Complex Functions PDF

Chapter 2: Complex Functions 2-1 Complex Functions Definition: Let S be a set of complex numbers. A function f defined on S is a rule that assigns to each z in S exactly one complex number w. The number w is called the value of f at z and is denoted by f (z ) ; that is, w =...

Chapter 2: Complex Functions 2-1 Complex Functions Definition: Let S be a set of complex numbers. A function f defined on S is a rule that assigns to each z in S exactly one complex number w. The number w is called the value of f at z and is denoted by f (z ) ; that is, w = f (z ). The set S is called the domain of definition of f , and is denoted by Dom ( f ). The range of f is the set f (z ) : z  S  , and is denoted by Ran ( f ). Remark: (1) If only one value of w corresponds to each value of z , we say that w is a single- valued function of z or that f (z ) is single-valued. For example, w = z 2. (2) If more than one value of w corresponds to each value of z , we say that w is a multi-valued function of z. For example, if w 2 = z (or w = z ), then to each value of z  0 there are two values of w. Hence w 2 = z defines a multi-valued function of z. Example: If f (z ) = z 2 − (2 + i )z , evaluate complex function f at the points i and 1 + i. Solution: f (i ) = (i ) 2 − (2 + i )(i ) = −1 − 2i + 1 = −2i. f (1 + i ) = (1 + i ) 2 − (2 + i )(1 + i ) = 1 + 2i − 1 − 2 − i 2 − 3i = −1 − i Real and Imaginary Parts of a Complex Function: It is often helpful to express the inputs and the outputs a complex function in terms of their real and imaginary parts. If w = f (z ) is a single-valued function, then the image of a complex number z = x + iy under f is a complex number w = u + iv. That is, by setting z = x + iy , we can express any complex function w = f (z ) in terms of two real functions as: f (z ) =u (x , y ) + iv (x , y ) (1-1) The functions u =u (x , y ) and v = v (x , y ) (1-2) Page 1 of 26 are called the real and imaginary parts of f, respectively, in terms of the real variables x and y. Remark: To represent w = f (z ) graphically, we take two Argand diagrams: one to represent the point z and the other to represent w. The former diagram is called the XOY -plane or the z-plane and the latter UOV -plane or the w-plane. Thus, given a point ( x , y ) in the z-plane, such as P in Figure (2-1), there corresponds a point (u ,v ) in the w-plane, say P  in Figure (2-2). The set of equations (1.2) [or the Figure (2-1) Figure (2-2) equivalent, w = f (z ) ] is called a transformation. We say that point P is mapped or transformed into point P  by means of the transformation and call P  the image of P. Remark: If w = f (z ) is multi-valued, a point (or curve) in the z-plane is mapped in general into more than one point (or curve) in the w-plane. Example: If , f (z ) = z 2 , then f (x + iy ) = (x + iy ) 2 = x 2 − y 2 + 2ixy. Hence the transformation is u (x , y ) = x 2 − y 2 and v (x , y ) = 2xy Then, the image of a point (1, 2) in the z-plane is the point (-3, 4) in the w-plane. Page 2 of 26 Example: Find the real and imaginary parts of the following functions (1) f (z ) = z 2 (2) g (z ) = z 2 − (2 + i )z (3) h (z ) = z + 2 Re(z ) Solution: (1) If f (z ) = z 2 , then f (x + iy ) = (x + iy ) 2 = x 2 − y 2 + 2ixy. Hence u (x , y ) = x 2 − y 2 and v (x , y ) = 2xy. (2) Since g (z ) = z 2 − (2 + i )z , then g (z ) = (x + iy ) 2 − (2 + i )(x + iy ) = x 2 − y 2 + 2ixy − 2x − 2iy − ix + y , g (z ) = x 2 − y 2 − 2x + y + i (2xy − x − 2 y ). Hence u (x , y ) = x 2 − y 2 − 2x + y and v (x , y ) = 2xy − x − 2 y. (3) Since h (z ) = z + 2 Re(z ) , then h (z ) = x + iy + 2Re(x + iy ) = x + iy + 2x = 3x + iy. Hence u (x , y ) = 3x and v (x , y ) = y. Complex function in polar form: Given a complex function w = f (z ) , if we replace the symbol z with r ( cos + i sin  ) , then we can write this function as: f (z ) =u (r , ) + iv (r , ) (1-3) We still call the real functions u (r , ) and v ( r , ) in (1-3) are called the real and imaginary parts of f, respectively. For example, replacing z with r ( cos + i sin  ) in the function f (z ) = z 2 , yields, by de Moivre’s formula, Page 3 of 26  f (z ) = ( r ( cos + i sin  ) ) = r 2 ( cos 2 + i sin 2 ) = r 2 cos 2 + ir 2 sin 2. 2 Hence u (r , ) = r 2 cos 2 and v (r , ) = r 2 sin 2. Page 4 of 26 (2-2) The Elementary Functions: (1) Polynomial Functions: The function P (z ) = an z n + an −1z n −1 + + a1z + a0 , an  0 Is called a polynomial function of degree n , where a0 , a1 , a2 , , an  and n is a non- negative integer. The domain of the function P ( z ) is. (2) Rational Algebraic Functions: The function P (z ) f (z ) = Q (z ) is a rational function, where P ( z ) , Q ( z ) are polynomial functions. The domain of the function f is z  : Q (z )  0. Example: Find the domain of the following functions: (1) f (z ) = z (2) g (z ) = iz z +12 z +z Solution: (1) f (z ) = z , z 2 + 1 = 0  z 2 = −1  z 2 = i 2  z = i z +12 Dom(f ) = z  : z 2 + 1  0 = − i , − i . (2) g (z ) = iz , z + z = 0  x + iy + x − iy = 0  x = 0 z +z   Dom(f ) = z  : z + z  0 = z  : Re ( z )  0. Page 5 of 26 (3) Complex Exponential Function: Definition: The function e z defined by e z = e x e iy = e x ( cos y + i sin y ) Is called the complex exponential function. Modulus, Argument, and Conjugate Complex Exponential Function: If we express the complex number w = e z in polar form: w = e x ( cos y + i sin y ) = r ( cos + i sin  ). Then we see that r = e x and  = y + 2n , n = 0,  1, 2, Because r is the modulus and  is an argument of w, we have: e z = e x and arg(z ) = y + 2n , n = 0,  1, 2,. We know from calculus that e x  0 for all real x, and so it follows from e z = e x that e z  0. This implies that e z  0 for all complex z. A formula for the conjugate of the complex exponential e z is found using properties of the real cosine and sine functions. Since the real cosine function is even, we have cosy =cos(−y) for all y, and since the real sine function is odd, we have −sin y = sin(−y) for all y, and so: e z = e x ( cos y + i sin y ) = e x ( cos y − i sin y ) = e x e −iy = e x −iy = e z. Therefore, for all complex z, we have shown: ez =ez. Algebraic Properties of e z : If z 1 and z 2 are complex numbers, then (1) e 0 = 1. (2) e z 1e z 2 = e z 1 + z 2. z1 (3) e z 2 = e z 1 −z 2. e Page 6 of 26  (4) (e z ) = e nz , n = 0,  1, 2, n. Periodicity: We say that a complex function f is periodic with period T if f(z + T) = f(z) for all complex z. The real exponential function is not periodic, but the complex exponential function is because it is defined using the real cosine and sine functions, which are periodic. If we have e z + 2 i = e z (cos 2 + i sin  ). Since cos2π = 1 and sin2π = 0, this implies to: e z + 2 i = e z. Then the complex exponential function e z is periodic with a pure imaginary period 2 i. Now by repeating this process we find that e z + 2 n i = e z for n = 0,  1, 2,. Thus , 2πi, 4πi, 6πi, and so on, are also periods of e z. Example: Find the value of the following: (1) e  i (2) e − i (3) e 2 i (4) e −2 i Solution: (1) e  i = cos  + i sin  = −1 (2) e − i = cos  − i sin  = −1 (3) e 2 i = cos 2 + i sin 2 = 1 (4) e −2 i = cos 2 − i sin 2 = 1. Page 7 of 26  Exercises In Problems 1–2, evaluate the given complex function f at the indicated points. (1) f (z ) = z 2 z − 2i at 2i , 1 + i. (2) f (z ) =| z |2 −2Re ( iz ) + z at 3 − 4i , 2 − i. In Problems 3–8, find the real and imaginary parts u and v of the given complex function f as functions of x and y. 2 (3) f (z ) = 6z − 5 + 9i (4) f (z ) = z 2 + z (5) f (z ) = z (6) f (z ) = z + 1 z +1 z (7) f (z ) = e 2 z +i (8) f (z ) = e z 2 In Problems 9–12, find the real and imaginary parts u and v of the given complex function f as functions of r and . (9) f (z ) = z (10) f (z ) =| z | (11) f (z ) = z 4 (12) f (z ) = e z In Problems 13–16, find the natural domain of the given complex function f. (13) f (z ) = 2Re ( z ) − iz 2 (14) f (z ) = iz z −1 (15) f (z ) = 3z + 2i (16) f (z ) = iz z + 4z + z 3 | z | −1 (17) Prove that: (i) e z 1e z 2 = e z 1 + z 2 (ii) e z = e x (iii) e z + 2 k  i = e z , k = 0,  1,  2, Page 8 of 26 (4) Complex Trigonometric (circular) Functions: If x isa real variable, then it follows that: e ix = cos x + i sin x and e −ix = cos x − i sin x. (2-1) By adding these equations and simplifying, we obtain an equation that relates the real cosine function with the complex exponential function: − ix cos x = e + e. ix (2-2) 2 In a similar manner, if we subtract the two equations in (2-1), then we obtain an expression for the real sine function: − ix sin x = e − e. ix (2-3) 2i The formulas for the real cosine and sine functions given in (2-2) and (2-3) can be used to define the complex sine and cosine functions. Namely, we define these complex trigonometric functions by replacing the real variable x with the complex variable z in (2-2) and (2-3). Definition: The complex sine and cosine functions are defined by: − iz − iz cos z = e iz + e and sin z = e iz − e. (2-4) 2 2i Furthermore, as in calculus we define − iz i (e iz + e − iz ) tan z = sin z = e iz − e − iz , iz cot z = cos z = iz , cos z i (e + e ) sin z e − e − iz sec z = 1 = 2 , csc z = 1 = iz 2i −iz. cos z e + e −iz iz sin z e − e Equation (2-4) also shows that Euler’s formula is valid in complex: e iz = cos z + i sin z for all z. Page 9 of 26 Periodicity of Trigonometric Functions: (1) sin z and cos z are periodic functions with period 2 , that is sin( z + 2 ) = sin z and cos(z + 2 ) = cos z for all z. Proof: i ( z + 2 ) sin(z + 2 ) = e − e −i ( z +2 ) = e iz +2 i − e −iz −2 i = e iz e 2 i − e −iz e −2 i 2i 2i 2i − iz  sin(z + 2 ) = e − e = sin z , where e 2 i = 1 and e −2 i = 1. iz 2i Similarly, we can show i ( z + 2 ) cos(z + 2 ) = e + e −i ( z +2 ) = e iz +2 i + e −iz −2 i = e iz + e −iz = cos z. 2 2 2 (2) tan z and cot z are periodic functions with period  , that is tan(z +  ) = tan z and cot( z + 2 ) = cot z for all z. (3) secz and cscz are periodic functions with period 2 , that is sec(z + 2 ) = sec z and csc(z + 2 ) = csc z for all z. Trigonometric Formulas for Complex Quantities: Many of the properties familiar in the case of real trigonometric functions also hold for the complex trigonometric functions. For example, we have: (1) sin(−z ) = − sin z (2) cos(−z ) = cos z (3) tan(−z ) = − tan z (4) sin 2 z + cos 2 z = 1 (5) 1 + tan 2 z = sec 2 z (6) 1 + cot 2 z = csc 2 z (7) sin(z 1  z 2 ) = sin z 1 cos z 2  sin z 2 cos z 1 (8) cos(z 1  z 2 ) = cos z 1 cos z 2 sin z 1 sin z 2 (9) sin 2z = 2sin z cos z (10) cos 2z = cos 2 z − sin 2 z = 2cos 2 z − 1 = 1 − 2sin 2 z tan z 1  tan z 2 (11) tan(z 1  z 2 ) = 1 tan z 1 tan z 2 Page 10 of 26 (12) tan 2z = 2 tan z2 1 − tan z Example: If z = x + iy , prove that (1) cos z = cos x cosh y − i sin x sinh y. (2) cos z = cos 2 x + sinh 2 y. 2 (3) The zeros of the equation cos z = 0 is z = (2n + 1)  , n . 2 Solution: i ( x +iy ) + e −i ( x +iy ) = e ix − y + e −ix + y = 1 e ix e − y + e −ix e y 2( ) (1) cos(x + iy ) = e 2 2  cos(x + iy ) = 1 e − y ( cos x + i sin x ) + 1 e y ( cos x − i sin x ) 2 2  y −y   e −y −e y  = cos x  e + e  + i sin x   = cos x cosh y − i sin x sinh y  2   2  −y −y where cosh y = e + e and sinh y = e − e. y y 2 2 (2) From part (1) we have cos z = cos x cosh y − i sin x sinh y = ( cos x cosh y ) + ( sin x sinh y ) 2 2 = cos 2 x cosh 2 y + sin 2 x sinh 2 y  cos z = cos 2 x cosh 2 y + sin 2 x sinh 2 y. 2 Since cosh 2 y − sinh 2 y = 1  cosh 2 y = sinh 2 y + 1 , then = cos 2 x (sinh 2 y + 1) + sin 2 x sinh 2 y = cos 2 x + sinh 2 y (cos 2 x + sin 2 x ) 2 cos z  cos z = cos 2 x + sinh 2 y , where cos 2 x + sin 2 x = 1. 2 Page 11 of 26 (3) If cos z = 0 , then cos z = 0 and so (from part (2)) cos 2 x + sinh 2 y = 0  cos 2 x = 0 and sinh 2 y = 0  cos x = 0 and sinh y = 0. Hence cos x = 0 and sinh y = 0  x = (2n + 1)  , n  and y = 0. 2 Since z = x + iy , then z = (2n + 1)  , n . 2 Page 12 of 26  Exercises (1) If z = x + iy , prove that (i) sin z = sin x cosh y + i cos x sinh y. (ii) sin z = sin 2 x + sinh 2 y. 2 (iii) The zeros of the equation sin z = 0 is z = n , n . In Problems 2–9, verify the given trigonometric identity: (2) sin(−z ) = − sin z (3) cos ( −z ) = − sin z (4) tan ( −z ) = − tan z (5) cos z = cos z (6) sin z = sin z ( ) (7) sin z −  = − cos z 2 (8) cos(z 1 + z 2 ) = cos z 1 cos z 2 − sin z 1 sin z 2 (9) cos(z 1 + z 2 ) = cos z 1 cos z 2 − sin z 1 sin z 2 Page 13 of 26 (5) Complex Hyperbolic Functions: The real hyperbolic sine and hyperbolic cosine functions are defined using the real exponential function as follows: −x −x cosh x = e + e and sinh x = e − e. x x 2 2 The complex hyperbolic sine and cosine functions are defined in an analogous manner using the complex exponential function. Definition: The complex hyperbolic sine and hyperbolic cosine functions are defined by: −z −z cosh z = e z + e and sinh z = e z − e. 2 2 The complex hyperbolic tangent, cotangent, secant, and cosecant are defined in terms of cosh z and sinh z : −z −z tanh z = sinh z = e z − e − z and coth z = cosh z = e z + e − z , z z cosh z e + e sinh z e − e sech z = 1 = 2 and csch z = 1 = z 2 − z. −z cosh z e + e z sinh z e − e Relation between complex trigonometric functions and hyperbolic functions: (1) cosh iz = cos z (2) sinh iz = i sin z (3) tanh iz = i tan z (4) coth iz = −i cot z (5) sech iz = sec z (6) csch iz = −i csc z Proof: (1) cosh iz = cos z −z If we replace z by iz in the relation cosh z = e z + e , we get 2 − iz cosh iz = e + e = cos z iz 2 (2) sinh iz = i sin z −z If we replace z by iz in the relation sinh z = e − e , we get z 2 − iz − iz sinh iz = e − e = i e − e = i sin z iz iz 2 2i Page 14 of 26 (3) tanh iz = i tan z tanh iz = sinh iz = i sin z = i tan z cosh iz cos z Similarly, we have the following relations: (1) cos iz = cosh z (2) sin iz = i sinh z (3) tan iz = i tanh z (4) cot iz = −i coth z (5) sec iz = sech z (6) csc iz = −i csch z Proof: (1) cos iz = cosh z − iz If we replace z by iz in the relation cos z = e + e , we get iz 2 −z cos iz = e + e = cosh z z 2 (2) sin iz = i sinh z − iz If we replace z by iz in the relation sin z = e − e , we get iz 2i −z −z sin iz = e − e = − 1 e − e = i sinh z z z 2i i 2i (3) tan iz = i tanh z tan iz = sin iz = i sinh z = i tanh z cos iz cosh z Hyperbolic Formulas for Complex Quantities: Many of the properties familiar in the case of real hyperbolic functions also hold for the complex hyperbolic functions. For example, we have: (1) sinh(− z ) = − sinh z (2) cosh(−z ) = cosh z (3) tanh(−z ) = − tanh z (4) cosh 2 z − sinh 2 z = 1 (5) 1 − tanh 2 z = sech 2 z (6) coth 2 z − 1 = csch 2 z (7) sinh(z 1  z 2 ) = sinh z 1 cosh z 2  sinh z 2 cosh z 1 (8) cosh(z 1  z 2 ) = cosh z 1 cosh z 2  sinh z 1 sinh z 2 (9) sinh 2z = 2sinh z cosh z (10) cosh 2z = cosh 2 z + sinh 2 z = 2cosh 2 z − 1 = 1 + 2sinh 2 z tanh z 1  tanh z 2 (11) tanh(z 1  z 2 ) = 1  tanh z 1 tanh z 2 Page 15 of 26 Example: Prove that (1) sinh z = sinh x cos y + i cosh x sin y. (2) sinh z = sinh 2 x + sin 2 y 2 Solution: (1) sinh z = sinh(x + iy ) = sinh x cosh iy + sinh iy cosh x Since cosh iy = cos y and sinh iy = i sin y , then sinh z = sinh x cos y + i sin y cosh x. (2) From part (1) we get sinh z = sinh 2 x cos 2 y + sin 2 y cosh 2 x sinh z = sinh 2 x cos 2 y + sin 2 y cosh 2 x = sinh 2 x cos 2 y + sin 2 y (1 + sinh 2 x ) 2 sinh z = sinh 2 x (cos 2 y + sin 2 y ) + sin 2 y = sinh 2 x + sin 2 y 2 Page 16 of 26  Exercises In Problems 1–6, verify the given hyperbolic identity: (1) sinh(− z ) = − sinh z (2) cosh(−z ) = cosh z (3) tanh(−z ) = − tanh z (4) cosh 2 z − sinh 2 z = 1 (5) sinh(z 1 + z 2 ) = sinh z 1 cosh z 2 + sinh z 2 cosh z 1 (6) cosh(z 1 + z 2 ) = cosh z 1 cosh z 2 + sinh z 1 sinh z 2 In Problems 7–12, prove that (7) coth iz = −i cot z (8) sech iz = sec z (9) csch iz = −i csc z (10) cot iz = −i coth z (11) sec iz = sech z (12) csc iz = −i csch z (12) Prove that (i) cosh z = cosh x cos y + i sinh x sin y. (ii) cosh z = sinh 2 x + cos 2 y 2 Page 17 of 26 (6) Complex Logarithmic Function: The natural logarithm of z = x + iy is denoted by ln z and is defined as the inverse of the exponential function; that is, w = ln z is defined for z  0 by the relation e w = z. (Note that z = 0 is impossible, since e w  0 for all w ) If we set w = u + iv and z = re i  , this becomes e u +iv = re i   e u = r and v =   u = ln r and v = . Therefore, given a nonzero complex number z we have shown that (w = u + iv = ln z ): If e w = z , then w = ln z = ln r + i . ( r =| z | 0 and  = arg(z ) ) (2-5) Since the argument of z is determined only up to integer multiples of 2 , the complex natural logarithm ln z (z  0) is multi-valued function (infinitely many-valued). Then the complex natural logarithm ln z (z  0) is defined by w = ln z = ln r + i ( + 2n ), n = 0,  1,  2, (z = re i  = re i ( + 2 n ) ). (2-6) Definition: The multi-valued function ln z (z  0) defined by: ln z = ln | z | +i arg(z ). (2-7) Is called the complex logarithm. From (2-5) we see that the complex logarithm can be used to find all solutions to the exponential equation e w = z when z isa nonzero complex number. Example: Find all complex solutions to each of the following equations: (1) e w = i (2) e w = 1 + i (3) e w = −2. Solution: For each equation e w = z , the set of solutions is given by w = ln z ; ln z is given by (2-7). Page 18 of 26 (1) If z = i , then z = 1 and arg(z ) =  + 2n , n = 0,  1,  2,. Hence from (2-7) we 2 have ( ) w = ln i = ln1 + i  + 2n , n = 0,  1,  2, 2. This implies to ( ) w = i  + 2n , n = 0,  1,  2, 2 , where ln1 = 0. Therefore, each value of w is a solution to e w = i. (2) If z = 1 + i , then z = 2 and arg(z ) =  + 2n , n = 0,  1,  2,. Hence, we have 4 ( ) w = ln(1 + i ) = ln 2 + i  + 2n , n = 0,  1,  2, 4. Therefore, each value of w is a solution to e w = 1 + i. (3) If z = −2 , then z = 2 and arg(z ) =  + 2n , n = 0,  1,  2, , and so w = ln(−2) = ln 2 + i ( + 2n ) , n = 0,  1,  2,. Therefore, each value of w is a solution to e w = −2. Principal Value of a Complex Logarithm: The value of ln z corresponding to the principal value Arg( z ) ( 0  Arg(z )  2 ) is denoted by Ln z called the principal value of ln z. Definition: The complex function Ln z defined by: Ln z = ln | z | +iArg(z ) , ( 0  Arg(z )  2 ) (2-8) Is called the principal value of the complex logarithm. Page 19 of 26 Example: Compute the principal value of the complex logarithm Ln z for: (1) z = i (2) z = 1 + i (3) z = −2. Solution: (1) If z = i , then z = 1 and Arg(z ) = . Hence from (2-8) we have 2 Ln i = ln1 +  i =  i. 2 2 (2) If z = 1 + i , then z = 2 and Arg(z ) =  , and so 4 Ln(1 + i ) = ln 2 +  i. 4 (3) If z = −2 , then z = 2 and Arg(z ) =  , and so Ln(−2) = ln 2 +  i. Algebraic Properties of ln z: If z 1 and z 2 are nonzero complex numbers and n is an integer, then (1) ln ( z 1z 2 ) = ln z 1 + ln z 2. z  (2) ln  1  = ln z 1 − ln z 2 z2  Remark: In general, ln z n  n ln z , for example ln i 2  2ln i ( ) ( ) ( ln i = ln | i | +i  + 2k  = 0 + i  + 2k  = i  + 2k  2 2 2 ) ( ) 1ln i = 2i  + 2k  = i ( + 4k  ) 2 ln i 2 = ln ( −1) = ln −1 + i ( + 2k  ) = 0 + i ( + 2k  ) = i ( + 2k  ). But, in case of the principal value of the complex logarithm, we have Ln z n = nLn z. Page 20 of 26  Exercises In Problems 1–4, find all complex values of the given logarithm. (1) ln ( −5 ) (2) ln ( 2 − 2i ) (3) ln ( 2 + 6i ) ( (4) ln − 3 + i ) In Problems 5–8, write the principal value of the logarithm in the form a + ib. (5) Ln ( 6 − 6i ) ( (6) Ln 1 + 3i ) ( (7) Ln −1 − 3i ) (8) Ln ( 3 −i ) Page 21 of 26 (7) Complex Powers: Definition: If  is a complex number and z  0 , then the complex power z  is defined to be: z  = e  ln z. (2-9) In general, (2-9) gives an infinite set of values because the complex logarithm ln z is multi-valued. Example: Find the values of the given complex power: (1) i 2i (2) (1 + i )i Solution: (1) Put z = i and  = 2i in (2-9), we get i 2i = e 2i ln i. ( Since ln i = ln1 + i  + 2n , n = 0,  1,  2, 2 ) , then i 2i =e (  2 ) 2 i  i  + 2 n   =e −(  + 4 n  ) =e −( 4 n +1) , n = 0,  1,  2,. (2) Put z = 1 + i and  = i in (2-9), we get (1 + i )i = e i ln(1+i ). ( Since ln(1 + i ) = ln 2 + i  + 2n , n = 0,  1,  2, 4 ) , then (1 + i )i = e  ( 4 ) i  ln 2 +i  + 2 n   =e ( ) i ln 2 − 8 n +1  4 , n = 0,  1,  2,. Page 22 of 26 Principal Value of a Complex Power: Definition: If  is a complex number and z  0 , then the function defined by: z  = e  Ln z. (2-10) Is called the principal value of the complex power z . Example: Find the principal value of the given complex power: (1) i 2i (2) (1 + i )i Solution: (1) Put z = i and  = 2i in (2-10), we get i 2i = e 2i Ln i. Since Ln i = ln1 +  i =  i , then 2 2 i 2i =e () 2 i  i    2  = e −. (2) Put z = 1 + i and  = i in (2-10), we get (1 + i )i = e i Ln (1+i ). Since Ln (1 + i ) = ln 2 +  i , then 4 (1 + i ) = e i ( i ln 2 +  i 4 ) = e i ln 2 − 4. Properties of Complex Power: For 1 , 2  , n  , we have (1) z 1 z  2 = z 1 + 2. 1 (2) z  2 = z 1 − 2. z (3) ( z  ) = z n. n Page 23 of 26  Exercises In Problems 1–6, find all values of the given complex power. (1) ( −1) (2) ( 2i ) 3i 1−i (3) ( 3) ( ) 2 i / i (4) 1 + 3i (5) ( −i ) (ei ) i 2 (6) In Problems 7–12, find the principal value of the given complex power. (7) ( −3) (8) (1 + i ) i / 1−i (9) 24i (10) i i / ( ) (1 + i ) 3i 2 −i (11) 1 + 3 (12) (13) Suppose that z = re i . Prove that z i = e − ( +2 k  ) ( cos ( ln r ) + i sin ( ln r ) ) , k = 0,  1,  2,. Page 24 of 26 (8) Inverse Trigonometric Functions. If z = sin w , then w = sin −1 z is called the inverse sine of z or arcsine of z ( arcsin z ).Similarly, we define other inverse trigonometric or circular functions cos −1 z , tan −1 z , etc. These functions, which are multiple-valued, can be expressed in terms of natural logarithms as follows. In all cases, we omit an additive constant 2k  i , k = 0,  1,  2, , in the logarithm: ( (1) sin −1 z = 1 ln iz + 1 − z 2 i ) (2) cos −1 z = 1 ln ( z + z2 − 1) i 2i ( ) (3) tan −1 z = 1 ln 1 + iz 1 − iz (4) cot −1 z = 1 ln ( z + i ) 2i z −i  2  (5) sec −1 z = 1 ln  1 + 1 − z  i  z    (6) csc −1 z = 1 ln  i + z − 1  2 i  z  (9) Inverse Hyperbolic Functions. If z = sinhw , then w = sinh −1 z is called the inverse hyperbolic sine of z. Similarly, we define other inverse hyperbolic functions cosh −1 z , tanh −1 z , etc. These functions, which are multiple-valued, can be expressed in terms of natural logarithms as follows. In all cases, we omit an additive constant 2k  i , k = 0,  1,  2, , in the logarithm: ( (1) sinh −1 z = ln z + z 2 + 1 ) (2) cosh −1 z = ln ( z + z2 − 1) ( ) (3) tanh −1 z = 1 ln 1 + z 2 1− z (4) coth −1 z = 1 ln ( z + 1 ) 2 z −1 Page 25 of 26   2  (5) sec −1 z = ln  1 + 1 − z   z    (6) csc −1 z = ln  1 + z + 1  2  z  Page 26 of 26