Chapter 3 Elementary Functions

Transcription

1 Chapter 3 Elementary Functions In this chapter, we will consier elementary functions of a complex variable. We will introuce complex exponential, trigonometric, hyperbolic, an logarithmic functions. 23. The exponential function We want to efine an exponential function of a complex variable z = x + iy. We require that f(x) = e x for all x R. f (z) = f(z). Definition. (The exponential function) We efine the exponential function of a complex variable z = x + iy to be Remark. e z = e x (cos y + i sin y). 1. The omain of the exponential function is the entire complex plane. 2. We sometimes use exp(z) to enote e z. 3. When z = iy, we have the Euler s formula e iy = cos y + i sin y Example 1. Express e 1+ π 2 i in the form a + ib. Proposition 1. The exponential function is an entire function an z ez = e z. Exercise. Remark. When z = 1 n, n = 2, 3,..., then e 1 n = n e. That is e 1 n is not the set of all n th roots of e. Proposition 2. Let z = x + iy. Then we have the following properties of e z : 1. e z = e x. 2. arg(e z ) = y + 2nπ, n Z. 3. e z 0 for all z C. 4. e z 1 e z 2 = e z 1+z e z = e z. 6. ez 1 e z 2 = ez 1 z (e z ) n = e nz for all n Z. 8. e z is a perioic function with perio of 2πi; that is e z+2πi = e z.

2 2 Exercise. Example 1. Solve the equation e z = 1. Example 2. Fin all the value of z such that e iz = 2. Exercises (page 67): 1 11, 14 15

3 3 24. Trigonometric functions Euler s formula gives e ix = cos x+i sin x an e ix = cos x i sin x. Thus we have cos x = 1 2 (eix +e ix ) an sin x = 1 2i (eix e ix ). Generalizing these formulas we have the following efinition. Definition. (The sine an cosine functions) We efine the sine an cosine functions of a complex variable z to be sin z = 1 2i (eiz e iz ) an cos z = 1 2 (eiz + e iz ). Example 1. Fin sin(iπ). Proposition The sine an cosine functions are entire functions. 2. z sin z = cos z an cos z = sin z. z 3. sin( z) = sin z an cos( z) = cos z. Exercise. Proposition 4. (Trigonometric ientities) (1) sin 2 z + cos 2 z = 1. (2) sin(z 1 ± z 2 ) = sin z 1 cos z 2 ± sin z 2 cos z 1. (3) cos(z 1 ± z 2 ) = cos z 1 cos z 2 sin z 1 sin z 2. Proposition 5. We have the following: (1) If z = iy is pure imaginary, then sin(iy) = i sinh y an cos(iy) = cosh y. (2) If z = x + iy, then cos z = cos x cosh y i sin x sinh y. (3) If z = x + iy, then sin z = sin x cosh y + i cos x sinh y. Example 1. Fin all the roots of sin z = 1. Solution: z = (4k + 1) π 2, k Z. Example 2. Show that cos(iz) = cos(iz). Example 3. Fin all the roots of sin z = cos z. Solution: z = π 4 + 2kπ, k Z. Proposition 6. We have the following:

4 4 (1) sin z 2 = sin 2 x + sinh 2 y. (2) cos z 2 = cos 2 x + sinh 2 y. (3) sin z = 0 if an only if z = nπ, n Z (4) cos z = 0 if an only if z = (2n + 1) π, n Z 2 Exercises. Example 1. Show that cos z cos x. The tangent, cotangent, secant, an cosecant functions Definition. We efine the tangent, cotangent, secant, an cosecant functions of a complex variable z to be Proposition 7. tan z = sin z cos z, cos z cot z = sin z, sec z = 1 cos z, csc z = 1 sin z. (1) tan z an sec z are analytic everywhere except at the singular points z = (2n + 1) π 2, n Z. (2) cot z an csc z are analytic everywhere except at the singular points z = nπ, n Z. (3) z tan z = sec2 z, z cot z = csc2 z, sec z = sec z csc z, z csc z = csc z cot z. z Exercises. Example 1. Show that tan(z + π) = tan z. Exercises(page 70): 1 11, 13 14, 16 17

5 5 25. Hyperbolic functions Definition. We efine the hyperbolic functions of a complex variable z as follows: sinh z = 1 2 (ez e z ), cosh z = 1 2 (ez + e z ), an Proposition 8. tanh z = sinh z cosh z, cosh z coth z = sinh z, sechz = 1 cosh z, cschz = 1 sinh z. 1. cosh z = cos(iz) an sinh z = i sin(iz). 2. cosh z an sinh z are entire functions. 3. z sinh z = cosh z an cosh z = sinh z. z 4. sinh( z) = sinh z an cosh( z) = cosh z. 5. cosh 2 z sinh 2 z = sinh(z 1 + z 2 ) = sinh z 1 cosh z 2 + sinh z 2 cosh z cosh(z 1 + z 2 ) = cosh z 1 cosh z 2 + sinh z 1 sinh z 2 8. sinh z 2 = sinh 2 x + sin 2 y. 9. cosh z 2 = sinh 2 x + cos 2 y 10. sinh z = 0 if an only if z = nπi, n Z 11. cosh z = 0 if an only if z = (2n + 1) πi 2, n Z Example 1. Show that sinh(z + iπ) = sinh z. Example 2. Fin all roots of the equation cosh z = i. z = ln 1 ± 2 + (± π 2 + 2nπ)i, n Z. Exercises (page 72): 1 10, 14

6 6 26. The logarithmic function an its branches In this section, we will efine the logarithmic function of a complex variable. As a first step we solve the equation e w = z, where z is a nonzero complex number. Writing w = u + iv an z = z e iargz, we have the solutions w = ln z + i(argz + 2nπ), n Z. Definition. We efine the logarithm of a nonzero complex number z to be or Example 1. Fin log( 1) an log(1 + i). Example 2. Fin log(e z ). Remark. log z = ln z + i(argz + 2nπ), log z = ln z + i arg z. n Z 1. logz is not a function. We will call it a multiple-value function. 2. e log z = z for all z log e z = z + 2nπi, n Z, for all z C. Definition. The principal value of the logarithm of a nonzero complex number z, log z, is efine to be Logz = ln z + iargz. Example 1. Fin Log(1 + 3i). Remark. 1. Logz is a function. 2. log z = Logz + 2nπi, n Z. Branches of log z As we mentione above, log z is a multiple-value function. Thus to efine a single-value function we restrict the omain an efine functions as log z = ln z + i arg z, α < arg z α + 2π, where α is any real number. However these functions are not continuous in their omains. To see this let us consier Logz. If we approach 1 along the curve z = e (π t)i, 0 t π, then lim z 1 Logz = lim (π t)i = πi. t 0 But if we approach 1 along the curve z = e (π t)i, 0 t π, then lim Logz = lim (π t)i = πi. z 1 t 0 Thus lim Logz oes not exist an hence Logz is not continuous at 1. The same arguments can z 1 be use to show that Log z is not continuous at any negative real number. Deleting these points from the omain of Logz, we obtain an analytic function Logz = ln z + iargz, z > 0, π < Argz < π.

7 7 Definition. Let f be a multiple-value function. A single-value function F is calle a branch of f if there is a omain D such that 1. F (z) f(z) for all z D. 2. F is analytic in D. Example 2. Logz = ln z + iargz, z > 0, π < Argz < π. is a branch of log z. It is calle the principal branch. Proposition 9. For any real number α, log z = ln z + i arg z, z > 0, α < arg z < α + 2π, is a branch of log z an z log z = 1, z > 0, α < arg z < α + 2π. z Definition. 1. A branch cut is a portion of a line or curve that is introuce to efine a branch of a multiplevalue function. 2. A point that is common to all branch cuts of a function f is calle a branch point of f. Example 1. The branch cut for the principal branch Logz is the origin an the ray θ = π an the branch point is the origin. Example 2. Fin the roots of the equation log z = 1. Solution: z = e. Example 3. Show that Log(z + 1) is analytic everywhere except on the half line y = 0, x 1.

8 8 27. Some ientities involving logarithms Many ientities involving logarithms of positive real numbers remain vali, with certain moification, for logarithms of nonzero complex numbers. Proposition 10. Let z 1 an z 2 be two nonzero complex numbers. Then log(z 1 z 2 ) = log z 1 + log z 2 an ( z1 ) log = log z 1 log z 2. z 2 log(z 1 z 2 ) = ln z 1 z 2 + i arg(z 1 z 1 ) = ln z 1 + ln z 2 + i(arg z 1 + arg z 2 ) = log z 1 + log z 2. Remark. The statements in the above proposition have the same meaning as the statements for arg(z 1 z 2 ) an arg(z 1 /z 2 ). That is it means that if two of the three logarithms are specifie, then there is a thir logarithm such that they are true. Example 1. Let z 1 = 1 an z 2 = i. Verify that log(z 1 z 2 ) = log z 1 + log z 2. Example 2. Let z 1 = 1 an z 2 = i. Show that Log(z 1 z 2 ) Logz 1 + Logz 2. Proposition 11. Let z be a nonzero complex number. Then z n = e n log z, n Z an z 1/n = e 1 n log z, n N. Example 1. Show that Log(i 3 ) 3Log i. Exercises (page 79): 1 13, 15 17

9 9 28. Complex Exponents In this section we will use the logarithms to efine complex exponents, z c an c z. Definition. Let z be a nonzero complex number an let c be any complex number. We efine z c as z c = e c log z, where log z = ln z + i arg z is the multiple-value function. Example 1. Fin the values of 1 i. Example 2. Fin the values of i 2+i. Example 3. Fin the values of (1 + 3i) 3/2. Remark. In general, z c is a multiple-value function. Proposition 12. If z 0 an c are complex numbers, the Exercise. 1 z c = z c. Proposition 13. Let α be a real number an let log z = ln z +i arg z, z > 0, α < arg z < α+2π, be a branch of log z. Then z c = e c log z is analytic in D = {(r, θ) : r > 0, α < θ < α + 2π}. Moreover z zc = cz c 1. Definition. The principal value of z c is z c = e clog z. Example 1. Fin the principal value of (1 + i) 2+3i. The square root function As a special case of the function z c we will stuy z 1/2. By efinition z 1/2 = e 1 2 log z = ± z e i 2 Arg z. This is a two-value function. The principal value of z 1 2 is z 1/2 = z e i 2 Arg z an the principal branch of z 1 2 is z 1/2 = z e i 2 Arg z, r > 0, π < Arg z < π. The principal branch of z 1 2 is analytic in D = {(r, θ) : r > 0, π < θ < π} an z z1/2 = 1 2z. 1/2 Example 1. Fin values of (1 + i) 1 2.

10 10 The function c z Definition. Let c be a nonzero complex number an let z be any complex number. We efine c z = e z log c, where log c = ln c + i arg c is the multiple-value function. Example 1. Consier 2 z an fin 2 i an 2 (i+1). Proposition 14. If a value of log c is specifie, then c z is an entire function an Exercise. z cz = c z log c. Example 1. If log i = Logi = π 2 i, then iz = e π 2 iz is an entire function.

11 Inverse trigonometric an hyperbolic functions Inverses of the trigonometric an hyperbolic functions can be efine in terms of logarithms. Consier the equation sin w = z. Substituting sin w = 1 2i (eiz e iz ) into this equation, we get e iz e iz = 2iz. Multiplying both sies by e iz an rearranging the equation, we have e 2wi 2ize iw = 1. Completing the square, we obtain Thus we have (e iw iz) 2 = 1 z 2. e iw = iz + (1 z 2 ) 1 2, where (1 z 2 ) 1 2 is ouble-value function. Taking logarithms of both sie, we have w = i log[iz + (1 z 2 ) 1 2 ]. Definition. ( Inverse of the sine function) We efine sin 1 z as sin 1 z = i log[iz + (1 z 2 ) 1 2 ], where (1 z 2 ) 1 2 = ± 1 z 2 e 1 2 Arg(1 z2). Remark. sin 1 z is a multiple-value function. Example 1. Fin the values of sin 1 2 Solution: sin 1 2 = (2n )π i ln 2 + 3, n Z. Definition. ( Inverse of the cosine an tangent functions) We efine cos 1 z = i log[z + i(1 z 2 ) 1 2 ] an where (1 z 2 ) 1 2 = ± 1 z 2 e 1 2 Arg(1 z2). tan 1 z = i ( i + z ) 2 log, i z Remark. When specific branches of the square root an logarithmic functions are use, the functions sin 1 z, cos 1 z, an tan 1 z become single-value an analytic in some omain because they are compositions of analytic functions. Moreover we have z sin 1 z = z cos 1 z = 1, (1 z 2 ) 1 2 1, (1 z 2 ) 1 2 z tan 1 z = z. 2

12 12 Definition. ( Inverse hyperbolic functions) We efine tanh 1 z = 1 ( 1 + z ) 2 log, 1 z an where (1 z 2 ) 1 2 = ± 1 z 2 e 1 2 Arg(1 z2). Example 1. Solve the equation sin z = 1. Example 2. Solve the equation cosh z = i. sinh 1 z = log[z + (z 2 + 1) 1 2 ], cosh 1 z = log[z + (z 2 1) 1 2 ], Exercises (page 85): 1 16