Section 7.2. The Calculus of Complex Functions

Transcription

1 Section 7.2 The Calculus of Complex Functions In this section we will iscuss limits, continuity, ifferentiation, Taylor series in the context of functions which take on complex values. Moreover, we will introuce complex extensions of a number of familiar functions. Since complex numbers behave algebraically like real numbers, most of our results efinitions will look like the analogous results for real-value functions. We will avoi going into much etail; the complete story of the calculus of complex-value functions is best left to a course in complex analysis. However, we will see enough of the story to enable us to make effective use of complex numbers in elementary calculations. We begin with a efinition of the limit of a sequence of complex numbers. Definition We say that the limit of a sequence of complex numbers {z n } is L, write lim z n = L, if for every ɛ > 0 there exists an integer N such that whenever n > N. z n L < ɛ Notice that the only ifference between this efinition the efinition of the limit of a sequence given in Section.2 is the use of the magnitue of a complex number in place of the absolute value of a real number. Even here, the notation is the same. The point is the same as it was in Chapter : the limit of the sequence {z n } is L if we can always ensure that the values of the sequence are within a esire istance of L by going far enough out in the sequence. Now if z n = x n + y n i L = a + bi, then lim z n = L if only if lim z n L = lim (xn a) 2 + (y n b) 2 = 0, the latter of which occurs if only if lim x n = a lim y n = b. Hence we have the following useful result. Proposition Let z n = x n + y n i L = a + bi. Then lim z n = L

2 2 The Calculus of Complex Functions Section 7.2 if only if lim x n = a lim y n = b. Thus to etermine the limiting behavior of a sequence {z n } of complex numbers, we nee only consier the behavior of the two sequences of real numbers, {R(z n )} {I(z n )}. Suppose for n =, 2,,.... Then z n = n 2n n + n i so lim R(z n n) = lim 2n + 2 = lim n n lim I(z n + n) = lim n = lim + n n lim z n = 2 + i. = 2 =, Suppose z n = n ( ( nπ cos ) ( nπ + sin ) ) i for n =, 2,,.... Then so lim R(z n) = lim lim I(z n) = lim lim z n = 0. cos ( ) nπ n sin ( ) nπ n Geometrically, since z n = n arg(z n) = nπ, the points in this sequence are converging to 0 along a spiral path, as seen in Figure Having efine the limit of a sequence of complex numbers, we may efine the limit of a complex-value function, as in Section 2., then efine continuity, as in Section 2.4. Definition Suppose f : C C, that is, f is a complex-value function of a complex variable. We say the limit of f(z) as z approaches a is L, written lim f(z) = L, z a if whenever {z n } is a sequence of points with z n a for all n lim z n = a, then lim f(z n) = L. = 0 = 0,

3 Section 7.2 The Calculus of Complex Functions Figure 7.2. Plot of the points z n = n - ( cos ( nπ ) ( + sin nπ ) ) i, n =, 2,, Definition We say the function f : C C is continuous at a if lim z a f(z) = f(a). As with real-value functions of a real variable, it is easy to show that algebraic functions of a complex variable are continuous wherever they are efine. In particular, complex polynomials, that is, functions P of the form P (z) = a n z n + a n z n + + a z + a 0, where n is a nonnegative integer the coefficients a 0, a,..., a n are complex numbers, are continuous at all points in the complex plane. Complex rational functions, that is, functions R of the form R(z) = P (z) Q(z), where both P Q are polynomials, are continuous at all points where they are efine. Since f(z) = z 2 iz + 4 5i is a polynomial, it is continuous at all points in the complex plane. In particular, lim f(z) = lim z i z i (z2 iz + 4 5i) = i 2 (i)(i) + 4 5i = 2 5i. Algebraic simplification may be useful in evaluating limits here as it was in Section 2.. For example, lim z i z i z 2 + = lim z i z i (z i)(z + i) = lim z i z + i = 2i = i 2i i = 2 i. Although this is not the time to go into any etail about the geometric meaning of the erivative of a function f : C C, the algebraic efinition manipulation of erivatives follows the pattern of the results for real-value functions in Chapter.

4 4 The Calculus of Complex Functions Section 7.2 Definition If f : C C, then the erivative of f at a, enote f (a), is given by provie the limit exists. f f(a + h) f(a) (a) = lim, (7.2.) h 0 h Note that h in this efinition is, in general, a complex number, not just a real number. Since the algebraic properties of the complex numbers are very similar to the algebraic properties of the real numbers, much of what we learne about ifferentiation in Chapter still hols true in our new situation. For example, if n is a nonzero rational number, then z zn = nz n. (7.2.2) Moreover, all the techniques we learne for computing erivatives in Sections..4, incluing the quotient, prouct, chain rules, still hol. If f(z) = z 5 + iz ( + 2i)z, then f (z) = 5z 4 + iz 2 2i. If then, using the quotient rule, g(w) = ( + i)w2 2w, g (w) = (2w )(6 + 2i)w ( + i)w2 (2) (2w ) 2 = (6 + 2i)(w2 w) (2w ) 2. From this point it is possible to follow the pattern of Chapter 5 evelop the theory of polynomial approximations using Taylor polynomials, efine in a manner analogous to the efinition in Section 5., as well as the theory of power series Taylor series. In particular, a power series a n (z a) n, (7.2.) where a 0, a, a 2,... a are complex numbers, is sai to converge absolutely at those points z for which the series a n z a n (7.2.4) converges. Since the latter series involves only real numbers, its convergence may be etermine using the tests evelope in Chapter 5. As before, absolute convergence implies convergence. Moreover, if the series (7.2.) converges at points other than a, then there exists an R, either a positive real number or, such that the series converges absolutely

5 Section 7.2 The Calculus of Complex Functions 5 for all z such that z a < R iverges for all z such that z a > R. However, note that in this case the set of all points in the complex plane such that z a < R is a isk of raius R centere at a, not an interval as it was in the real number case. Since the series Consier the power series z n. (7.2.5) z n is a geometric series, it converges for all values of z for which z <. Hence converges for all z for which z <, that is, for all z insie the unit circle centere at the origin of the complex plane. Thus the raius of convergence of (7.2.5) is R =. Using the same argument as we use in Section., we can show that for all z with z <. For example, z n z n = z ( ) n i = 2 i 2 = 2 2 i = 2(2 + i) (2 i)(2 + i) = i. Consier the power series z n n!. (7.2.6) To etermine its raius of convergence, we apply the ratio test to the series z n n! = z n n!, (7.2.7) obtaining ρ = lim z n+ (n + )! z n n! z = lim n + = 0

6 6 The Calculus of Complex Functions Section 7.2 for all values of z. Since ρ = 0 for any value of z, (7.2.7) converges for all z in the complex plane. That is, the raius of convergence of (7.2.6) is R =. Of course, we also know that (7.2.7) converges for all z because, from our work in Section 6., it is equal to e z. The power series in the last example is the extension to complex numbers of the series we use to efine the exponential function in Section 6.. With it, we can efine the complex exponential function. Definition The complex exponential function, with value at z enote by exp(z), is efine for all points in the complex plane by exp(z) = z n n!. (7.2.8) Of course, this efinition agrees with our ol efinition when z is real. In Chapter 6 we use the exponential function to give meaning to exponents which were not rational numbers. Similarly, the complex exponential function may be use to efine complex exponents. However, we will only consier the case of raising e to a complex power. Definition If z is a complex number with I(z) 0, then we efine e z = exp(z). With this efinition we now have e z = exp(z) for all z in the complex plane, the case when I(z) = 0, that is, when z is real, having been treate in Section 6.. Although we will not repeat them here, the arguments from Section 6. come over to establish the following proposition. Proposition For any complex numbers w z, e w+z = e w e z (7.2.9) e w z = ew e z. (7.2.0) Also, as in Section 6., irect ifferentiation of (7.2.8) yiels the following result. Proposition z ez = e z. (7.2.) Using the prouct chain rules, ( z 2 e z2) = z 2 ( 2z)e z2 + 2ze z2 = 2z( z 2 )e z2. z

7 Section 7.2 The Calculus of Complex Functions 7 r θ z = re iθ Figure Plot of the point re iθ in the complex plane The exponential of a pure imaginary number is particularly interesting. To see why, let θ be a real number consier e iθ = (iθ) n n! = + iθ + (iθ)2 2! + (iθ)4 4! + (iθ)5 5! + = + iθ θ2 2! iθ + θ4! 4! + iθ5 ) 5! ) = ( θ2 2! + θ4 4! + + i (θ θ! + θ5 5! = cos(θ) + i sin(θ). Proposition For any real number θ, e iθ = cos(θ) + i sin(θ). (7.2.2) As a consequence, if θ is a real number, then e iθ = arg(e iθ ) = θ. That is, e iθ is a point in the complex plane on the unit circle centere at the origin, a istance of θ raians away, in a counterclockwise irection, along the circle from (, 0). Moreover, if z is a nonzero complex number with z = r arg(z) = θ, then z = r(cos(θ) + i sin(θ)) = re iθ. (7.2.) This exponential notation provies a compact way to isplay any nonzero complex number in polar form. See Figure If z = i, then z = 2 Arg(z) = π 4, so z = 2e i π 4.

8 8 The Calculus of Complex Functions Section 7.2 Moreover, z = 2e i π 4 z 2 = 2e 2i π 4 = 2e i π 2 = 2i. If w = e i π z = 5e i π 8, then wz = ( e i π ) ( 5e i π 8 ) = 5e i( π + π 8 ) = 5e i π 24 w z = ei π 5e i π 8 Since for any real number θ, = 5 ei( π π 8 ) 5π = ei e iθ = cos(θ) + i sin(θ), it follows that e iθ = cos( θ) + i sin( θ) = cos(θ) i sin(θ). Hence e iθ e iθ = cos(θ) + i sin(θ) (cos(θ) i sin(θ)) = 2i sin(θ). (7.2.4) Solving (7.2.4) for sin(θ), we have sin(θ) = 2i (eiθ e iθ ). Similarly, e iθ + e iθ = cos(θ) + i sin(θ) + cos(θ) i sin(θ) = 2 cos(θ), (7.2.5) from which we obtain cos(θ) = 2 (eiθ + e iθ ). Proposition For any real number θ, sin(θ) = 2i (eiθ e iθ ) (7.2.6) cos(θ) = 2 (eiθ + e iθ ). (7.2.7) These formulas are very similar to the formulas we use to efine the hyperbolic sine cosine functions in Section 6.7. We will now use these formulas to efine the complex sine cosine functions; at the same time, we will exten the efinitions of the hyperbolic sine cosine functions. In oing so, we will see just how closely relate the circular hyperbolic trigonometric functions really are.

9 Section 7.2 The Calculus of Complex Functions 9 Definition The complex sine function, with value at z enote by sin(z), the complex cosine function, with value at z enote by cos(z), are efine for all z in the complex plane by sin(z) = 2i (eiz e iz ) (7.2.8) cos(z) = 2 (eiz + e iz ). (7.2.9) The complex hyperbolic sine function, with value at z enote by sinh(z), the complex hyperbolic cosine function, with value at z enote by cosh(z), are efine for all z in the complex plane by sinh(z) = 2 (ez e z ) (7.2.20) cosh(z) = 2 (ez + e z ). (7.2.2) Note that these functions are efine so that they agree with their original versions when evaluate at real numbers. With these efinitions it is a simple matter to prove that sin(z) = cos(z), (7.2.22) z For example, cos(z) = sin(z), (7.2.2) z sinh(z) = cosh(z), (7.2.24) z cosh(z) = sinh(z). (7.2.25) z z cos(z) = ( ) z 2 (eiz + e iz ) = 2 (ieiz ie iz ) = i 2 (eiz e iz ) = i2 2i (eiz e iz ) = 2 (eiz e iz ) = sin(z).

10 0 The Calculus of Complex Functions Section 7.2 Note that sin(i) = 2i (ei2 e i2 ) = 2i (e e ) = i sinh() = i i 2 sinh() = i sinh(). Using the prouct chain rules, we have sin(2z) cos(z) = sin(2z)( sin(z))() + cos(z) cos(2z)(2) z = sin(2z) sin(z) + 2 cos(2z) cos(z). The final complex-value function we will efine is the complex logarithm function. Analogous to our other efinitions in this section, we woul like this function to share the basic characteristic properties of the orinary logarithm function to agree with that function when evaluate at a positive real number. In particular, if we let Log(z) enote the complex logarithm of a complex number z log(r) enote the real logarithm of a positive real number r, then for a nonzero complex number z with z = r Arg(z) = θ we woul like to have Log(z) = Log(re iθ ) = Log(r) + Log(e iθ ) = log(r) + iθ. (7.2.26) Moreover, using (7.2.26) to efine the complex logarithm function will guarantee that our new function agrees with the orinary logarithm function when evaluate at positive real numbers, for if z is a positive real number, then z = z Arg(z) = 0, giving us Log(z) = log(z). Definition The complex logarithm function, with value at z enote by Log(z), is efine for all nonzero complex numbers z with z = r Arg(z) = θ by where log(r) is the orinary real-value logarithm of r. Log(z) = log(r) + iθ, (7.2.27) Note that we have use the principal value of arg(z), that is, Arg(z), in the efinition of Log(z) in orer to give Log(z) a unique value. Moreover, note that this efinition gives meaning to the logarithm of a negative real number, although it still oes not efine the logarithm of 0. Since 2 2i = 8 Arg(2 2i) = π 4, we have Log(2 2i) = log( 8) π 4 i = 2 log(8) π 4 i = 2 log(2) π 4 i.

11 Section 7.2 The Calculus of Complex Functions Since 4 = 4 Arg( 4) = π, we have Log( 4) = log(4) + πi = 2 log(2) + πi. Problems. For each of the following, fin lim z n. plane. (a) z n = n n (c) z n = e i π n + n + 2n + i 2. Evaluate each of the following limits. Also, plot z, z 2, z,..., z 5 in the complex (b) 2 n n 2 () z n = e i π(n ) n ( ) i n (a) lim z i (4z 6z + ) (b) lim z i (z2 z) (c) z lim w i z i () lim z i z 4 z 2 +. Fin the erivative of each of the following functions. (a) f(z) = z 2 6z 5 + 8i (b) g(w) = (c) f(z) = (z 4i)e z2 4. (a) Show that for all z with z <. w 6i + w + i () h(s) = (s 2 + ) exp(s 2 si) + z 2 = ( ) n z 2n (b) How oes (a) help explain why, for real values of x, the Taylor series converges only on the interval (, )? 5. (a) If z = x + yi, show that + x 2 = ( ) n x 2n R(e z ) = e x cos(y) I(e z ) = e x sin(y). (b) If z = x + yi, fin e z arg(e z ). 6. Show that e iπ + = 0.

12 2 The Calculus of Complex Functions Section Verify the ifferentiation formulas for sin(z), sinh(z), cosh(z). 8. (a) Show that 2 x x = log(2). (b) Some computer algebra systems evaluate the integral in (a) as 2 x x = Log( ) Log( 2). Reconcile this answer with the answer in (a). 9. Let z w be complex numbers. Verify the following two properties of the complex logarithm. (a) Log(wz) = Log(w) + Log(z) (b) Log ( ) w z = Log(w) Log(z) 0. For a positive integer n, an nth root of unity is a complex number z with the property that z n =. Show that for m = 0,,..., n, z m = e i 2mπ n is an nth root of unity. Plot these points in the complex plane for n = 0.. (a) Use the fact that sin(z) = 2i (eiz e iz ) to fin the complex power series representation for sin(z). (b) Use the fact that cos(z) = 2 (eiz + e iz ) to fin the complex power series representation for cos(z). 2. Define a complex version of the tangent function show that tan(z) = i ( e iz e iz e iz + e iz ).. (a) Show that sin(ix) = i sinh(x) for every real number x. (b) Show that cos(ix) = cosh(x) for every real number x. 4. Let z = x + yi. (a) Show that R(sin(z)) = sin(x) cosh(y)

13 Section 7.2 The Calculus of Complex Functions (b) Show that I(sin(z)) = cos(x) sinh(y). R(cos(z)) = cos(x) cosh(y) I(sin(z)) = sin(x) sinh(y). 5. (a) Show that for any nonzero complex number z, e Log(z) = z. (b) If z is a nonzero complex number, oes it necessarily follow that Log(e z ) = z?