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Contents 17.1 Complex Numbers 17.2 Powers and Roots 17.3 Sets in the Complex Plane 17.4 Functions of a Complex Variable 17.5 Cauchy-Riemann Equations 17.6 Exponential and Logarithmic Functions 17.7 Trigonometric and Hyperbolic Functions 17.8 Inverse Trigonometric and Hyperbolic Functions Ch17_1 17.1 Complex Numbers DEFINITION 17.1 Complex Number A complex number is any number of the z = a + ib where a and b are real numbers and i is the imaginary units. z = x + iy, the real number x is called the real part and y is called the imaginary part: Re(z) = x, Im(z) = y Ch17_2 DEFINITION 17.2 Complex Number Complex number z1 x1 iy1 and z2 x2 iy2 are equal, z1 z2, if Re( z1 ) Re( z2 ) and Im( z1 ) Im( z2 ) x + iy = 0 iff x = 0 and y = 0. Ch17_3 Arithmetic Operations Suppose z1 x1 iy1, z2 x2 iy2 z1 z2 ( x1 x2 ) i ( y1 y2 ) z1 z2 ( x1 x2 ) i ( y1 y2 ) z1 z2 ( x1x2 y1 y2 ) i ( y1x2 x1 y2 ) z1 x1x2 y1 y2 y1x2 x1 y2 2 i 2 2 2 z2 x2 y2 x2 y2 Ch17_4 Complex Conjugate Suppose z x iy, z x iy, and z1 z2 z1 z2 z1 z2 z1 z2 z1 z2 z1 z2 z1 z2 z1 z2 Ch17_5 Two important equations z z ( x iy ) ( x iy ) 2 x (1) zz ( x iy)( x iy) x 2 i 2 y 2 x 2 y 2 (2) z z ( x iy ) ( x iy ) 2iy and (3) zz zz Re( z ) , Im( z ) 2 2i Ch17_6 Geometric Interpretation Fig 17.1 is called the complex plane and a complex number z is considered as a position vector. Ch17_7 DEFINITION 17.3 Modulus or Absolute Values The modulus or absolute value of z = x + iy, denoted by │z│, is the real number | z | x 2 y 2 zz Ch17_8 Example 3 If z = 2 − 3i, then z 22 (3)2 13 As in Fig 17.2, the sum of the vectors z1 and z2 is the vector z1 + z2. Then we have z1 z2 z1 z2 (5) The result in (5) is also known as the triangle inequality and extends to any finite sum: z1 z2 ... zn z1 z2 ... zn (6) Using (5), z1 z2 ( z2 ) z1 z2 z2 z1 z2 z1 z2 (7) Ch17_9 Fig 17.2 Ch17_10 17.2 Powers and Roots Polar Form Referring to Fig 17.3, we have z = r(cos + i sin ) (1) where r = |z| is the modulus of z and is the argument of z, = arg(z). If is in the interval − < , it is called the principal argument, denoted by Arg(z). Ch17_11 Fig 17.3 Ch17_12 Example 1 Express1 3i in polar form. Solution See Fig 17.4 that the point lies in the fourth quarter. r z 1 3i 1 3 2 3 5 tan , arg( z ) 1 3 5 5 z 2 cos i sin 3 3 Ch17_13 Example 1 (2) In addition, choose that − < , thus = −/3. z 2 cos( ) i sin( ) 3 3 Ch17_14 Fig 17.4 Ch17_15 Multiplication and Division Suppose z1 r1 (cos 1 i sin 1 ) z2 r2 (cos 2 i sin 2 ) Then z1 z2 r1r2 [(cos 1 cos 2 sin 1 sin 2 ) i (sin 1 cos 2 cos 1 sin 2 )] for z2 0, z1 r1 [(cos 1 cos 2 sin 1 sin 2 ) z2 r2 i (sin 1 cos 2 cos 1 sin 2 )] (2) (3) Ch17_16 From the addition formulas from trigonometry, z1z2 r1r2[cos(1 2 ) i sin(1 2 )] (4) z1 r1 [cos(1 2 ) i sin(1 2 )] z2 r2 (5) Thus we can show z1 | z1 | | z1z2 | | z1 | | z2 | , , (6) z2 | z2 | z1 arg ( z1z2 ) arg z1 arg z2 , arg arg z1 arg z2 (7) z2 Ch17_17 Powers of z z r (cos 2 i sin 2 ) 2 2 z r (cos 3 i sin 3 ) 3 3 z n r n (cos n i sin n ) (8) Ch17_18 Demoivre’s Formula When r = 1, then (8) becomes (cos i sin )n cos n i sin n (9) Ch17_19 Roots A number w is an nth root of a nonzero number z if wn = z. If we let w = (cos + i sin ) and z = r (cos + i sin ), then n (cos n i sin n ) r (cos i sin ) n r , r1 / n cos n cos , sin n sin 2k , k 0,1,2,..., n 1 n The root corresponds to k=0 called the principal nth root. Ch17_20 Thus the n roots of a nonzero complex number z = r (cos + i sin ) are given by wk r 2k 2k cos n i sin n 1/ n (10) where k = 0, 1, 2, …, n – 1. Ch17_21 17.3 Sets in the Complex Plane Terminology z x iy, z0 x0 iy0 z z0 ( x x0 ) 2 ( y y0 ) 2 If z satisfies |z – z0| = , this point lies on a circle of radius centered at the point z0. Ch17_22 Example 1 (a) |z| = 1 is the equation of a unit circle centered at the origin. (b) |z – 1 – 2i|= 5 is the equation of a circle of radius 5 centered at 1 + 2i. Ch17_23 If z satisfies |z – z0| < , this point lies within (not on) a circle of radius centered at the point z0. The set is called a neighborhood of z0, or an open disk. A point z0 is an interior point of a set S if there exists some neighborhood of z0 that lies entirely within S. If every point of S is an interior point then S is an open set. See Fig 17.7. Ch17_24 Fig 17.7 Ch17_25 Fig 17.8 The graph of |z – (1.1 + 2i)| < 0.05 is shown in Fig 17.8. It is an open set. Ch17_26 Fig 17.9 The graph of Re(z) 1 is shown in Fig 17.9. It is not an open set. Ch17_27 Example 2 Fig 17.10 illustrates some additional open sets. Ch17_28 Example 2 (2) Ch17_29 If every neighborhood of z0 contains at least one point that is in a set S and at least one point that is not in S, z0 is said to be a boundary point of S. The boundary of S is the set of all boundary points. If any pair of points z1 and z2 in an open set S can be connected by a polygonal line that lies entirely in S is said to be connected. See Fig 17.11. An open connected set is called a domain. Ch17_30 Fig17.11 Ch17_31 A region is a domain in the complex plane with all, some or none of its boundary points. Since an open connected set does not contain any boundary points, it is a region. A region containing all its boundary points is said to be closed. Ch17_32 17.4 Functions of a Complex Variable Complex Functions w f ( z ) u ( x, y ) iv ( x, y ) (1) where u and v are real-valued functions. Also, w = f(z) can be interpreted as a mapping or transformation from the z-plane to the w-plane. See Fig 17.12. Ch17_33 Fig 17.12 Ch17_34 Example 1 Find the image of the line Re(z) = 1 under f(z) = z2. Solution f ( z ) z 2 ( x iy ) 2 u ( x, y ) x 2 y 2 , v( x, y ) 2 xy Now Re(z) = x = 1, u = 1 – y2, v = 2y. y v / 2, then u 1 v / 4 2 See Fig 17.13. Ch17_35 Fig 17.13 Ch17_36 DEFINITION 17.4 Limit of a Function Suppose the function f is defined in some neighborhood of z0, except possibly at z0 itself. Then f is said to possess a limit at z0, written lim f ( z ) L z z0 if, for each > 0, there exists a > 0 such that f ( z ) L whenever 0 | z z0 | . Ch17_37 THEOREM 17.1 Limit of Sum, Product, Quotient Suppose limzz0 f ( z ) L1 and limzz0 g ( z ) L2 . Then (i) (ii) lim [ f ( z ) g ( z )] L1 L2 z z0 lim f ( z ) g ( z ) L1L2 z z0 f ( z ) L1 (iii) lim , L2 0 z z0 g ( z ) L2 Ch17_38 DEFINITION 17.5 Continuous Function A function f is continuous at a point z0 if lim f ( z ) f ( z0 ) z z0 A function f defined by f ( z ) an z n an1z n1 a2 z 2 a1z a0 , an 0 (2) where n is a nonnegative integer and ai, i = 0, 1, 2, …, n, are complex constants, is called a polynomial of degree n. Ch17_39 DEFINITION 17.6 Derivative Suppose the complex function f is defined in a neighborhood of a point z0. The derivative of f at z0 is f ( z0 z ) f ( z0 ) f ( z0 ) lim (3) z 0 z provided this limit exists. If the limit in (3) exists, f is said to be differentiable at z0. Also, if f is differentiable at z0, then f is continuous at z0. Ch17_40 Rules of differentiation Constant Rules: d d c0, cf ( z ) cf ( z ) dz dz Sum Rules: d [ f ( z ) g ( z )] f ( z ) g ( z ) dz Product Rule: d [ f ( z ) g ( z )] f ( z ) g ( z ) g ( z ) f ( z ) dz (4) (5) (6) Ch17_41 Quotient Rule: d f ( z ) g ( z ) f ( z ) f ( z ) g ( z ) d z g ( z) [ g ( z )]2 Chain Rule: d f ( g ( z )) f ( g ( z )) g ( z ) dz Usual rule d n z nz n1 , n an integer dz (7) (8) (9) Ch17_42 Example 3 2 z 4 3 . Differentiate (a ) f ( z ) 3 z 5 z 2 z , (b) f ( z ) 4z 1 Solution (a) f '( z ) 12 z 3 15 z 2 2 (4 z 1)2 z z 4 4 z 2 z (b) f '( z ) 2 (4 z 1) (4 z 1) 2 2 2 Ch17_43 Example 4 Show that f(z) = x + 4iy is nowhere differentiable. Solution With z x iy, we have f ( z z ) f ( z ) ( x x) 4i ( y y ) x 4iy And so f ( z z ) f ( z ) x 4iy lim lim z 0 z 0 x iy z (10) Ch17_44 Example 4 (2) Now if we let z0 along a line parallel to the x-axis then y=0 and the value of (10) is 1. On the other hand, if we let z0 along a line parallel to the y-axis then x=0 and the value of (10) is 4. Therefore f(z) is not differentiable at any point z. Ch17_45 DEFINITION 17.7 Analyticity at a Point A complex function w = f(z) is said to be analytic at a point z0 if f is differentiable at z0 and at every point in some neighborhood of z0. A function is analytic at every point z is said to be an entire function. Polynomial functions are entire functions. Ch17_46 17.5 Cauchy-Riemann Equations THEOREM 17.2 Cauchy-Riemann Equations Suppose f(z) = u(x, y) + iv(x, y) is differentiable at a point z = x + iy. Then at z the first-order partial derivatives of u and v exists and satisfy the Cauchy-Riemann equations u v u v and (1) x y y x Ch17_47 THEOREM 17.2 Proof Proof Since f ’(z) exists, we know that f ( z z ) f ( z ) f ( z ) lim z 0 (2) z By writing f(z) = u(x, y) + iv(x, y), and z = x + iy, form (2) f ( z ) (3) u ( x x, y y ) iv ( x x, y y ) u ( x, y ) iv ( x, y ) lim z 0 x iy Ch17_48 THEOREM 17.2 Proof (2) Since the limit exists, z can approach zero from any direction. In particular, if z0 horizontally, then z = x and (3) becomes u ( x x, y ) u ( x, y ) f ( z ) lim x 0 x (4) v( x x, y ) v( x, y ) i lim x 0 x By the definition, the limits in (4) are the first partial derivatives of u and v w.r.t. x. Thus u v f ( z ) i (5) x x Ch17_49 THEOREM 17.2 Proof (3) Now if z0 vertically, then z = iy and (3) becomes u ( x , y y ) u ( x , y ) f ( z ) lim y 0 i y iv( x, y y ) iv( x, y ) lim y 0 i y which is the same as u v f ( z ) i y y (6) (7) Then we complete the proof. Ch17_50 Example 1 The polynomial f(z) = z2 + z is analytic for all z and f(z) = x2 − y2 + x + i(2xy + y). Thus u = x2 − y2 + x, v = 2xy + y. We can see that u v 2x 1 x y u v 2 y y x Ch17_51 Example 2 Show that f(z) = (2x2 + y) + i(y2 – x) is not analytic at any point. Solution u v 4 x and 2y x y u v 1 and 1 y x We see that u/y = −v/x but u/x = v/y is satisfied only on the line y = 2x. However, for any z on this line, there is no neighborhood or open disk about z in which f is differentiable. We conclude that f is nowhere analytic. Ch17_52 THEOREM 17.3 Criterion for Analyticity Suppose the real-valued function u(x, y) and v(x, y) are continuous and have continuous first-order partial derivatives in a domain D. If u and v satisfy the Cauchy-Riemann equations at all points of D, then the complex function f(z) = u(x, y) + iv(x, y) is analytic in D. Ch17_53 Example 3 x y For the equation f ( z ) 2 i 2 , we have 2 2 x y x y u y 2 x2 v 2 2 2 x ( x y ) y u 2 xy v 2 2 2 y x (x y ) That is, the Cauchy-Riemann equations are satisfied except at the point x2 + y2 = 0, that is z = 0. We conclude that f is analytic in any domain not containing the point z = 0. Ch17_54 From (5) and (7), we have u v v u f ( z ) i i x x y y (8) This is a formula to compute f ’(z) if f(z) is differentiable at the point z. Ch17_55 DEFINITION 17.8 Harmonic Functions A real-valued function (x, y) that has continuous second-order partial derivatives in a domain D and satisfies Laplace’s equation is said to be harmonic in D. THEOREM 17.4 A Source of Harmonic Functions Suppose f(z) = u(x, y) + iv(x, y) is analytic in a domain D. Then the functions u(x, y) and v(x, y) are harmonic functions. Ch17_56 THEOREM 17.4 Proof we assume u and v have continuous second order derivative u v u v , , then x y y x 2u 2 v 2 xy x 2u 2v and 2 yx y 2u 2u Thus 2 0 2 x y Similarly we have 2v 2v 2 0 2 x y Ch17_57 Conjugate Harmonic Functions If u and v are harmonic in D, and u(x,y)+iv(x,y) is an analytic function in D, then u and v are called the conjugate harmonic function of each other. Ch17_58 Example 4 (a) Verify u(x, y) = x3 – 3xy2 – 5y is harmonic in the entire complex plane. (b) Find the conjugate harmonic function of u. Solution 2 2 u u u u 2 2 (a) 3 x 3 y , 2 6 x, 6 xy 5, 2 6 x x y x y 2u 2u 2 6x 6x 0 2 x y Ch17_59 Example 4 (2) v u v u 2 2 (b) 3 x 3 y and 6xy 5 y x x y Integratin g the first one, v( x, y ) 3 x 2 y y 3 h( x) v and 6 xy h' ( x), h' ( x) 5, h( x) 5 x C x Thus v( x,y ) 3 x 2 y y 3 5 x C Ch17_60 17.6 Exponential and Logarithmic Functions Exponential Functions Recall that the function f(x) = ex has the property f ( x) f ( x) and f ( x1 x2 ) f ( x1 ) f ( x2 ) and the Euler’s formula is eiy cos y i sin y, Thus y : a real number (1) (2) e xiy e x (cos y i sin y) Ch17_61 DEFINITION 17.9 e e z Exponential Functions x iy e (cos y i sin y) x (3) Ch17_62 Example 1 Evaluate e1.7+4.2i. Solution e1.7 4.2i e1.7 (cos 4.2 i sin 4.2) 2.6873 4.7710i Ch17_63 Also we have dez ez dz e e e z1 z2 z1 z2 z1 e , z2 e z1 z2 e Ch17_64 Periodicity e z i 2 e z ei 2 e (cos 2 i sin 2 ) e z z Ch17_65 Polar From of a Complex number z r (cos i sin ) re i Ch17_66 Logarithm Function Given a complex number z = x + iy, z 0, we define w = ln z if z = ew (5) Let w = u + iv, then x iy eu iv eu (cos v i sin v) eu cos v ieu sin v We have x eu cos v, y eu sin v and also e2u x 2 y 2 r 2 | z |2 , u log e | z | y tan v , v 2n , arg z, n 0, 1, 2,... x Ch17_67 DEFINITION 17.10 Logarithm of a Complex Number For z 0, and = arg z, ln z log e | z | i ( 2n ) , n 0, 1, 2, (6) Ch17_68 Example 2 Find the values of (a) ln (−2) (b) ln i, (c) ln (−1 – i ). Solution (a ) arg( 2) , log e | 2 | 0.6932 ln( 2) 0.6932 i ( 2n ) (b) arg( i ) 2 , log e 1 0 ln(i ) i ( 2n ) 2 5 (c) arg( 1 i ) , log e | 1 i | log e 2 0.3466 4 5 ln( 1 i ) 0.3466 i ( 2n ) 4 Ch17_69 Example 3 Find all values of z such that e z 3 i. Solution z ln( 3 i ), | 3 i | 2, arg( 3 i ) 6 z ln( 3 i ) log e 2 i ( 2n ) 6 0.6931 i ( 2n ) 6 Ch17_70 Principal Value Ln z log e | z | iArg z (7) Since Arg z ( , ]is unique, there is only one value of Ln z for which z 0. Ch17_71 Example 4 The principal values of example 2 are as follows. (a) Arg (2) Ln (2) 0.6932 i (b) Arg (i ) 2 , Ln (i ) i 2 5 (c) Arg (1 i ) is not the principal value. 4 3 Let n 1, then Ln (1 i ) 0.3466 i 4 Ch17_72 Example 4 (2) Each function in the collection of ln z is called a branch. The function Ln z is called the principal branch or the principal logarithm function. Some familiar properties of logarithmic function hold in complex case: ln( z1 z2 ) ln z1 ln z2 z1 ln( ) ln z1 ln z2 z2 (8) Ch17_73 Example 5 Suppose z1 = 1 and z2 = −1. If we take ln z1 = 2i, ln z2 = i, we get ln( z1 z2 ) ln( 1) ln z1 ln z2 3i z1 ln( ) ln( 1) ln z1 ln z2 i z2 Ch17_74 Analyticity The function Ln z is not analytic at z = 0, since Ln 0 is not defined. Moreover, Ln z is discontinuous at all points of the negative real axis. Since Ln z is the principal branch of ln z, the nonpositive real axis is referred to as a branch cut. See Fig 17.19. Ch17_75 Fig 17.91 Ch17_76 It is left as exercises to show d 1 Ln z dz z (9) for all z in D (the complex plane except those on the non-positive real axis). Ch17_77 Complex Powers In real variables, we have x e ln x . If is a complex number, z = x + iy, we have z e ln z , z 0 (10) Ch17_78 Example 6 Find the value of i2i. Solution With z i, arg z / 2, 2i, from (9), i e 2i 2 i[log e 1i ( / 2 2 n )] e (1 4 n ) where n 0, 1, 2,... Ch17_79 17.7 Trigonometric and Hyperbolic Functions Trigonometric Functions From Euler’s Formula, we have eix eix sin x 2i eix eix cos x 2 (1) Ch17_80 DEFINITION 17.11 Trigonometric Sine and Cosine For any complex number z = x + iy, eiz e iz sin z 2i eiz e iz cos z 2 Four additional trigonometric functions: sin z 1 tan z , cot z , cos z tan z 1 1 sec z , csc z cos z sin z (2) (3) Ch17_81 Analyticity Since eiz and e-iz are entire functions, then sin z and cos z are entire functions. Besides, sin z = 0 only for the real numbers z = n and cos z = 0 only for the real numbers z = (2n+1)/2. Thus tan z and sec z are analytic except z = (2n+1)/2, and cot z and csc z are analytic except z = n. Ch17_82 Derivatives d d eiz e iz eiz e iz sin z cos z dz dz 2i 2 Similarly we have d sin z cos z dz d tan z sec 2 z dz d sec z sec z tan z dz d cos z sin z dz d (4) cot z csc 2 z dz d csc z csc z cot z dz Ch17_83 Identities sin( z ) sin z cos ( z ) cos z cos 2 z sin 2 z 1 sin( z1 z2 ) sin z1 cos z2 cos z1 sin z2 cos( z1 z2 ) cos z1 cos z2 sin z1 sin z2 sin 2 z 2 sin z cos z cos 2 z cos 2 z sin 2 z Ch17_84 Zeros If y is real, we have y y y y e e e e sinh y and cosh y 2 2 From Definition 11.17 and Euler’s formula ei ( xiy ) e i ( xiy ) sin z 2i e y e y e y e y sin x( ) i cos x( ) 2 2 (5) Ch17_85 Thus we have sin z sin x cosh y i cos x sinh y (6) cos z cos x cosh y i sin x sinh y (7) and From (6) and (7) and cosh2y = 1 + sinh2y | sin z |2 sin2 x sinh2 y (8) | cos z |2 cos 2 x sinh2 y (9) Ch17_86 Example 1 From (6) we have sin( 2 i ) sin 2 cosh 1 i cos 2 sinh 1 1.4301 0.4891i Ch17_87 Example 2 Solve cos z = 10. iz iz Solution e e cos z 10 2 e 2iz 20eiz 1 0, eiz 10 3 11 iz log e (10 3 11) 2ni Since log e (10 3 11) log e (10 3 11) we have z 2n i log e (10 3 11) Ch17_88 DEFINITION 17.12 Hyperbolic Sine and Cosine For any complex number z = x + iy, e z e z sinh z 2 e z e z cosh z 2 Additional functions are defined as sinh z 1 tanh z coth z cosh z tanh z 1 1 sec h z csc h z cosh z sinh z (10) (11) Ch17_89 Similarly we have d sinh z cosh z and d cosh z sinh z dz dz (12) sin z i sinh( iz ) , cos z cosh(iz ) (13) sinh z i sin( iz ) , cosh z cos( iz ) (14) Ch17_90 Zeros sinh z i sin iz i sin( y ix ) i[sin( y ) cosh x i cos( y ) sinh x] Since sin(−y) = − sin y, cos(−y) = cos y, then sinh z sinh x cos y i cosh x sin y (15) cosh z cosh x cos y i sinh x sin y (16) It also follows from (14) that the zeros of sinh z and cosh z are respectively, z = ni and z = (2n+1)i/2, n = 0, 1, 2, …. Ch17_91 Periodicity From (6), sin( z 2i ) sin( x iy 2 ) sin( x 2 ) cosh y i cos( x 2 ) sinh y sin x cosh y i cos x sinh y sin z The period is then 2. Ch17_92 17.8 Inverse Trigonometric and Hyperbolic Functions Inverse Sine We define z sin w From (1), if w sin 1 z eiw eiw 2 iw iw z, e 2ize 1 0 2i eiw iz (1 z 2 )1/2 (1) (2) Ch17_93 Solving (2) for w then gives sin 1 z i ln[iz (1 z 2 )1/ 2 ] Similarly we can get (3) cos 1 z i ln[z i(1 z 2 )1/ 2 ] (4) i iz tan z ln 2 iz (5) 1 Ch17_94 Example 1 Find all values of sin 1 5. Solution From (3), sin 1 5 i ln[ 5i (1 ( 5) 2 )1/ 2 ] (1 ( 5) ) 2 1/ 2 (4) 1/ 2 2i sin 1 5 i ln[( 5 2)i ] i[log e ( 5 2) ( 2n )i ], 2 n 0, 1, 2,... Ch17_95 Example 1 (2) Noting that 1 log e ( 5 2) log e log e ( 5 2). 52 Thus for n 0, 1, 2,... sin 1 5 2 2n i log e ( 5 2) (6) Ch17_96 Derivatives If we define w = sin-1z, z = sin w, then d d dw 1 z sin w gives dz dz dz cos w Using cos 2 w sin 2 w 1, cos w (1 sin 2 w)1/ 2 (1 z 2 )1/ 2 , thus d 1 1 sin z dz (1 z 2 )1/ 2 d 1 1 cos z dz (1 z 2 )1/ 2 d 1 1 tan z dz 1 z2 (7) (8) (9) Ch17_97 Example 2 Find the derivative of w = sin-1 z at z = 5. Solution (1 ( 5) 2 )1/ 2 (4)1/ 2 2i dw dz z 5 1 1 1 i 2 1/ 2 2i 2 (1 ( 5) ) Ch17_98 Inverse Hyperbolic Functions Similarly we have sinh1 z ln[z ( z 2 1)1/ 2 ] cosh 1 z ln[z ( z 2 1)1/ 2 ] (10) (11) 1 1 z tanh z ln 2 1 z (12) d 1 sinh 1 z 2 1/ 2 dz ( z 1) (13) 1 Ch17_99 1 d 1 cosh z 2 1/ 2 ( z 1 ) dz (14) d 1 1 tanh z dz 1 z2 (15) Ch17_100 Example 3 Find all values of cosh-1(−1). Solution From (11), cosh 1 (1) ln( 1) log e 1 ( 2n )i ( 2n )i (2n 1)i n 0, 1, 2,... Ch17_101