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 iy 1 and z 2 x 2 iy 2 are equal, z1 z 2, if Re( z1 ) Re( z 2 ) and Im( z1 ) Im( z 2 ) x + iy = 0 iff x = 0 and y = 0. Ch17_3 Arithmetic Operations z1 x1 iy 1 , z 2 x 2 iy 2 Suppose z1 z 2 ( x1 x 2 ) i ( y1 y 2 ) z1 z 2 ( x1 x 2 ) i ( y1 y 2 ) z1 z 2 ( x1 x 2 y1 y 2 ) i ( y1 x 2 x1 y 2 ) z1 z2 x1 x 2 y1 y 2 2 x2 y2 2 i y1 x 2 x1 y 2 2 x2 y2 2 Ch17_4 Complex Conjugate S u p p o se z x iy , z x iy , an d z1 z 2 z1 z 2 z1 z 2 z1 z 2 z1 z 2 z1 z 2 z1 z1 z2 z2 Ch17_5 Two important equations z z ( x iy ) ( x iy ) 2 x z z ( x iy )( x iy ) x i y x y 2 2 2 2 z z ( x iy ) ( x iy ) 2 iy and Re( z ) z z 2 , Im( z ) (1) 2 (2) (3) zz 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 y 2 2 zz Ch17_8 Example 3 If z = 2 − 3i, then z 2 ( 3) 2 2 13 As in Fig 17.2, the sum of the vectors z1 and z2 is the vector z1 + z2. Then we have z1 z 2 z 1 z 2 (5) The result in (5) is also known as the triangle inequality and extends to any finite sum: z1 z 2 ... z n z1 z 2 ... z n (6) Using (5), z1 z 2 ( z 2 ) z1 z 2 z 2 z1 z 2 z1 z 2 (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 Express 1 3i in polar form. Solution See Fig 17.4 that the point lies in the fourth quarter. r z 1 tan 3i 3 1 3 2 , arg( z ) 1 5 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 z1 r1 (cos 1 i sin 1 ) Suppose z 2 r2 (cos 2 i sin 2 ) Then z1 z 2 r1 r2 [(cos 1 cos 2 sin 1 sin 2 ) i (sin 1 cos 2 cos 1 sin 2 )] (2) for z2 0, z1 z2 r1 r2 [(cos 1 cos 2 sin 1 sin 2 ) i (sin 1 cos 2 cos 1 sin 2 )] (3) Ch17_16 From the addition formulas from trigonometry, z1z2 r1r2 [cos(1 2 ) i sin(1 2 )] z1 z2 r1 r2 [cos(1 2 ) i sin(1 2 )] (4) (5) Thus we can show | z1 z2 | | z1 | | z2 | , z1 z2 , | z1 | | z2 | (6) z1 arg ( z1 z2 ) 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 r (cos n i sin n ) n n (8) Ch17_18 Demoivre’s Formula When r = 1, then (8) becomes (cos i sin ) cos n i sin n 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 (cos n i sin n ) r (cos i sin ) n r, r n 1/ n cos n cos , sin n sin 2 k , 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 i sin n n 1/ n (10) where k = 0, 1, 2, …, n – 1. Ch17_21 17.3 Sets in the Complex Plane Terminology z x iy , z 0 x 0 iy 0 z z0 ( x x0 ) ( y y 0 ) 2 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 ( x iy ) 2 2 u ( x , y ) x y , v ( x , y ) 2 xy 2 2 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 z 0 | . Ch17_37 THEOREM 17.1 Suppose lim Then (i) (ii) z z0 Limit of Sum, Product, Quotient f ( z ) L1 and lim z z 0 g ( z ) L 2 . lim [ f ( z ) g ( z )] L1 L 2 z z0 lim f ( z ) g ( z ) L1 L 2 z z0 (iii) lim z z0 f (z) g (z) L1 L2 , L2 0 Ch17_38 DEFINITION 17.5 Continuous Function A function f is continuous at a point z0 if lim f ( z ) f ( z 0 ) z z0 A function f defined by f ( z ) a n z a n 1 z n n 1 a 2 z a1 z a 0 , a n 0 2 (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 ) lim f ( z0 z ) f ( z0 ) z 0 z (3) 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 c0, dz d cf ( z ) cf ( z ) (4) dz Sum Rules: d [ f ( z ) g ( z )] f ( z ) g ( z ) (5) dz Product Rule: d [ f ( z ) g ( z )] f ( z ) g ( z ) g ( z ) f ( z ) (6) dz Ch17_41 Quotient Rule: d f ( z ) g ( z ) f ( z ) f ( z ) g ( z ) 2 d z g ( z) [ g ( z )] (7) Chain Rule: d f ( g ( z )) f ( g ( z )) g ( z ) (8) dz Usual rule d z nz n n 1 , n an integer (9) dz Ch17_42 Example 3 Differentiate ( a ) f ( z ) 3 z 4 5 z 3 2 z , ( b ) f ( z ) z 2 4z 1 . Solution ( a ) f '( z ) 12 z 15 z 2 3 2 (4 z 1)2 z z 4 2 ( b ) f '( z ) (4 z 1) 2 4z 2z 2 (4 z 1) 2 Ch17_43 Example 4 Show that f(z) = x + 4iy is nowhere differentiable. Solution With z x i y , we have f ( z z) f ( z) ( x x ) 4 i ( y y ) x 4 iy And so lim z 0 f ( z z) f ( z) z lim z 0 x 4 i y x i y (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 x v y and u y v x (1) Ch17_47 THEOREM 17.2 Proof Proof Since f ’(z) exists, we know that f ( z ) lim f ( z z) f ( z) z (2) By writing f(z) = u(x, y) + iv(x, y), and z = x + iy, form (2) z 0 f ( z ) lim z 0 (3) u ( x x , y y ) iv ( x x , y y ) u ( x , y ) iv ( x , y ) x i y 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 f ( z ) lim u ( x x, y ) u ( x, y ) x x 0 i lim (4) v( x x, y ) v( x, y ) x x 0 By the definition, the limits in (4) are the first partial derivatives of u and v w.r.t. x. Thus f ( z ) u x i v x (5) Ch17_49 THEOREM 17.2 Proof (3) Now if z0 vertically, then z = iy and (3) becomes f ( z ) lim u ( x, y y ) u ( x, y ) i y y 0 lim iv ( x , y y ) iv ( x , y ) (6) i y y 0 which is the same as f ( z ) i u y v y (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 x u y 2x 1 2 y v y v x Ch17_51 Example 2 Show that f(z) = (2x2 + y) + i(y2 – x) is not analytic at any point. Solution u x u y 4 x and 1 and v y v x 2y 1 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 f (z) For the equation u x u y y x 2 2 (x y ) 2 2 2 2 xy (x y ) 2 2 2 x x y 2 2 i y x y 2 2 , we have v y v x 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 f ( z ) u x i v x v y i u 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 u 2 x 2 v 2 and xy u 2 Thus u 2 x 2 u y 2 v 2 yx 2 y 2 0 Similarly we have v 2 x 2 v 2 y 2 0 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 (a ) u x 2 3x 3 y , 2 u 2 x u 2 2 u x 2 6 x, u y u 2 6 xy 5 , y 2 6 x 2 y 2 6x 6x 0 Ch17_59 Example 4 (2) (b ) v y u x 3x 3 y 2 2 and v x u y 6xy 5 Integratin g the first one, v ( x, y ) 3 x y y h ( x ) 2 and v x 3 6 xy h ' ( x ), h ' ( x ) 5 , h ( x ) 5 x C Thus v ( x,y ) 3 x y y 5 x C 2 3 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 x 2 ) f ( x1 ) f ( x 2 ) (1) and the Euler’s formula is e iy cos y i sin y , Thus e x iy y : a real num ber (2) e (cos y i sin y ) x 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 e 1 .7 4 .2 i e 1 .7 (cos 4 . 2 i sin 4 . 2 ) 2 . 6873 4 . 7710 i Ch17_63 Also we have de z e z dz z1 e e z2 e z1 z2 , e z1 e z2 e z1 z2 Ch17_64 Periodicity e z i 2 e e z i 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 e u iv e (cos v i sin v ) e cos v ie sin v u u u We have x e cos v , y e sin v u u and also e 2u x y r | z | , u log e | z | tan v 2 2 y 2 2 , v 2 n , 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 ( 2 n ) ( b ) arg( i ) 2 , log e 1 0 2n ) 2 5 ( c ) arg( 1 i ) , log 4 ln( i ) i ( ln( 1 i ) 0 . 3466 i ( 5 4 e | 1 i | log e 2 0 . 3466 2n ) Ch17_69 Example 3 Find all values of z such that e z Solution z ln( 3 i ), | 3 i | 2 , arg( 3 i. 3 i) 6 z ln( 3 i ) log e 2 i ( 0 . 6931 i ( 2n ) 6 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 ) , Ln ( i ) i 2 ( c ) Arg ( 1 i ) 2 5 is not the principal value. 4 Let n 1, then Ln ( 1 i ) 0 . 3466 3 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 z 2 ) ln z1 ln z 2 ln( z1 z2 ) ln z1 ln z 2 (8) Ch17_73 Example 5 Suppose z1 = 1 and z2 = −1. If we take ln z1 = 2i, ln z2 = i, we get ln( z1 z 2 ) ln( 1) ln z1 ln z 2 3 i ln( z1 z2 ) ln( 1) ln z1 ln z 2 i 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 dz Ln z 1 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 ln z z e , z0 (10) Ch17_78 Example 6 Find the value of i2i. Solution With z i , arg z / 2 , 2 i , from (9), i 2i e 2 i [log e 1 i ( / 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 e e ix sin x 2i ix e e ix cos x ix (1) 2 Ch17_80 DEFINITION 17.11 Trigonometric Sine and Cosine For any complex number z = x + iy, sin z e iz e iz e e iz cos z 2i iz (2) 2 Four additional trigonometric functions: tan z sec z sin z , cot z 1 cos z tan z 1 1 cos z , csc z , (3) sin z 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 e e iz sin z dz dz iz e e iz 2i iz cos z 2 Similarly we have d dz d dz d dz sin z cos z tan z sec z 2 sec z sec z tan z d dz d dz d cos z sin z cot z csc z 2 (4) csc z csc z cot z dz Ch17_83 Identities sin( z ) sin z cos 2 z sin 2 cos ( z ) cos z z 1 sin( z1 z 2 ) sin z1 cos z 2 cos z1 sin z 2 cos( z1 z 2 ) cos z1 cos z 2 sin z1 sin z 2 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 e e y sinh y y e e y and cosh y 2 y (5) 2 From Definition 11.17 and Euler’s formula sin z e i ( x iy ) e i ( x iy ) 2i e e y sin x ( 2 y e e y ) i cos x ( y ) 2 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 | sin 2 | cos z | cos 2 x sinh 2 2 x sinh 2 y 2 y (8) (9) Ch17_86 Example 1 From (6) we have sin( 2 i ) sin 2 cosh 1 i cos 2 sinh 1 1 . 4301 0 . 4891 i Ch17_87 Example 2 Solve cos z = 10. Solution e e iz cos z iz 10 2 e 2 iz 20 e 1 0 , e iz iz 10 3 11 iz log e (10 3 11 ) 2 n i Since log e (10 3 11 ) log e (10 3 11 ) we have z 2 n i log e (10 3 11 ) Ch17_88 DEFINITION 17.12 Hyperbolic Sine and Cosine For any complex number z = x + iy, e e z sinh z z e e z cosh z 2 z (10) 2 Additional functions are defined as tanh z sinh z cosh z sec h z 1 cosh z 1 coth z tanh z csc h z (11) 1 sinh z Ch17_89 Similarly we have d dz sinh z cosh z and d cosh z sinh z (12) dz 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 2 i ) 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 w sin if 1 (1) z From (1), e iw e iw z, e 2 iw 2 ize iw 1 0 2i e iw iz (1 z ) 2 1/ 2 (2) Ch17_93 Solving (2) for w then gives sin 1 z i ln[iz (1 z ) 2 1/ 2 (3) ] Similarly we can get cos 1 z i ln[ z i (1 z ) 2 1/ 2 tan 1 z i 2 ln iz iz ] (4) (5) Ch17_94 Example 1 Find all values of sin Solution From (3), sin 1 (1 ( 5 ) ) 1 5. 5 i ln[ 2 1/ 2 sin 1 (4) 1/ 2 5 i ln[( 5i (1 ( 5 ) ) 2 1/ 2 ] 2i 5 2 )i ] i[log e ( 5 2 ) ( 2 n ) i ], 2 n 0 , 1, 2 ,... Ch17_95 Example 1 (2) Noting that log e ( 5 2 ) log 1 e 52 log e ( 5 2 ). Thus for n 0 , 1, 2 ,... sin 1 5 2 2 n i log e ( 5 2 ) (6) Ch17_96 Derivatives If we define w = sin-1z, z = sin w, then d dz z d sin w g ives dz dw dz U sin g co s w sin w 1, 2 (1 z ) 2 1/ 2 , 2 1 co s w co s w (1 sin w ) 2 1/ 2 th u s d sin 1 z dz d cos dz d dz 1 tan 1 z z 1 (1 z ) 2 1/ 2 1 2 1/ 2 (8) 2 (9) (1 z ) 1 1 z (7) Ch17_97 Example 2 Find the derivative of w = sin-1 z at z = 5 . Solution (1 ( 5 ) ) 2 1/ 2 dw dz z 5 (4) 1/ 2 2i 1 (1 ( 5 ) ) 2 1/ 2 1 2i 1 i 2 Ch17_98 Inverse Hyperbolic Functions Similarly we have sinh cosh 1 z ln[ z ( z 1) ] 1 z ln[ z ( z 1) ] 2 1 tanh z 2 1 ln 2 d dz sinh 1 z 1 z 1 z 1 ( z 1) 2 1/ 2 1/ 2 1/ 2 (10) (11) (12) (13) Ch17_99 d cosh 1 z dz d dz tanh 1 z 1 ( z 1) 2 1/ 2 1 1 z (14) (15) 2 Ch17_100 Example 3 Find all values of cosh-1(−1). Solution From (11), cosh 1 ( 1) ln( 1) log e 1 ( 2 n ) i ( 2 n ) i ( 2 n 1) i n 0 , 1, 2 ,... Ch17_101