Part D. Complex Analysis

Transcription

1 Part D. Comple Analsis Chapter 3. Comple Numbers and Functions. Man engineering problems ma be treated and solved b using comple numbers and comple functions. First, comple numbers and the comple plane are discussed in Secs. 3. and 3.. Some basic concept on comple analtic functions is given in Sec. 3.3 and the Cauch-Riemann equation to check the analticit of a comple function in Sec In the remaining sections of Chapter 3 we stud the most important elementar comple functions (eponential function, trigonometric functions, logarithmic functions, etc.) 3. Comple Numbers. Comple Plane B definition, a comple number is an ordered pair (, ) of real numbers and, written where = Re, = Im. = (, ) In practice, comple numbers are more frequentl written = + i or = + j. with i = -. If =, then = i ; called pure imaginar. Comple numbers satisf the same commutative, associative, and distributive laws as real numbers. 64

2 Addition, Multiplication Addition of two comple numbers = + i and = + i is defined b + = ( + i ) + ( + i ) = + + i( + ). Multiplication is defined b = ( + i )( + i ) = + i( + ) + i = - + i( + ). EXAMPLE. Real part, imaginar part, sum and product of comple numbers = 8 + 3i ==> Re = 8, Im = 3. = 9 - i ==> Re = 9, Im = -. + = (8 + 3 i) + (9 - i) = 7 + i. = (8 + 3 i)(9 - i) = 78 + i. Subtraction, Division The difference of two comple numbers is defined b - = ( + i ) - ( + i ) = - + i( - ). Division is defined b i ( i )( i ) = = + = + - = + + i - i i i + ( + )( - ) + +. EXAMPLE. Difference and quotient of comple numbers = i, = 9 - i ==> - = (8 + 3 i) - (9 - i) = i i (8 + 3 i)(9 + i) i = = = = + i. 9 - i (9 - i)(9 + i)

3 Comple Plane We choose two perpendicular coordinate aes, the horiontal -ais, called the real ais, and the vertical - ais, called the imaginar ais. We now plot a given comple number = + i as the point P with coordinates,. The -plane in which the comple numbers are represented in this wa is called the comple plane. Addition and subtraction can now be visualied in the comple plane as illustrated below. 66

4 Comple Conjugate Numbers The comple conjugate (or = + i is defined b * ) of a comple number * = - i. We can epress the real part and the imaginar part of i = + as We also have = = + = = - i * * Re ( ), Im ( ). ( + ) = +, ( - ) = -. * * * * * * * * * * * æ ö = ç = * ( ),. è ø 3. Polar Form of Comple Numbers, Powers and Roots We can substantiall increase the usefulness of the comple plane if we emplo the polar coordinates b using = r cos q, = r sin q. The polar form of is i = + i = r(cosq + isin q ) = re q where r is called the absolute value or modulus of given b r = = + and q is called the argument of given b q = = - arg tan. 67

5 Geometricall, q is the angle from the positive -ais, which is measured in radians. It should be noted that the value of q is multiple-valued, repeated at integer multiples of p. The value of q that lies in the interval - p < q p is called the principal value of the argument of and is denoted b Arg. EXAMPLE. Polar form of comple numbers. Principal argument 4 = + i ==> = (cos 4 p + isin 4 p ) = e. Hence r = =, arg = p ± np ( n =,,, L ), Arg = p. 4 4 Caution! We must pa attention to the quadrant in which lies. Let = + i, = -- i and q = arg( + i), q = arg( -- i). Then p 3p tanq = tanq =. However, Arg =, Arg = Triangle inequalit For an comple numbers we have the important triangle inequalit p i 68

6 + +. B induction, we obtain the generalied triangle inequalit + + L + n + + L + n. EXAMPLE. Triangle inequalit = + i, = - + 3i ==> i = 7 = = + 3 = 5.. Multiplication and Division in Polar Form The polar form of comple numbers is especiall useful in multiplication or division. Let r i re q i i = (cosq + sin q ) = and = r q + i q = r e q (cos sin ). Multiplication. [(cosq cosq sinq sin q ) (sinq cosq cosq sin q )] [ cos( q q ) isin( q q )] = rr - + i + = rr ==> A simple multiplication of two numbers [ cos( q q ) sin( q q )] iq iq i( q+ q ) re r e rr e rr i = = = ==> A multiplication using polar form 69

7 In general, =, arg( ) = arg + arg. Division. iq re r i( q-q ) r e i r e r r In general, [ i ] = = = cos( q - q) + sin( q - q). q =, arg = arg - arg. EXAMPLE 4. Integer powers. De Moivre s formula Let = = L = = r(cosq + isin q ) = re iq, then we obtain Thus, Roots n n n n n in n = r (cosq + isin q ) = r e q = r (cosnq + isin nq ). n (cosq + isin q ) = cosnq + isin nq ; De Moivre s formula n For = w ( n =,, L ), there are n distinct values of w satisfing the equalit. Each of these values is called an nth root of, and we write The n values of polar forms. Let n w w n w = = n. n = can easil be obtained b using = r(cosq + isin q ) and w = R(cosf + isin f). n = ==> R (cosnf + isin nf ) = r(cosq + isin q ). Thus, we have n R = r or R = r n 7

8 and nf = q + kp or q kp f = +, k is an integer. n n For k =,, L, n -, we get n distinct values. Consequentl, n ( ¹ ) has n distinct values n n æ q + kp q + kp ö w = = r ç cos + isin, è n n ø k =,, L, n -. In particular, taking =, we have r = and Arg =. Then n kp kp = cos + isin, k =,, L, n -. n n ==> called nth roots of unit. 7

9 3.3 Derivative. Analtic Function Circles and Disks. Half-Planes; Sets of The unit circle = and the circle - a = r are shown below. - a < r; open circular disk - a r; closed circular disk An open circular disk - a < r with an arbitraril small r is called the neighborhood of a. An open annulus (circular ring) r < - a < r and a closed annulus r - a r B the upper half-plane we mean the set of all points = + i such that > and the condition < defines the lower half-plane. 7

10 Concepts on Sets in the Comple Plane A point set (or set) is a collection of comple numbers. A set S is called open if ever point of S has a neighborhood that belongs to S. ==> The points in the interior of a circle or a square form an open set. An open set S is called connected if an two of its points can be joined b line segments whose points belong to S. An open connected set is called a domain; for eample, open disk or open annulus. An open square with a diagonal removed is not a domain. (Wh?) The complement of a set S in the comple plane is the set of all points of the comple plane that do not belong to S. A region is a set consisting of a domain plus some or all of its boundar points. Comple Function Let S be a set of comple numbers. Then a function f defined on S is a rule that assigns to ever in S a comple number w, written b w = f ( ). : comple variable, S: domain of definition of f. Eample: w = f ( ) = + is a comple function defined for all ==> Its domain S is the whole comple plane. The set of all values of a function f is called the range of f. 73

11 Since w is a comple function, we ma write w = f ( ) = u(, ) + iv(, ). EXAMPLE. Function of a comple variable Let w = f ( ) = + 3 = u(, ) + iv(, ). Find u(, ) and v(, ) and calculate the value of f at = + 3 i. Solution. w = ( + i) + 3( + i) = i( + 3 ). Also ==> u(, ) = - + 3, v(, ) = + 3. f i i i i ( + 3 ) = ( + 3 ) + 3( + 3 ) = EXAMPLE. Function of a comple variable * Let w = f ( ) = i + 6. Find u(, ) and v(, ) and calculate the value of f at = + 4 i. Solution. w = i( + i) + 6( - i) = i( - 6 ). Also ==> u(, ) = 6 -, v(, ) = - 6. f ( + 4 i) = i( + 4 i) + 6( - 4 i) = -5-3 i. Limit, Continuit A function f ( ) is said to have the limit l as approaches a point, if f ( ) - l < e when - < d for arbitraril small e and d, written lim f ( ) = l. 74

12 In the real case, can approach onl along the real line. In the comple case, however, can approach from an direction in the comple plane. A function f ( ) is said to be continuous at = if f ( ) is defined, and lim f ( ) = f ( ). Derivative The derivative of f ( ) at a point is defined b f '( ) = lim D f ( + D) - f ( ) D or f f ( ) - f ( ) '( ) = lim. - Note. Differentiabilit at means that, along whatever path approaches, its value should be the same. 75

13 EXAMPLE 3. Differentiabilit. Derivative The function f ( ) = is differentiable for all because f f ( + D) - f ( ) ( + D) - '( ) = lim = lim D D D D = lim( + D ) =. D The differentiation rules are the same as in real calculus such that æ f ö f ' g - fg ' ( cf )' = cf ', ( f + g)' = f ' + g ', ( fg)' = f ' g + fg ', ç = g è ø g where c is a constant. ' EXAMPLE 4. * not differentiable Let D = D + id. Then * * * f ( + D) - f ( ) ( + D) - ( D) D - id = = =. D D D D + id If D first, then f '( ) =. If D first, then f '( ) = -. Hence, b definition, * is not differentiable. Analtic Functions A function f ( ) is said to be analtic in a domain D if it is defined and differentiable at all points of D. A function f ( ) is said to be analtic at a point = if it has a derivative at ever point in some neighborhood of. 76

14 3.4 Cauch-Riemann Equations. Laplace s Equation The Cauch-Riemann equations are of most importance since the provide a criterion for the analticit of a comple function w = f ( ) = u(, ) + iv(, ). THEOREM. (Cauch-Riemann equations) A comple function f ( ) is analtic in a domain D if and onl if the first partial derivatives of u and v satisf two equations u = v, u = - v everwhere in D. These equations are called Cauch- Riemann equations. PROOF. If f ( ) is analtic, f '( ) eists such that f f ( + D) - f ( ) '( ) = lim. D D Let D = D + id. Then f [ u( + D, + D ) + iv( + D, + D) ] - [ u(, ) + iv(, ) ] '( ) = lim. D D + id We can let D approach ero along an path in a neighborhood of. Choosing path I such that D first and then D, we have [ u( + D, ) - u(, ) ] [ v( + D, ) - v(, ) ] f '( ) = lim + i lim. D D D D Since f '( ) eists, the above two limits eist. Hence f '( ) = u + iv. 77

15 Similarl, if we choose path II such that D first and then D, we have [ u(, + D) - u(, ) ] [ v(, + D) - v(, ) ] f '( ) = lim + i lim id D D = - iu + v. Equating the two equations, we have u = v, u = - v. id EXAMPLE. Cauch-Riemann equations () Show that a function Solution. = - + i ==> f ( ) = is analtic for all. u v (, ) = -, (, ) =. ==> u = = v, u = - = - v. ==> Analtic for all. () Show that a function f ( ) * = is not analtic. * Solution. = - i ==> u(, ) =, v(, ) = -. ==> u = ¹ v = -, u = - v =. ==> Not analtic. 78

16 EXAMPLE. Cauch-Riemann equations Check if f ( ) 3 = is analtic. Solution. Since = ( + i) = i(3 - ), we have u v 3 3 (, ) = - 3, (, ) = 3 -. ==> u v u v = 3-3 =, = - 6 = -. Hence f ( ) 3 = is analtic for ever. EXAMPLE 3. An analtic function of constant absolute value is constant. Show that if f ( ) is analtic in a domain D and f ( ) = k = const in D, then f ( ) = const in D. Solution. B assumption, u + v = k. B differentiation, uu + vv =, uu + vv =. Using v we have = - u in the first equation and v u uu - vu =, uu + vu =. = in the second, Multipl the first b u and the second b v and add. Then ( + ) =. u v u Similarl, multipl the first b - v and the second b u and add. Then ( u + v ) u =. If If v u + v = k =, then u = v =. ==> f ( ) = ; const. u + v ¹, then u = u =. Also since f ( ) is analtic, = v =. Thus u = const and v = const ==> f ( ) = const. 79

17 THEOREM. (Laplace s equation) If f ( ) = u(, ) + iv(, ) is analtic in a domain D, then u and v satisf Laplace s equation and Ñ u = u + u = Ñ v = v + v = respectivel, in D and have continuous second partial derivatives in D. ( Ñ read "nabla squared") PROOF. Differentiating u = v w.r.t. and u = - v w.r.t., we have u = v, u = - v. Thus u + u =. The derivative of an analtic function is itself analtic. ==> u and v have continuous partial derivatives of all order. Similarl, differentiating u = v w.r.t. and u = - v w.r.t., we obtain v + v =. 3.5 Eponential Function In the remaining sections we discuss the elementar comple functions, eponential function, trigonometric function, logarithmic function, and so on. We begin with the comple eponential function e or ep( ), 8

18 which is defined b + i e = e = e (cos + isin ). Also its derivative is ( e )' = u + iv = e cos + ie sin = e ==> analtic for all Further Properties e e = e + for an = + i and = + i. Indeed, e e = e + i e + i = e = e (cos sin ) (cos sin ) [ cos( ) sin( )] + e i + + i( + ) +. = i Let e = e = e (cos + isin ). Then e = e and arge = ± np ( n =,,, L ). Also e ¹ ==> e ¹. Periodicit of e with period p i. ± np i Since e = e for all, all the values of repeated at ± np ( n =,,, L ). w = e is The horiontal strip of width p - p < p is called the fundamental region of e. 8

19 EXAMPLE. Function values. Solution of equations ().4-.6i e = e.4 (cos.6 - isin.6) = i. () Solve the equation e = i. + i e = e = e (cos + isin ) = i. ==> e = 5, = ln 5 =.69, - æ ö = tan ç =.97. è 3 ø 4 \ = i ± np i ( n =,,, L ). 3.6 Trigonometric Functions, Hperbolic Functions Just as e etends e to comple, the comple trigonometric functions etend the real trigonometric functions. The idea is the use of Euler formulas i -i e = cos + isin, e = cos - isin. ==> i -i i -i cos = ( e + e ), sin = ( e - e ). i The Euler formula is also valid in comple: i e = cos + isin for all, 8

20 from which the following formulas also hold for comple value = + i : i -i i -i cos = ( e + e ), sin = ( e - e ). i Furthermore, we define Also, sin cos tan =, cot = cos sin sec =, csc =. cos sin (cos )' = - sin, (sin )' = cos, (tan )' = sec. EXAMPLE. Real and imaginar parts. Absolute value. Periodicit Show that Solution. cos = cos cosh - isin sinh sin = sin cosh + icos sinh cos = ( e + e ) = é ëe + e i - i i( + i) - i( + i) = e (cos + isin ) + e (cos - isin ) - = ( e + e )cos - ( e - e )sin - i - = cos cosh - isin sinh. sin = ( e - e ) = é ëe - e i - i i( + i) - i( + i) i i = e (cos + isin ) - e (cos - isin ) - i i = ( e + e )sin + ( e - e )cos - i - = sin cosh + icos sinh. ù û ù û 83

21 EXAMPLE. Solution of equations. Zeros of cos and sin Solve (a) cos = 5, (b)cos =, (c)sin =. Solution. i i (a) cos = ( e + e - ) = 5 ==> e i i - e + =. Thus i i e = e - + = 5 ± 5 - = or.. ==> e - i = or., e =. ==> = ±.9, = ± np ( n =,,, L ). \ = ± np ±.9 i ( n =,,, L ). (b) cos = cos cosh - isin sinh = ==> cos =, sinh =. ==> =, = ± (n + ) p ( n =,,, L ). \ = ± ( n + ) p ( n =,,, L ). (c) sin = sin cosh + icos sinh = ==> sin =, sinh =. ==> =, = ± np ( n =,,, L ). \ = ± np ( n =,,, L ). General formulas for real trigonometric functions also hold for comple values. cos( ± ) = cos cos m sin sin sin( ± ) = sin cos ± cos sin. cos + sin =. 84

22 Hperbolic Functions The comple hperbolic cosine and sine are defined b - - cosh = ( e + e ), sinh = ( e - e ). sinh cosh tanh =, coth =. cosh sinh Also sech =, csch =. cosh sinh (cosh )' = sinh, (sinh )' = cosh. Comple trigonometric and hperbolic functions coshi = cos, sinhi = isin. cosi = cosh, sini = isinh. 3.7 Logarithm, General Power We now introduce the comple logarithm, which is more complicated than the real logarithm. The natural logarithm of = + i is denoted b w = ln and defined b the inverse of the eponential function, that is w w = ln <==> e =. Substituting = re iq, we have iq ln = ln( re ) = ln r + iq ( r =, q = arg ) Since q = arg is multiple-valued, ln is infinitel manvalued. 85

23 The value of ln corresponding to the principal argument Arg is denoted b Ln and is called the principal value of ln. That is, Ln = ln r + iarg. ==> ln = Ln ± np i. EXAMPLE. Natural logarithm. Principal value ln = ± np i = ± np i ( n =,,, L) Ln = ln( - ) = + pi ± np i ( n =,,, L) Ln( - ) = pi ln i = + i ± np i ( n =,,, L) Lni = i p p ln( - 4 i) = i ± np i ( n =,,, L) Ln( - i) = p i p The relationships for the real natural logarithm also hold for comple, that is, But ln( ) = ln + ln, ln( / ) = ln - ln. ( ) Ln ¹ Ln + Ln. EXAMPLE. Let = = e pi = -. Then ln( ) = ln = ± np i = ± np i ( n =,, L ) ln = pi ± np i, ln = pi ± np i ( n =,,, L ) ==> ln + ln = ± np i ( n =,,, L ) But ( ) ( ) \ ln = ln + ln. L n = Ln = ¹ Ln + Ln = pi. Similarl to real e ln r = r, we have e ln =. 86

24 Since arg( e ) = ± np ( n =,,, L ) is multiple-valued, so is ln( e ) = ln( e ) + iarg( e ) = ln( e ) + i( ± np ) = + i ± npi We also have = ± np i ( n =,,, L). (ln )' =. PROOF. Let ln = u + iv, where u = r = = + v = = ± n ln ln ln( ), arg arctan p. u u = = v =, + + ( / ) æ ö = = - v = - ç ( / ) è ø ==> analtic Also æ ö (ln )' = u + iv = + i ç ( / ) è ø - i = =. + General Powers General powers of a comple number = + i are defined b the formula c c ln e c = ( : comple, ¹ ). Since ln is infinitel man-valued, multiple-valued. c will be The particular value = e c cln 87

25 c is called the principal value of. We also have a = e ln a. EXAMPLE 3. General power p ( p ) i i ln i -( / ) n i e ep( iln i) ep i i n i e p m = = = éë ± ùû = p. [ ] é { 4p p } -i ( + ) = ep ( - )ln( + ) = ep ( - ) ln + ± i i i i i n i ë [ i ] = e sin( ln ) + cos( ln ). p / 4± np ù û 88