Math 0B Supplement Bill Helton September 3, 004 1
Supplement to Appendix G Supplement to Appendix G and Chapters 7 and 9 of Stewart Calculus Edition 5: August 003 Contents 1 Complex Exponentials: For Appendix G Stewart Edition 5 3 1.1 Complex Exponentials Yield trigonometric Identities.................. 4 1. Exercises.......................................... 5 Integration of Functions which Take Complex Values: For Chapter 7. Stewart Edition 5 6.1 Integrating Products of Sines, Cosines and Exponentials................ 6. Exercises.......................................... 9 3 The Fundamental Theorem of Algebra: For Chapter 7.4 Stewart Edition 5 10 3.1 Zeroes and their multiplicity................................ 10 3. Real Coefficients...................................... 11 3.3 Rational Functions and Poles............................... 1 3.4 Exercises.......................................... 13 4 Partial Fraction Expansions PFE): For Chapter 7.4 Stewart Edition 5 14 4.1 A Shortcut when there are no Repeated Factors..................... 14 4. The Difficulty with Repeated Factors........................... 16 4.3 Every Rational Function has a Partial Fraction Expansion............... 16 4.4 The Form of the PFE................................... 17 4.5 Exercises.......................................... 18 5 Improving on Euler s Method: For Chapter 9. Stewart Ed. 5 19 5.1 Exercises.......................................... 1 6 Appendix: Differentiation of Complex Functions 6.1 Deriving the Formula for e z Using Differentiation....................
Supplement to Appendix G 3 1 Complex Exponentials: For Appendix G Stewart Edition 5 This material is a supplement to Appendix G of Stewart. You should read the appendix, except maybe the last section on complex exponentials, before this material. How should we define e a+bi where a and b are real numbers? In other words, what is e z when z is a complex number? We would like the nice properties of the exponential to still be true. Probably, some of the most basic properties are that for any complex numbers z and w we have e z+w = e z e w and d dx ewx = we wx. 1.1) It turns out that the following definition produces a function with these properties. Definition of complex exponential: e a+bi = e a cos b + i sin b) = e a cos b + ie a sin b In particular, for any real number x, Euler s formula holds true: We now prove the first key property in 1.1). Theorem 1.1 If z and w are complex numbers, then Proof. Let z = a + ib and w = h + ik, then e ix = cos x + i sin x. 1.) e z+w = e z e w. e z e w = e a cos b + i sin b)e h cos k + i sin k) = e a e h [cos b cos k sin b sin k) + icos b sin k + sin b cos k)] = e a+h [cosb + k) + i sinb + k)] = e [a+h+ib+k)] = e z+w. We leave checking the second property to the exercises. For those who are interested there is an appendix, Section 6, which discusses what we mean by derivative of a function of complex variables and explains how to obtain the second property as well. It s easy to get formulas for the trigonometric functions in terms of the exponential. Look at Euler s formula 1.) with x replaced by x: e ix = cos x i sin x. We now have two equations in cos x and sin x, namely cos x + i sin x = e ix cos x i sin x = e ix.
Supplement to Appendix G 4 Adding and dividing by gives us cos x whereas subtracting and dividing by i gives us sin x: Exponential form of sine and cosine: cos x = eix + e ix, sin x = eix e ix Setting x = z = a + bi gives formulas for the sine and cosine of complex numbers. We can do a variety of things with these formula. Here are some we will not pursue: Since the other trigonometric functions are rational functions of sine and cosine, this gives us formulas for all the trigonometric functions. The hyperbolic and trigonometric functions are related: cos x = coshix) and i sin x = sinhix). 1.1 Complex Exponentials Yield trigonometric Identities The exponential formulas we just derived, together with e z+w = e z e w imply the identities sin α + cos α = 1 sinα + β) = sin α cos β + cos α sin β cosα + β) = cos α cos β sin α sin β. These three identities are the basis for deriving trigonometric identities. Hence we can derive trigonometric identities by using the exponential formulas and e z+w = e z e w. We now illustrate this with some examples. Example 1. Show that cos x + sin x = 1. Indeed, we have e ix + e ix ) e ix + e ix ) + = 1 e ix ) + + e ix ) + eix ) + e ix ) ) i 4 i wherein we have used the fact that i = 1. Example 1.3 sinx) = eix e ix = 1 4 + ) = 1, i = eix e ix ) i = sin x cos x. = 1 i e ix ) e ix ) ) e ix + e ix ) i
Supplement to Appendix G 5 1. Exercises 1. Use the relationship between the sine, cosine and exponential functions to express cos 3 x) as a sum of sines and cosines.. Show that e πi +1 = 0. This uses several basic concepts in mathematics, such as π, e, addition, multiplication and exponentiation of complex numbers in one compact equation. 3. What are the cartesian coordinates x and y of the complex number x + iy = e +3i? 4. Use the fact that and the product rule to prove that d ) ) cosbx) + i sinbx) = b sinbx) + i cosbx) dx d e a+ib)x) = a + ib)e a+ib)x. dx This is the key differentiation property for complex exponentials.
Supplement to Appendix G 6 Integration of Functions which Take Complex Values: For Chapter 7. Stewart Edition 5 This supplements Chapter 7. of Stewart Ed. 5. Now we turn to the issue of integrating functions which take complex values. Of course this is bound up with what we mean by antiderivatives of complex functions. A function, such as fx) = 1 + i)x + i3x, may have complex values but the variable x is only allowed to take on real values and we only define definite integrals for this type of functions. In this case nothing differs from what we already learned about integrals of real valued functions. The Riemann sum definition of an integral still applies. The Fundamental Theorem of Calculus is still true. The properties of integrals, including substitution and integration by parts still work. For example, 0 1 + i)x + 3ix ) dx = 0 x dx + i 0 x dx + 3i = 1 + i) + 8i = + 1i. 0 x dx = 1 + i)x + ix 3 On the other hand, we can t evaluate 1 0 x + i) 1 dx right now. Why is that? We would expect to write x + i) 1 dx = lnx + i) + C and use the Fundamental Theorem of Calculus, but this has no meaning 1 because we only know how to compute logarithms of positive numbers..1 Integrating Products of Sines, Cosines and Exponentials In Section 7. products of sines and cosines were integrated using trigonometric identities. There are other ways to do this now that we have complex exponentials. Examples will make this clearer. Example.1 Let s integrate 8 cos3x) sin x. e 3ix + e 3ix ) e ix e ix ) 8 cos3x) sin x = 8 i = i It is not difficult to integrate this, namely, 8 cos3x) sin x dx = i e 4ix + e ix e ix e 4ix). e 4ix + e ix e ix e 4ix ) dx 0 = i e 4ix 4i e ix i eix i ) + e 4ix + C. 4i 1 Some of you might suggest that we write ln x+i instead of lnx+i). This does not work. Since x+i = x + 1, the function fx) = ln x + i only takes on real values when x is real. Its derivative cannot be the complex number x + i) 1 since fx + h) fx))/h is real.
Supplement to Appendix G 7 Sort this by powers of e ±x to get e 4ix + e 4ix e ix i 4i 4i i ) eix + C = cos4x) + cosx) + C. i Example. Let s integrate e x sin x. Problems like this were solved in Section 7.1 by using integration by parts twice. Here is another way. Using the formula for sine and integrating we have e x sin x dx = 1 e x e ix e ix ) dx = 1 e +i)x e i)x ) dx i i ) = 1 e +i)x e i)x + C i + i i = iex e ix ) i) e ix + i) + C. 5 5 Sort by powers of e ±x to get ie x [ e ix e ix ) ie +ix + e ix ) ] + C 10 = e x 5 sin x + 1 ) 5 cos x + C. This method works for integrals of products of sines, cosines and exponentials, and often for quotients of them, though this requires more advanced methods, such as partial fractions). The advantage of using complex exponentials is that it takes the guess out of computing such integrals. The method, however, could be messier than the one presented in the book, though it is often simpler. We also point out that Example.1 went beyond those illustrated in the book. Here are some more examples that illustrate this method. Example.3 Let s integrate cos 3 x) cos7x). We have e cos 3 ix + e ix ) 3 x) = = ei3x + 3e ix + 3e ix + e i3x, cos7x) = ei7x + e i7x. 8 Therefore cos 3 x) cos7x) dx = 1 16 = 1 16 = 1 16 = 1 16 e i3x + 3e ix + 3e ix + e i3x )e i7x + e i7x ) dx e i10x + e i4x + 3e i8x + 3e i6x + 3e i6x + 3e i8x + e i4x + e i10x ) dx cos10x) + cos4x) + 6 cos8x) + 6 cos6x)) dx 10 sin10x) + 4 sin4x) + 6 8 sin8x) + 6 ) 6 sin6x) + C = 1 80 sin10x) + 3 64 sin8x) + 1 16 sin6x) + 1 3 sin4x) + C.
Supplement to Appendix G 8 Example.4 Let s integrate sin x)e 5x. e sin ix e ix ) x) = = 1 i 4 eix + e ix ). It follows that sin x)e 5x dx = 1 4 e ix + e ix )e 5x dx = 1 e x 5+i) 5x e + e x 5 i) ) dx 4 ) = 1 e x 5+i) e 5x + ex 5 i) + C 4 5 + i 5 5 i = e 5x e ix 5 i) 4 9 = e 5x 4 + e ix ) 5 + i) + C 5 9 ) 5 9 eix + e ix ) i 9 eix e ix ) + C 5 ) = e 5x 5 18 cosx) + 1 9 sinx) 5 10 + C. Example.5 Let s integrate x cos 3 x). Therefore e cos 3 ix + e ix ) 3 x) = = ei3x + 3e ix + 3e ix + e i3x. 8 cos 3 x) dx = 1 xe i3x + 3e ix + 3e ix + e i3x ) dx 8 = 1 x cos3x) + 6 cos x) dx 8 = 1 ) x cos3x) dx + 6 x cos x dx. 8 Using integration by parts, we get x cos3x) dx = 1 3 x sin3x) + 1 9 cos3x) + C, x cos x dx = x sin x + cos x + C. Hence x cos 3 x) dx = 1 1 x sin3x) + 1 36 cos3x) + 3 x sin x + cos x) + C. 4
Supplement to Appendix G 9. Exercises Compute the following integrals using complex exponentials. π 1.. 3. 4. π 7 sin5x) cos3x) dx e i7x cosx) dx cos x) e 3x dx sin 3 x) cos10x) dx
Supplement to Appendix G 10 3 The Fundamental Theorem of Algebra: For Chapter 7.4 Stewart Edition 5 A polynomial p of degree n is a function of the form px) = p 0 + p 1 x + p x + + p n x n. 3.1) where the coefficients p j can be either real or complex numbers. The following is a basic fact which is hard to prove and we shall not attempt a proof here). Fundamental Theorem of Algebra: Any nonconstant polynomial can be factored as a product of linear factors with complex coefficients times a constant. Linear factors are of the form x β. This tells us that we can factor a polynomial of degree n into a product of n linear factors. For example, 3x + x 1 = 3x 1 3 )x + 1) n = here) x 3 8 = x )x + α)x + α), where α = 1 ± i 3 n = 3 here) x + 1) = x + i) x i) n = 4 here). Thus, if we allow complex numbers, partial fractions can be done with only linear factors. When we only allowed real numbers as coefficients of the factors, we obtained both linear and quadratic factors. 3.1 Zeroes and their multiplicity Notice that a very important feature of the factorization is: Each factor x β of p corresponds to a number β which is a zero of the polynomial p, namely, pβ) = 0. To see this, let us consider the factorization of p evaluated at β, namely pβ) = kβ β 1 )β β )... β β n ). This is equal to zero if and only if one of the factors is 0; say the jth factor is zero, which gives β β j = 0. Thus β = β j for some j. For some polynomials a factor x β j will appear more than once, for example, in the polynomial px) = 7x ) 5 x 3)x 8)
Supplement to Appendix G 11 the x factor appears 5 times, the x 3 factor appears once, the x 8 factor appears twice. The jargon for this is is a zero of p of multiplicity 5 3 is a zero of p of multiplicity 1 The general form for a factored polynomial is 8 is a zero of p of multiplicity. px) = kx β 1 ) m 1 x β ) m... x β l ) m l 3.) where β j is a zero of multiplicity m j and k is a constant. 3. Real Coefficients All polynomials which you see in Math 0B have real coefficients, so a useful fact is: If all the coefficients q j of the polynomial q are real numbers, then qβ) = 0 implies qβ) = 0. To see this, think of x as being a real number. Suppose x β) k is a factor of q, then we claim that x β) k is also a factor. Indeed, a) Since x β) k is a factor of qx), we have qx) = x β) k rx) for some polynomial rx). b) Taking complex conjugates, we get qx) = x β) k rx). c) Since qx) has real coefficients, qx) = qx) and so by b), x β) k is a factor of qx). A polynomial p with real coefficients has a factorization px) = x β 1 )x β 1 )... x β k )x β k )x r 1 )... x r l ), 3.3) where β 1,..., β k are complex numbers, and r 1,... r l are real numbers. Equivalently, px) = x + b 1 x + c 1 )... x + b k x + c k )x r 1 )... x r l ) 3.4) where b 1,..., b k, respectively c 1,..., c k and r 1,..., r l are real numbers. In fact you can check that b j = Reβ j ) and c j = β j = Reβ j )) + Imβ j )). The advantage of the first representation is that all terms are linear in x and the disadvantage is that some of them may contain numbers β j which are not real. The advantage of the second representation is that all numbers in the factoring are real.
Supplement to Appendix G 1 3.3 Rational Functions and Poles The quotient of two polynomials p q is called a rational function. Given a rational function f, any number β for which fx) is not bounded as x β is called a pole of f. For example, gx) = x 7 x 1) x 9) 3 has poles at 1, 9 and. You might think calling a pole peculiar, but lim definition requires. Poles have multiplicity; in this case 1 is a pole of p of multiplicity x gx) = as the 9 is a pole of p of multiplicity 3 is a pole of p of multiplicity. Notice that the growth rate of g near of a high multiplicity pole exceeds that of g near a low multiplicity pole.
Supplement to Appendix G 13 3.4 Exercises 1. Expand p = x )x 3)x + 1) in the form 3.1).. Show that if p is a polynomial and p5) = 0, then px) x 5 is a polynomial. 3 3. a)how many poles does the rational function rx) = 6+x+5x 3? b)what are the pole locations and their multiplicities for rx) = have? Does it have a pole at 3 x x )x +5x+7)? 4. The following is the simplest mathematical model used for a building hit by an earthquake. If the bottom of the building is displaced horizontally from rest by a distance bt) at time t, then the roof of the building is displaced from vertical by a distance rt). The issue is to describe the relationship between b and r in a simple way. Fortunately, there is a rational function Qs) called the transfer function of the building with the property that when b is a pure sine wave bt) = sin wt at frequency w π, then r is a sine wave of the same frequency and with amplitude Qiw). While earthquakes are not pure sine waves, they can be modelled by combinations of sine waves. If Qs) = s s + 3i +.01)s 3i +.01)s + 7i +.1)s 7i +.1), then at approximately what frequency does the building shake the most, the second most? 5. Electric circuits behave similarly and are typically described by their transfer function Q. If ct), a sinusoidal current of frequency w/π is imposed, and vt) is the voltage one measures, then vt) is a sine wave of the same frequency with amplitude Qiw). If Qs) = 1 s + 3i +.01)s 3i +.01) + s 10, then approximately how much accuracy do we lose in predicting the amplitude for our output with the simpler mathematical model Qs) = 1 s + 3i +.01)s 3i +.01) when a sine wave at frequency w π is put in? Hint: You may use the fact that Qs) Qs) Qs) Qs), even though we have not proved it. r has the form rt) = Qiw) sinwt + ψiw)).
Supplement to Appendix G 14 4 Partial Fraction Expansions PFE): For Chapter 7.4 Stewart Edition 5 4.1 A Shortcut when there are no Repeated Factors The partial fraction expansion can be found by the method in the text. There are easier methods for computing the constants involved. You especially save a lot of time when there are no repeated factors in the denominator. We ll tell you the general principle and then do some specific examples. Suppose that where α 1,..., α n are all distinct. Suppose also that the degree of px) is less than n. Then px) x α 1 ) x α n ) = C 1 + + C n, 4.1) x α 1 x α n where the constants C 1,..., C n need to be determined to find the partial fraction expansion. Multiply both sides of 4.1) by x α i and then set x = α i. The left side is some number. On the right side, we are left with only C i because all the other terms have a factor of x α i which is 0 when x = α i. We now illustrate this method with some examples. Example 4.1 Partial fractions with no repeated factors. I) x Let s expand + x 1)x+)x+3) by partial fractions. We have fx) = x + x 1)x + )x + 3) = C 1 x 1 + C x + + C 3 x + 3. Multiply by x 1 to eliminate the pole at x = 1 and get Set x = 1 and get x 1)fx) = Hence C 1 = 1 4. Similarly, x + x + )x + 3) = C 1 + C x 1) + C 3x 1). x + x + 3 1 + 1 + )1 + 3) = C 1. x + C = x + )fx) = = 4 + x= x 1)x + 3) x= 3)1 = and, respectively, C 3 = x + 3)fx) = x= 3 x + = x 1)x + ) x= 3 9 + 4) 1) = 11 4.
Supplement to Appendix G 15 Example 4. Partial fractions with no repeated factors. II) x+1 Let s find a PFE of. Note that x 3 +x fx) = x + 1 x 3 + x has two natural forms of partial fraction expansions corresponding to whether we factor the denominator x 3 + x in the form 3.4) or 3.3). Stewart s textbook Chapter 7.4 Ed. 5) uses 3.4), so we emphasize and recommend that one, namely fx) = A x + Bx + C x + 1. We proceed like Stewart, but save a little time with A = xfx) = 1 x=0 1 = 1. Next multiply both sides in 4.) by xx + 1) to get x + 1 = xx + 1)fx) = x + 1 + xbx + C). Cancel 1 on both sides and then divide by x, to get 1 = x + Bx + C. Set x = 0 to get C = 1 and so B = 1. While we do not recommend the use of 3.3) form of expansion, for the sake of the curious, let us show how it is done. Since x = x 0, it follows fx) = x + 1 x 3 + x) = x + 1 xx i)x + i) = C 1 x + C x i + C 3 x + i. Also, x i)fx) = C = i + 1 x=i ii) = 1 i xfx) = C 1 = x=0 and 1 i)i = 1. 4.) x + i)fx) = C 3 = i + 1 x= i i) i) = 1 + i. Note that C 3 = C and we can get the first PFE from this particular PFE by Hence, fx) = 1 x + C x i + C 3 x + i = 1 x + C x + i) + C 3 x i) x = 1 + 1 x + ReC ) x + )ImC ) x. + 1 which is what we got before. fx) = 1 x + x + 1 x + 1,
Supplement to Appendix G 16 4. The Difficulty with Repeated Factors Let us apply the previous method to rx) = 1 x 1) x 3) whose partial fraction expansion we know by Stewart s book) has the form We can find C quickly from rx) = A x 1) + B x 1) + C x 3. 4.3) C = x 3)rx) = x=3 1 3 1) = 1 4 and A from A = x 1) rx) = 1 x=1 1 3 = 1. However, B does not succumb to this technique; you must use other means to find it. What we have gotten from our method are just the coefficients of the highest terms at each pole. To find B many ways will do. For example, the one in Stewart will do and we have made it go much faster by finding A and C already. Another way to find the missing number B is to plug in one value of x, say x = 0 and get Hence B = 1 1. 1 1) 3) = r0) = 1 B 1 3 1, which yields 4 B = 1 3 + 1 + 1 ). 1 4.3 Every Rational Function has a Partial Fraction Expansion Now we mention a pleasant fact. Theorem 4.3 Every rational function f = p q has a partial function expansion. The core of the reason is the fundamental Theorem of Algebra, which can be used to factor q as in formula 3.). This produces fx) = px) x β 1 ) m 1 x β ) m x βl ) m l. If f has real coefficients, then f can always be written in the form fx) = px) x + b 1 x + c 1 ) x + b k x + c k )x r 1 ) x r l ). This is the factoring behind the various cases treated in Stewart Chapter 7.4 Ed 5. We emphasize that the terms in f may repeat. One then needs to write out the appropriate form for the PFE and then identify the coefficients as it has been explained in Stewart Ed.5 Chapter 7.4 and to some extent in these notes.
Supplement to Appendix G 17 4.4 The Form of the PFE Here is one way to look at the form of the PFE of a rational function f. We just give the rough idea which might be too vague to be very helpful. Recall that a high multiplicity pole has a faster growth rate that a lower multiplicity pole. Thus it can overshadow the lower multiplicity pole. Example 4.4 The function fx) = 1 x 1) x 3) 1 1 x 1) 1 3) = 1 but when we subtract this pole from f we get ex) = fx) 1 ex) = 1 x 1)x 3) has a multiplicity pole at 1 whose strength is 1 x 1), 1 x 1) = 1 x 1) 1 x 3 + 1 ) which still has a pole at 1, though now it is a pole of multiplicity 1. Thus we must include this first order pole in the PFE, and this gives some intuition behind the correct form for the PFE of f. fx) = Similar intuition tells us that gx) = x 7 x 1) x 9) 3 gx) = A x 1) + B x 1) + C x 3 has a PFE of the form A x 1) + B x 1) + C x 9) 3 + D x 9) + E x 9 + F x + Gx + H, where the coefficients A,... H are to be found then by the method explained before.
Supplement to Appendix G 18 4.5 Exercises 1. Use partial fraction techniques to solve Exercises 17 through 38 in Section 7.4 Edition 5.. Find the partial fraction expansion of rx) = x+1 x 1) x+). 3. Let rx) = 3 x 1)x ). What value of A makes rx) A x 1 4. Find the partial fraction expansion of rx) = x 3 + xx +1)x +4). 5. Find the partial fraction expansion of rx) = x 3 + xx 1)x 4). have its only pole located at? 6. Consider the PFE of the function r given in 4.3). We claim that the number d dx x 1) rx) ) x=1 is either A, B, or C in the partial fraction expansion. a) Which one is it? b) Does such a formula hold for any rational function with a second order pole? Justify why). c) Find a similar formula for a rational function with a third order pole.
Supplement to Appendix G 19 5 Improving on Euler s Method: For Chapter 9. Stewart Ed. 5 This section supplements Chapter 9. of Stewart Edition 5. Suppose we are given the differential equation y = F x, y) with initial condition yx 0 ) = y 0. Euler s method, discussed in Section 9., produces a sequence of approximations y 1, y,... to yx 1 ), yx ),... where x n = x 0 + nh are equally spaced points. Note that this is almost the left-endpoint approximation in numerical integration Chapter 7 of Stewart Ed. 5). To see this, suppose that we have an approximation y n 1 for yx n 1 ), and that we want an approximation for yx n ). Integrate y = F x, y) from x n 1 to x n and use the left endpoint approximation, to get xn yx n ) yx n 1 ) = F x, y) dx x n 1 hf x n 1, yx n 1 )). Now we have a problem that did not arise in numerical integration: We don t know yx n 1 ). What can we do? We replace yx n 1 ) with the approximation y n 1 to obtain yx n ) y n 1 hf x n 1, y n 1 ). Rearranging terms and denoting by y n the approximation to yx n ) thus obtained, we have Euler s method: y n = y n 1 + hf x n 1, y n 1 ). 5.1) We know that the left endpoint approximation is a poor way to estimate integrals and that the Trapezoidal Rule is better. Can we use it here? Adapting the argument that led to 5.1) for use with the Trapezoidal Rule gives us the formula y n = y n 1 + h ) F x n 1, y n 1 ) + F x n, y n ). 5.) You should carry out the steps. Unfortunately, 5.) can t be used: We need y n on the right side in order to compute it on the left! Here is a way around this problem: First, use 5.1) to estimate predict ) the value of y n and call this prediction y n. Second, use y n in place of y n in the right side of 5.) to obtain a better estimate, called the correction. The formulas are predictor) yn = y n 1 + hf x n 1, y n 1 ) 5.3) corrector) y n = y n 1 + h ) F x n 1, y n 1 ) + F x n, y n). This is an example of a predictor-corrector method for differential equations. The following table contains results for Example 9..3 in Stewart, namely, for the differential equation y = x + y with initial condition y0) = 1:
Supplement to Appendix G 0 step size y1) by 5.1) y1) by 5.3) 0.50.500000 3.8150 0.0.976640 3.405416 0.10 3.187485 3.4816 0.05 3.306595 3.43438 0.0 3.383176 3.43607 0.01 3.40968 3.436474 The correct value is 3.436564, so 5.3) is much better than Euler s method for this problem.
Supplement to Appendix G 1 5.1 Exercises 1. Write down a predictor-corrector method based on Simpson s Rule for numerical integration. Hint: a bit tricky is that we consider not two, but three grid points x n, x n 1, x n and assume we know f n and f n 1. The problem for you is to give an algorithm for producing f n.
Supplement to Appendix G 6 Appendix: Differentiation of Complex Functions Suppose we have a function fz) whose values are complex numbers and whose variable z may also be a complex number. We can define limits and derivatives as Stewart did for real numbers. Just as for real numbers, we say the complex numbers z and w are close if z w is small, where z w is the absolute value of a complex number. 3 We say that lim z α fz) = L if, for every real number ɛ > 0 there is a corresponding real number δ > 0 such that fz) L < ɛ whenever 0 < z α < δ. The derivative of f at α is defined by f fz) fα) α) = lim. z α z α Our variables will usually be real numbers, in which case z and α are real numbers. Nevertheless the value of a function can still be a complex number because our functions contain complex constants; for example, fx) = 1 + i)x + 3ix. Since our definitions are the same, the formulas for the derivative of the sum, product, quotient and composition of functions still hold. Of course, before we can begin to calculate the derivative of a particular function, we have to know how to calculate the function. What functions can we calculate? Of course, we still have all the functions that we studied with real numbers. So far, all we know how to do with complex numbers is basic arithmetic. Thus we can differentiate a function like fx) = 1 + ix x + i or a function like gx) = 1 + i e x since fx) involves only the basic arithmetic operations and gx) involves a complex) constant times a real function, e x, that we know how to differentiate. On the other hand, we cannot differentiate a function like e ix because we don t even know how to calculate them. 6.1 Deriving the Formula for e z Using Differentiation Two questions left dangling in Section 1 are the following How did you come up with the definition of complex exponential? How do you know it satisfies the simple differential equation properties in 1.1)? We consider each of these in turn. In Appendix G Stewart uses Taylor series to come up with a formula for e a+bi. Since you haven t studied Taylor series yet, we take a different approach. Using the first of formula in 1.1) with α = a and β = b, it follows that e a+bi should equal e a e bi. Thus we only need to know how to compute e bi when b is a real number. 3 The definitions are nearly copies of Stewart Sections.4 and.8. We have used z and α instead of x and a to emphasize the fact that they are complex numbers and have called attention upon the fact that δ and ɛ are real numbers.
Supplement to Appendix G 3 Think of b as a variable and write fx) = e xi = e ix. By the second property in 1.1) with α = i, we have f x) = ifx) and f x) = if x) = i fx) = fx). It may not seem like we re getting anywhere, but in fact we are! Let us look at the equation f x) = fx). There s not a complex number in sight, so let s forget about them for a moment. Do you know of any real functions fx) with f x) = fx)? Yes. Two such functions are cos x and sin x. In fact, If fx) = A cos x + B sin x, then f x) = fx). We need constants probably complex) so that it is reasonable to let e ix = A cos x + B sin x. How can we find A and B? When x = 0, we get e ix = e 0 = 1. Since A cos x + B sin x = A cos 0 + B sin 0 = A, we want A = 1. We can get B by looking at the derivative e ix ) at x = 0. You should check that this gives B = i. Remember that we want the derivative of e ix to equal ie ix.) Thus we get Euler s formula: e ix = cos x + i sin x Putting it all together we finally have our definition for e a+bi. We still need to verify that our definition for e z satisfies 1.1). The verification that e α+β = e α e β is left as an exercise. Next, we will prove that e z ) = e z for complex numbers. Then, by the Chain Rule, it will follow that e αx ) = e αx αx) = αe αx, which is what we wanted to prove. Example 6.1 A proof that e z ) = e z ) By the definition of derivative and the fact that e α+β = e α e β with α = z and β = w, we have e z ) = lim w 0 e z+w e z w e z e w 1) = lim = e z e w 1 lim w 0 w w 0 w. Let w = x + iy where x and y are small real numbers. Then, using the definition of complex exponential, we get e w 1 = ex cos y + i sin y) 1. w x + iy Since x and y are small, we can use linear approximations 4 for e x, cos y and sin y, namely 1 + x, 1 and y, respectively. For example, the approximation cos y 1 comes from cos y) = 0 at y = 0.) is approximately equal to Thus ew 1) w 1 + x)1 + iy) 1 x + iy = 1 + x) + i1 + x)y 1 x + iy = x + iy) + ixy x + iy = 1 + ixy x + iy. When x and y are very small, their product is much smaller than either one of them. ixy e x+iy = 0 and so lim w 1) w = 1. This shows that e z ) = e z. lim w 0 w 0 Thus 4 Linear approximations are discussed in Section 3.11 of Stewart.