《复变函数论 Theory of Complex Variable Functions》课程授课教案（教材讲义，复分析 Complex Analysis）Chapter Ⅲ Elementary Functions

Chapter IIIElementaryFunctionsIn this chapter, we will generalize various elementary functions to corresponding functions of acomplex variable.To be specific, we define analytic functions of a complex variablez thatreduce to the elementary functions in calculus when z=x+io.We startby definingthecomplexexponentialfunctionandthenuseittodeveloptheotherss3.1.TheExponential Function1.PropertiesofexponentialfunctionsIf wewrite=x+iyandz2=x+iy2then=(ee)(e")=(ee)(e)=e+(+)=e+2eie'-2e2 =ee"i-=2e"2de"=e'dzIf e'=βei where β=e and Φ=y,then leier andArg(e")= y + 2nπ (n =0,±1,+2,.)2.ExampleExample1.Thereare values of zsuch thate" =-1(3.1.9)To find them, we write equation (3.1.9) as e*e = le*. Then, by Proposition 1.8.1, we havee=1andy=元+2元(n=0,±l+2...)Thus,x=O,and we find that

Chapter Ⅲ Elementary Functions In this chapter, we will generalize various elementary functions to corresponding functions of a complex variable. To be specific, we define analytic functions of a complex variable that reduce to the elementary functions in calculus when z = + ixz 0 . We start by defining the complex exponential function and then use it to develop the others. §3.1. The Exponential Function 1.Properties of exponential functions If we write 111 = + iyxz and 222 = + iyxz then ))(())(( . 21 2211 21 21 2121 21 iyxiyxzz iyiyxx )( zzyyixx eeeeeeeeeeeee ++ + = = = = 221 1zzzz = eee − or 21 2 1 zz z z e e e − = . zz ee dz d = If where and φ β iz = ee x β = e φ = y ,then and xz || = ee nnye ±±=π+= K),2,1,0(2)(Arg z . 2. Example Example 1. There are values of z such that −= 1 z e . (3.1.9) To find them, we write equation (3.1.9) as . Then, by Proposition 1.8.1, we have iiyx π = 1eee = 1 x e and = π + π nny = ± ± K),2,1,0(2 . Thus, x = 0, and we find that

s3.2.TheLogarithmicFunction1.Definition of logarithm of a complex numberIf wsatisfied(3.2.1)o=where z is anynonzero complex number,then w is calleda logarithm ofthe number z2.Definition of a logarithmic functionThe setLogz = (lnz|+i(argz +2n元): neZ)=ln+iArgz, Vz eCl (0) (3.2.2)is called the logarithm of z.Usually,we writeLogz = In|z|+i(argz+2n)(n = 0, ±1, ±2,.),and then get a simple relationelg=(0)(3.2.3)Thus, we get a multi-valued functionLog :CI(0) →C,called the logarithmic function.3.ExamplesExample1.If z =-1-3i, then r = 2 and 0=-2元/3.Hence1_2+2n元)=1n2+2|nLog(-1- /3i) = In2 +i(n=0,±1,±2...)33Equality (3.2.3)isvalidforall nonzerocomplexnumber,but theequality Loge=zisnottrue.To find Loge=z,we use the definition of e'(Sec.3.1)and see that[e=e* and Arg(e")=y+2n (n=0, ±1,±2,.)when z= x+iy.Hence, we know thatLog(e')=Inlei+iArg(e')= In(e*)+i(y+2nπ)=(x+iy)+2nmi (n=0, ±1, ±2,..)ThereforeLog(ei)= z+2nπi (n=0, ±1, ±2,..)(3.2.4)The principal value of Logz is the value obtained from equation (3.2.2) when n =0there and is denoted by log z. Thuslogz = Inr+iargz(3.2.5)Note that logz is well defined and single-valued when z + O and thatLogz=logz+2nπi(n=0,±1±2,...)(3.2.6)Clearly, logz reduces to the usual logarithm in calculus when z is a positive real numberz=r.To see this, one need only write z=reio, in which case equation (3.2.5)becomeslogz=Inr,That is, logr=Inr.Example 2. From expression (3.2.2), we find thatLogl=lnl+i(0+2n元)=2nπi(n=0,±1,±2,...)As expected, log1=0Our final example here reminds us that, although we were unable to find logarithms ofnegative real numbers in calculus,we can now do soExample 3. Observe thatLog(-1)=ln1+i(π+2n元)=(2n+1)πi(n=0,±1,±2,.)and that log(-1)= πi

§3.2. The Logarithmic Function 1.Definition of logarithm of a complex number If w satisfied zew = (3.2.1) where z is any nonzero complex number,then w is called a logarithm of the number z . 2. Definition of a logarithmic function The set { } π Z zziznnzizz ∈∀+=∈++= C }0{\,Argln:)2(arg||lnLog (3.2.2) is called the logarithm of z . Usually, we write = + + nzizz π )2(arg||lnLog n = ± ± K),2,1,0( , and then get a simple relation }{ Log ze z = z ≠ )0( . (3.2.3) Thus, we get a multi-valued function }0{\:Log → CC , called the logarithmic function. 3. Examples Example 1. If −−= 31 iz , then r = 2 and θ = − π 3/2 . Hence nii π πin π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ −+=⎠ ⎞ ⎜ ⎝ ⎛ +−+=−− 3 1 22ln2 3 2 2ln)31(Log n = ±± K),2,1,0( . Equality (3.2.3) is valid for all nonzero complex number, but the equality is not true. To find , we use the definition of (Sec. 3.1) and see that zez Log = zez Log = z e xz = ee and nye π+= z 2)(Arg n = ± ± K),2,1,0( when z x += iy . Hence, we know that iniyxnyieeieez z z x π ++=++=+= 2)()2()ln()(Argln)(Log π n ±±= K),2,1,0( . Therefore, inzez += 2)(Log π n = ± ± K),2,1,0( . (3.2.4) The principal value of is the value obtained from equation (3.2.2) when there and is denoted by . Thus Logz n = 0 log z = + arglnlog zirz . (3.2.5) Note that log z is well defined and single-valued when z ≠ 0 and that = + 2logLog πinzz n = ± ± K),21,0( . (3.2.6) Clearly, log z reduces to the usual logarithm in calculus when z is a positive real number z = r . To see this, one need only write , in which case equation (3.2.5) becomes . That is, . i0 = rez = lnlog rz = lnlog rr Example 2. From expression (3.2.2), we find that += + π = 2)20(1ln1Log πinni n = ± ± K),2,1,0( . As expected, = 01log . Our final example here reminds us that, although we were unable to find logarithms of negative real numbers in calculus, we can now do so. Example 3. Observe that − += π + π = + )12()2(1ln)1(Log πinni n = ± ± K),2,1,0( and that )1log( =− πi

s3.3.BranchesandDerivativesofLogarithms1. Properties of the branche Lα(z)PutD.=(reie:r>0,α0,α0,α<0<α+2),(5)dzZ2.Definition of a branch of a multi-valued functionA branch of a multi-valued function F defined on D is any single-valued functionf:E→C suchthat(I) f is analytic on the domain E;(2) EcD; and(3) VzeE, f(z)eF(z)

§3.3. Branches and Derivatives of Logarithms 1. Properties of the branche α z)(L Put ,0:{ παθα }2 , a function defined by θ α rreD += i : DL αα → C L ) α = ln)( + irz θ ,0,( 2παθα θ rrez += i . (3.3.2) From this definition, we can prove that the function zL )( has the following properties. α (1) )( ; )( α α Dzze zL ∈∀= (2) = ∈∀ Z , whenever − nzLzfn n ),()( )12( π − π = i . , 2. Definition of a branch of a multi-valued function A branch of a multi-valued function defined on is any single-valued function such that F D : Ef → C (1) f is analytic on the domain E ; (2) E ⊂ D ; and (3) z ∈∀ E , ∈ zFzf )()(

s3.4.SomeIdentitiesonLogarithmsAs suggested by relations (3.2.3) in Sec.3.2, as well as Exercises 3, 4, and 5 with Sec.3.3, someidentities involving logarithms in calculus carry over to complex analysis and others do not. In thissection, wederive a fewthat do carry over, sometimes with qualifications as to how they are to beinterpreted. A reader who wishes to pass to Sec. 3.2 can simply refer to results here when needed.1.Operations of LogzIf z, and z2 denote any two nonzero complex numbers, then(3.4.2)Log(z,-2)= Log +Log2, Vz,2 0Log = Log-, - Log2(3.4.3)Z22. Properties of LogzWeincludetwootherpropertiesof Logzthatwillbeofspecial interestinz"=emLog(n=0,±1,±2,...)(3.4.4)When n =1, this reduces, of course, to relation (3.2.3), Sec. 3.2. Equation (3.4.4) is readilyverified by writing z = reie and noting that each side becomes r"eine.Also,21/" =exp(-Logz) (n=1, 2,..)(3.4.5)nThat is, the term on the right here has n distinct values, and those values are the nth roots ofz.Toprovethis,wewrite z=rexp(i),where istheprincipal valueof Argz.Then,inview of definition (3.2.2), Sec.3.2, of Logz,(1(0+2k元)exp(Logz)=^expl-Inr+kETnn(nThus,fromthedefinitionoftheexponential function,weobtain that(.0+2k元）1/n:keZ(3.4.6)exp(=Logz)=3"/rexplnnThis establishes property (3.4.5), which is also valid for every negative integer n too (seeExercise 5)

§3.4. Some Identities on Logarithms As suggested by relations (3.2.3) in Sec. 3.2, as well as Exercises 3, 4, and 5 with Sec. 3.3, some identities involving logarithms in calculus carry over to complex analysis and others do not. In this section, we derive a few that do carry over, sometimes with qualifications as to how they are to be interpreted. A reader who wishes to pass to Sec. 3.2 can simply refer to results here when needed. 1.Operations of Logz If and denote any two nonzero complex numbers, then 1 z 2 z 21 1 LogLog)(Log 2 = + zzzz , 0 ∀ zz 21 ≠ . (3.4.2) Log ,LogLog 1 2 2 1 zz z z −= (3.4.3) 2. Properties of Logz We include two other properties of Logz that will be of special interest in ),2,1,0( = Logznn nez ±±= K . (3.4.4) When , this reduces, of course, to relation (3.2.3), Sec. 3.2. Equation (3.4.4) is readily verified by writing and noting that each side becomes . Also, n = 1 iθ = rez inn θ er ),2,1()Log 1 exp( /1 = nz = K n z n (3.4.5) That is, the term on the right here has distinct values, and those values are the roots of . To prove this, we write n nth z = irz θ )exp( , where θ is the principal value of . Then, in view of definition (3.2.2), Sec. 3.2, of , Argz Logz ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ∈ ⎠ ⎞ ⎜ ⎝ ⎛ + = + k Z n ki r n z n : )2( ln 1 exp)Log 1 exp( πθ . Thus, from the definition of the exponential function, we obtain that : . 2 exp)Log 1 exp( n /1 n zk n k irz n = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ∈ ⎠ ⎞ ⎜ ⎝ ⎛ + = Z πθ (3.4.6) This establishes property (3.4.5), which is also valid for every negative integer too (see Exercise 5). n

s3.5.ComplexPowerFunctions1.Definitionof acomplexpowerfunctionWhen cisanycomplexnumber,thecomplexpowerzofanonzerocomplexnumberzisdefinedbymeansoftheequation- =eclog, z+0(3.5.1)Thus,we obtain amultiple-valuedfunction w= z(z+O),calleda complexpowerfunction2.ExamplesExample 1. Powers of z are, in general, multiple-valued, as illustrated by writing7-2i = exp(-2iLogi)and then1Logi = In1 +2n+(n =0, ±1, ±2,...)+2n元22This shows that1-2 = exp[(4n+1)元](n = 0, ±1, ±2,.) .(3.5.2)Note that these values of i-2iiareall realnumbers.Since the exponential function has the property 1/e=e-", one can see that11= exp(-cLog-)= 2-czcexp(cLog)and, in particular, that 1/i2 = i-2i. According to expression (3.5.2), then,(3.5.3)=exp[(4n+ )](n =0, 1, 2..),.The principal value of z° occurs when Logz is replaced by log z in definition (3.5.1):P.V. 2 = elog= =(-')-x(3.5.5)Example 2. The principal value of (-i)’isexp[ilog(-i)] = expl / In1.expThat is,元P.V.(-i)" = exp(3.5.6)Example 3. The principal branch of ≥2/3 can be witten(2222argz=3/r2exp-lnr+_iarg=expllogz=exp333.3ThusP.V./=cos2ag+Fsin,2argz(3.5.7)33w,asonecanseedirectlyfromTheorem2.12.1This function is analytic in thedomainD

§3.5. Complex Power Functions 1.Definition of a complex power function When is any complex number, the complex power of a nonzero complex number is defined by means of the equation c c z z zcc ez Log = , z ≠ 0. (3.5.1) Thus, we obtain a multiple-valued function zzw ≠= )0( , called a complex power function. c 2. Examples Example 1. Powers of z are, in general, multiple-valued, as illustrated by writing )Log2exp( 2 i ii i −= − and then ),2,1,0( 2 1 22 2 1lnLog ⎟ ±±= K ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ +=⎠ ⎞ ⎜ ⎝ ⎛ ++= π π nninii π . This shows that ),2,1,0]()14exp[( i −2i π nn ±±=+= K . (3.5.2) Note that these values of are all real numbers. i i −2 Since the exponential function has the property , one can see that zz ee − /1 = c c zzc zcz − = )Logexp( =−=)Logexp( 11 and, in particular, that . According to expression (3.5.2), then, ii ii 22 /1 − = ).,2,1,0]()14exp[( 1 2 nn ±±=+= K i i π (3.5.3) The principal value of occurs when is replaced by in definition (3.5.1): c z Logz log z == −π .P.V )( log czcc zez . (3.5.5) Example 2. The principal value of is i −i)( 2 exp 2 1lnexp)]log(exp[ ππ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =− − iiii . That is, . 2 exp)(.P.V π =− i i (3.5.6) Example 3. The principal branch of can be written 3/2 z . 3 arg2 arg exp 3 2 ln 3 2 explog 3 2 exp 3 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ = + ⎠ ⎞ ⎜ ⎝ ⎛ z z irzir Thus . 3 arg2 sin 3 arg2 .P.V cos 3/2 3 2 3 2 z ri z = rz + (3.5.7) This function is analytic in the domain , as one can see directly from Theorem 2.12.1. D π−

s3.6.TrigonometricFunctions1.Definition of sine and cosine functions of a complex variableThesineand cosinefunctionsofacomplexvariablezasfollows:e-e-ie"+e-iVZECsinz=(3.6.1)COSz=22iThese functions are entire since they are linear combinations of the entire functions e= and e-Knowing thederivatives ofthoseexponential functions,wefind from equations(3.6.1)thatddcosz=-sinz,Vze(3.6.2)sinz=coszdzdzIt is easy to seethatsin(-z)=-sinz and cos(-z)=cosz,VzEC:(3.6.3)and a variety of other identities from trigonometry are validwith complexvariableExample. In order to show that2 sinz, cosz, = sin(z, +z,)+sin(z, -z,), Vz,z, EC,(3.6.4)usingdefinitions(3.6.1)andpropertiesoftheexponentialfunction,wefirstwriteei -e-ir eF2 +e-22sinz,cosz,22iMultiplication then reduces the right-hand side here toei() +=2) e-i(+=2)-e-i(=1--2)ei(=1--2)2i2ithat is, sin(=, + z2)+ sin(z, -z2); and identity (3.6.4) is established2.Usefuleqautions(3.6.5)sin(z, +z,)= sinz, cosz, +cosz, sinz2,(3.6.6)cos(z) +z2)= cosz, cosz2-sinz, sin22sin? z+ cos? z =1,(3.6.7)sin2z=2sinzcosz,cos2z=cos2z-sin2z,(3.6.8)元一.元）sin=cosz,sinz-cOsz.(3.6.9)22sin(iy)=isinh y and cos(iy)=cosh y,(3.6.10)e'+e-y-e'where sinhy:,coshy=22sin z=sin xcoshy+icosxsinh y,(3.6.11)cosz=cosxcoshy-isinxsinhy,(3.6.12)where z=x+iysin(z+2元)= sinz, sin(z+元)=-sinz,(3.6.13)(3.6.14COs(z+2元)=COSz，COs(z+元)=-COSzsin z= sin'x+ sinh y,(3.6.15)cosz=cosx+sinhy(3.6.16)3.Definition of zero of an analytic functionAzero ofan analytic function f is a number zo suchthatf(zo)=0sinz=0ifandonlyif z=n元(n=0,±1,±2,..),0 ifamd onlyif ==+n(n=0,±1,±2..)COSz=O2

§3.6. Trigonometric Functions 1.Definition of sine and cosine functions of a complex variable The sine and cosine functions of a complex variable z as follows: i ee z iziz 2 sin − − = , 2 cos iziz ee z − + = , ∀z ∈C. (3.6.1) These functions are entire since they are linear combinations of the entire functions and . Knowing the derivatives of those exponential functions, we find from equations (3.6.1) that iz e iz e − zz dz d = cossin , zz dz d −= sincos , ∀z ∈C. (3.6.2) It is easy to see that =− −sin)sin( zz and − = cos)cos( zz , ∀z ∈C; (3.6.3) and a variety of other identities from trigonometry are valid with complex variable. Example. In order to show that )sin()sin(cossin2 21 21 21 = + + − zzzzzz , ∀ ,zz 21 ∈C, (3.6.4) using definitions (3.6.1) and properties of the exponential function, we first write ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − − 2 2 2cossin2 1 1 2 2 21 iziziz iz ee i ee zz . Multiplication then reduces the right-hand side here to i ee i ee zzizzizzizzi 2 2 )()()()( + 21 +− 21 − 21 −− 21 − + − , that is, )sin()sin( 21 21 + + − zzzz ; and identity (3.6.4) is established. 2.Useful eqautions 21 21 21 + = + sincoscossin)sin( zzzzzz , (3.6.5) 21 21 21 + = − sinsincoscos)cos( zzzzzz (3.6.6) 1cossin 2 2 zz =+ , (3.6.7) = cossin22sin zzz , zzz , (3.6.8) 2 2 −= sincos2cos z cos z 2 sin ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ + π , z cosz 2 sin ⎟ −=⎠ ⎞ ⎜ ⎝ ⎛ − π . (3.6.9) = sinh)sin( yiiy and = cosh)cos( yiy , (3.6.10) where 2 sinh yy ee y − − = , 2 cosh yy ee y − + = = coshsinsin + sinhcos yxiyxz , (3.6.11) = coshcoscos − sinhsin yxiyxz , (3.6.12) where z x += iy . z + π = sin)2sin( z , z + π = −sin)sin( z , (3.6.13) z + π = cos)2cos( z , z + π = −cos)cos( z . (3.6.14 yxz 2 2 2 += sinhsinsin , (3.6.15) yxz 2 2 2 cos cos += sinh . (3.6.16) 3.Definition of zero of an analytic function A zero of an analytic function f is a number such that 0 z 0)(zf 0 = . z = 0sin if and only if z = π nn = ± ± 2,1,0( ,K) , z = 0cos if and only if ),2,1,0( 2 π nnz ±±=+= K π

4.Definitions of other trigonometric functionssin zCOSz(3.6.17)tanz =cotz=COSzsinz11secz=(3.6.18)CSCz=COSzsinz5.Derivatives of other trigonometric functionsddtanz=sec*z,cotz=-csc22,(3.6.19)dzdzdd(3.6.20)-secz=secztanz,-CSCz=-CSCzCotzdzdz

4.Definitions of other trigonometric functions z z z cos sin tan = , z z z sin cos cot = . (3.6.17) z z cos 1 sec = , z z sin 1 csc = . (3.6.18) 5. Derivatives of other trigonometric functions zz dz d 2 = sectan , zz dz d 2 −= csccot , (3.6.19) zzz dz d = tansecsec , zzz dz d −= cotcsccsc . (3.6.20)

s3.7.HyperbolicFunctions1.Definitions of hyperbolic functionsThe hyperbolic sine and the hyperbolic cosine of a complex variable are defined aste-e-, coshz=e'+e"sinhz=(3.7.1)222. Derivatives of hyperbolic functionsdα sinh = = cosh = ,coshz = sinh z,(3.7.2)dzdz3.Useful equations(3.7.3)-isinh(iz)= sin z, cosh(iz)=cosz,(3.7.4)-isin(iz)=sinhz,cos(iz)=coshzsinh(-2)= -sinh z, cosh(-z)= cosh z(3.7.5)cosh2 z - sinh2 z=1,(3.7.6)sinh(z +z2)= sinhz, coshz +coshz, sinhz2.(3.7.7)cosh(z +z2)=coshz,coshz+sinhz,sinhz2(3.7.8)sinhz=sinhxcosy+icoshxsin y,(3.7.9)(3.7.10)coshz=coshxcosy+isinhxsiny,sinh z= sinh?x+sin’ y,(3.7.11)cosh == sinh x+cos y,(3.7.12)where z=x+iysinhz =sin(-y+ix),(3.7.13)where z=x+iy.sin(x+iy) = sin x+ sinh" ysinhz=0 if and only if z=n元i(n =0, ±1,±2,...)coshz =0 ifand only if ≥=(")+n元)i(n=0,±1,±2...)24.Definitionofotherhyperbolicfunctionsinhztanhz :(3.7.16)coshzdd-tanh z= sech'zcoshz=-csch22(3.7.17)dzdzddsechz = -sechz tanh zcschz=-cschzcoth=.(3.7.18)dzdzs3.8.InverseTrigonometricand HyperbolicFunctions1.Definitions of inverse trigonometric functionssin-' z = -iLog[iz +(1 - 22)l/2](3.8.2)ZExample. Expression (3.8.2) tells us that

§3.7. Hyperbolic Functions 1.Definitions of hyperbolic functions The hyperbolic sine and the hyperbolic cosine of a complex variable are defined as t 2 sinh zz ee z − − = , 2 cosh zz ee z − + = . (3.7.1) 2. Derivatives of hyperbolic functions zz dz d sinh = cosh , zz dz d cosh = sinh , (3.7.2) 3.Useful equations − = sin)sinh( zizi , = cos)cosh( ziz , (3.7.3) − = sinh)sin( zizi , = cosh)cos( ziz . (3.7.4) − = −sinh)sinh( zz , − = cosh)cosh( zz . (3.7.5) cosh 1sinh 2 2 zz =− , (3.7.6) 21 1 2 1 2 + = coshsinh)sinh( + sinhcosh zzzzzz , (3.7.7) 21 1 2 1 2 cosh( + = coshcosh) + sinhsinh zzzzzz (3.7.8) = cossinhsinh + sincosh yxiyxz , (3.7.9) cosh = + sinsinhcoscosh yxiyxz , (3.7.10) yxz 2 2 2 sinh += sinsinh , (3.7.11) yxz 2 2 2 cosh sinh += cos , (3.7.12) where z x += iy . 2 2 sinh +−= ixyz )sin( , (3.7.13) where z x += iy . yxiyx 2 2 2 +=+ sinhsin)sin( . z = 0sinh if and only if = π ninz = ± ± K),2,1,0( z = 0cosh if and only if ),2,1,0() 2 ( π ninz ±±=+= K π . 4.Definition of other hyperbolic function z z z cosh sinh tanh = (3.7.16) zz dz d 2 tanh = sech , zz dz d 2 cosh −= csch , (3.7.17) zzz dz d sech −= tanhsech , zzz dz d csch −= cothcsch . (3.7.18) §3.8. Inverse Trigonometric and Hyperbolic Functions 1.Definitions of inverse trigonometric functions sin ])1([Log 1 2/12 −+−= ziziz − . (3.8.2) z . Example. Expression (3.8.2) tells us that

sin-'(-i)= -iLog(1± /2)ButLog(1+ 2)= In(1+ 2)+2nm (n = 0, ±1, ±2,..)andLog(1- V2)= In(V2 -1)+(2n+1)m (n =0, ±1, ±2,)1Since In(/2-1)= In-"1+ /2=-In( + /2),Then, the numbers(-1)"In(1+ ~2)+ nπi (n=0, ±1, ±2,...)constitute the values of Log(1± 2). Thus, in rectangular form,sin-(-i)= n元 +i(-1)** n(1+ /2) (n = 0, ±1,±2,..).cos-' z = -iLog[= + (1 - 2)]/2](3.8.3)ii+ztan"' z =(3.8.4)-Log2i-z2.Derivatives of inverse trigonometric functionsd1sin-Z =(1- 23)/2 dzd-1Cos-"z=(1- 22)1/2dz1dtan- ≥ =1+23,dz3.Derivatives of inversehyperbolic functionssinh-' z = Log[= + (2? +1)1/2],cosh-l z= Log[= + (-? -1)/2].1+ zItanh-' z=↓Log1-z2

)21(Log)(sin 1 ±−=− − ii . But ++=+ 2)21ln()21(Log πin n = ± ± K),2,1,0( and ++−=− )12()12ln()21(Log πin n = ± ± K),2,1,0( . Since ln )21( 21 1 ln)12ln( −= + + =− , Then, the numbers in n )21ln()1( ++− π n = ± ± K),2,1,0( constitute the values of ± )21(Log . Thus, in rectangular form, )21ln()1()(sin 1 1 +−+=− − n+ π ini n = ± ± K),2,1,0( . cos ])1([Log 1 2/12 −+−= ziziz − (3.8.3) zi zii z − + = − Log 2 tan 1 . (3.8.4) 2. Derivatives of inverse trigonometric functions 2/12 1 )1( 1 sin z z dz d − = − , 2/12 1 )1( 1 cos z z dz d − − = − . 2 1 1 1 tan z z dz d + = − , 3. Derivatives of inverse hyperbolic functions sinh ])1([Log 1 2/12 ++= − zzz , cosh ])1([Log 1 2/12 −+= − zzz z z z − + = − 1 1 Log 2 1 tanh 1

s3.8.InverseTrigonometricandHyperbolicFunctionsInverses ofthetrigonometricand hyperbolicfunctions canbedescribed in terms of logarithms.Inordertodefinetheinversesinefunction sin-,wewriteW=sin-"zwhen z=sinw.Thatis, w= sin''z wheneh-e-hrZ=2iIf we put this equation in the form(e)-2iz(e)-1=0,which is quadratic in e, and solve for ew [see Exercise 8 (a), Sec. 1.9], we find thateh=iz+(1-22)/2(3.8.1)where (1-2)/2 is, of course, a double-valued function of z. Taking logarithms ofeach sideofequation (3.8.1) and recalling that w = sin-' z, we arrive at the expressionsin- z = -iLog[iz +(1 - 22)/21.(3.8.2)The following example illustrates the fact that sin-'z is a multiple-valued function, withinfinitelymanyvaluesat eachpointzExample. Expression (3.8.2) tells us thatsin-'(-i) = -Log(1± 2)ButLog(1+ 2) = In(1 + /2)+ 2nmi(n=0,±1,±2,...)andLog(1- /2)= In(V2 -1)+(2n+1)元i(n=0,±1,±2,...)1Since In(V2 -1)= Inn1+ /2 = -In(1+ /2),Then, the numbers(-1)" In(1+/2)+nmi(n=0,±l,±2,...)constitute the set of values of Log(1± 2). Thus, in rectangular form,sin-'(-i)= n元 + i(-1)+ In(1 + /2)(n=0.±1. ±2....)One can apply the technique used to derive expression (3.8.2) for sin-'z to show thatcos- 2 = -iLog[2+ i(1- 22)1/2](3.8.3)and thati+zitan-l _=(3.8.4)Log2i-zThe functions cos-'zand tan-'=are also multiple-values.When specific branches ofthe square root and logarithmic functions are used, all three inverse functions becomesingle-valued and analytic because they are then compositions of analytic functions.The derivatives ofthesethreefunctionsare readily obtained from theaboveexpressions.Thederivatives of the first two depend on the values chosen for the square roots1d-1d-sin~cos-(1- 22)1/2(1- 22)/2 ,dzdzd1tan-Thederivativeof thelast onedzwhichthefunction ismade single-valuedInverse hyperbolic functions can be treated in a corresponding manner. It turns out that

§3.8. Inverse Trigonometric and Hyperbolic Functions Inverses of the trigonometric and hyperbolic functions can be described in terms of logarithms. In order to define the inverse sine function , we write 1 sin − zw 1 sin− = when = sin wz . That is, zw when 1 sin− = i ee z iw iw 2 − − = . If we put this equation in the form 01)(2)( 2 =−− iw iw eize , which is quadratic in , and solve for [see Exercise 8 (a), Sec. 1.9], we find that iw e iw e 2/12 zize )1( iw −+= , (3.8.1) where is, of course, a double-valued function of . Taking logarithms of each side of equation (3.8.1) and recalling that , we arrive at the expression 2/12 − z )1( z zw 1 sin− = sin ])1([Log 1 2/12 −+−= ziziz − . (3.8.2) The following example illustrates the fact that is a multiple-valued function, with infinitely many values at each point z 1 sin − z . Example. Expression (3.8.2) tells us that )21(Log)(sin 1 ±−=− − i . But ++=+ 2)21ln()21(Log πin n = ± ± K),2,1,0( and ++−=− )12()12ln()21(Log πin n = ± ± K),2,1,0( . Since ln )21( 21 1 ln)12ln( −= + + =− , Then, the numbers in n )21ln()1( ++− π n = ± ± K),2,1,0( constitute the set of values of ± )21(Log . Thus, in rectangular form, )21ln()1()(sin 1 1 +−+=− − n+ π ini n = ± ± K),2,1,0( . One can apply the technique used to derive expression (3.8.2) for z to show that 1 sin − cos ])1([Log 1 2/12 −+−= ziziz − (3.8.3) and that zi zii z − + = − Log 2 tan 1 . (3.8.4) The functions and are also multiple-values. When specific branches of the square root and logarithmic functions are used, all three inverse functions become single-valued and analytic because they are then compositions of analytic functions. z 1 cos− z 1 tan − The derivatives of these three functions are readily obtained from the above expressions. The derivatives of the first two depend on the values chosen for the square roots: 2/12 1 )1( 1 sin z z dz d − = − , 2/12 1 )1( 1 cos z z dz d − − = − . The derivative of the last one 2 1 1 1 tan z z dz d + = − , does not, however, depend on the manner in which the function is made single-valued. Inverse hyperbolic functions can be treated in a corresponding manner. It turns out that