (z) = Az 2 , where A is a complex constant. Give a formula for A in terms of the constants a, b and c. (a) Show that u is harmonic if and only if a = -c. ...
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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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