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3 Complexnumbersand hyperbolic functions This chapter is concernedwith the representationand manipulation of complex numbers. Complex numbers pervade this book, underscoringtheir wide application in the mathematics of the physical sciences.The application of complex numbers to the description of physical systemsis left until later chapters and only the basic tools are presentedhere. 3.1 The need for complex numbers Although complex numbers occur in many branchesof mathematics,they arise most directly out of solvingpolynomial equations.We examinea specificquadratic equation as an example. Consider the quadratic equation (3.1 ) z " - 4 2 *5 : 0 . Equation (3.1) has two solutions,zy a.nd22,such that (z - zr)Q - z) :0. (3.2) Using the lamiliar formula for the roots of a quadratic equation,(1.4),the solutions z1 and 22,written in brief as 2r.2,are 4+ J;$ -4 1 1 " ' (3.3) Both solutions contain the squareroot of a negativenumber. However, it is not true to say that there are no solutionsto the quadratic equation.Thefundamental theorem of algebra states that a quadratic equation will always have two solutions and these are in fact given by (3.3). The second term on the RHS of (3.3) is called an imaginaryterm sinceit contains the square root of a negativenumber; 83 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS Figure 3.1 The functionf (r) -- "' - 4z * 5. the first term is called a real term. The full solution is the sum of a real term and an imaginary term and is called a complex number. A plot of the function z2 -42 * 5 is shown in figure 3.1.It will be seenthat the plot does not IQ): intersect the z-axis, correspondingto the fact that the equation /(z) :0 has no purely real solutions. The choice of the symbol z for the quadratic variable was not arbitrary; the conventional representationof a complex number is z, where z is the sum of a real part x and i times an imaginary part y, i.e. z : x *i y , where i is usedto denote the squareroot of - 1. The real part x and the imaginary part y arc usually denoted by Rez andlmz respectively.We note at this point that some physical scientists,engineersin particular, use j instead of i. However, for consistency,we will use i throughout this book. :2J-t - 2i, and hencethe two solutions of In our particular example,lZ (3.1) are z rz : 2 t z j: z t i. Thus , her ex : 2 and, y : *1 . For compactnessa complex number is sometimeswritten in the form z : (x,y), where the componentsof z may be thought ofas coordinatesin an xy-plot. Such a plot is called an Argand diagram and is a common representationof complex numbers; an example is shown in figure 3.2. 3.2 MANIPULATION OF COMPLEX NUMBERS Figure 3.2 The Argand diagram. Our particular example of a quadratic equation may be generalisedreadily to polynomials whose highest power (degree)is greater than2, e.g.cubic equations (degree3), quartic equations(degree4) and so on. For a generalpolynomial /(z), of degreen, the fundamental theorem of algebrastatesthat the equation f(z) : O will have exactly n solutions.We will examine casesof higher-degreeequations in subsection3.4.3. The remainder of this chapter deals with: the algebra and manipulation of complex numbers; their polar representation,which has advantagesin many circumstances ; complexexponentialsand logarithms; the useof complex numbers in finding the roots of polynomial equations;and hyperbolic functions. 3.2 Manipulation of complex numbers This section considersbasic complex number manipulation. Some analogy may be drawn with vector manipulation (seechapter 7) but this section stands alone as an introduction. 3.2,1 Addition and subtraction The addition of two complex numbers, 21 and 22, in general gives another complex number. The real componentsand the imaginary componentsare added separatelyand in a like manner to the familiar addition of real numbers: 4 + 22:(xr * iyr)* (xz* iy) : (x1+x2)+ i(y I yz), 85 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS lrn z 4 +2 2 Figure3.3 The addition of two complexnumbers. or in component notation z1 | z2 - (xr"yr) * 6z,y) : (xr * xz,yr * yz). The Argand representationof the addition of two complex numbers is shown in figure 3.3. By straightforward application of the commutativity and associativity of the real and imaginary parts separately, we can show that the addition of complex numbers is itself commutative and associative,i.e. z 1 *2 2 : 2 2 +z t , z r *k z *4 ) : Q r l z z ) l z t . Thus it is immaterial in what order complex numbers are added. r sum the compl* nambas 1 + 2,, 3 - 4i, -2 + ,. Summingthe real termswe obtain l +3 - 2 : 2 . and summingthe imaginarytermswe obtain 2 i - 4 i +i : - i . Hence (1 + 2D+ (3 -4') + (-2 + i) : 2 - i. < The subtraction of complex numbers is very similar to their addition. As in the case of real numbers, if two identical complex numbers are subtracted then the result is zero. 86 3.2 MANIPULATION OF COMPLEX NUMBERS Figure 3.4 The modulus and argument of a complex number 3.2.2 Modulus and argument The modulus of the complex number z is denoted by lzl and is defined as (3.4) lt l: \ , F r t . Hence the modulus of the complex number is the distanceof the corresponding point irom the origin in the Argand diagram, as may be seenin figure 3.4. The argument of the complex number z is denoted by arg z and is defined as (3.5) a,sz-*1l-'(i) It can be seen that argz is the angle that the line joining the origin to z on the Argand diagram makes with the positive x-axis. The anticlockwise directton is taken to be positive by convention. The angle arg z is shown in figure 3.4. Account must be taken of the signs of x and y individually in determining in which quadrant arg z lies. Thus, for example,if x and y are both negative then arg z lies in the range -n < arg z < -n/2 rather than in the first quadrant (0 < arg z <n/2), though both casesgive the same value for the ratio of y to x. of the canplex rupiber z :2 >Fuil tlw mdubs aul -ii. Using (3.4), the modulus is given by Ej: JY * 1-3Y: Using (3.5),the argumentis givenby '81. ary z : tan-t (-)) . The two angleswhosetangentsequal-1.5 are -0.9828rad and 2.1588rad. Sincex : 2 and y : -3, z clearlylies in the fourth quadrant;thereforeatg z : -0.9828is the appropriate answer.< COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS 3.2.3 Muhiplication Complex numbers may be multiplied together and in general give a complex number as the result. The product of two complex numbers z1 &nd z2 is found by multiplying them out in full and remembering that i2 : -1, t-e. z1z2-- (x1+ iyt)(x2+ iy2) : xfiz * ixflz + iyrxz+ i2yryz : (4xz - t.:r) * i(4y2 * ytx). (3.6) the complexwnbers zr * 3 * 2i otd z2: -l - 4i Bv direct multiolication we find 4 2 2 : ( 3 +2 i ) ( - l - 4 i ) :-3-2i-l2i-8i2 :5-14i.< (3.7) The multiplication of complex numbers is both commutative and associative, i.e. ztz2: z2zl, (3.8) (z1z)23: ztQzzi. (3.e) The product of two complex numbers also has the simple properties laz2l: lz1llz2l, arg(zP) : arg zt I arg 22. (3.10) (3.1I ) These relations are derived in subsection3.3.1. >Vertfy tlnt (3.14) hoills , Wiluct of 4 :3 + A uid zz: *l - From (3.7) : l5- l4tl: .rFT elry : lz1z2l We also find "m 1a1:J3aY:1V' tzr:JerY+(4Y:",tm, and hence : Jm : Etzzl. < lzllz2l: "ryi"m We now examine the effect on a complex number z of multiplying it by +1 and *i. These four multipliers have modulus unity and we can seeimmediately from (3.10)that multiplying z by another complex number of unit modulus gives a product with the same modulus as z. We can also see from (3.11) that if we 88 3.2 MANIPULATION OF COMPLEX NUMBERS Figure 3.5 Multiplication of a complex number by *l and *i. multiply z by a complex number then the argument of the product is the sum of the argument of z and the argument of the multiplier. Hence multiplying z by unity (which has argument zero) leaves z unchanged in both modulus and argument, i.e. z is completely unaltered by the operation. Multiplying by -1 (which has argument z) leads to rotation, through an angle z, of the line joining the origin to z in the Argand diagram. Similarly,multiplication by i or -i leads to correspondingrotations of n12 or -nf2 respectively.This geometrical interpretation of multiplication is shown in figure 3.5. >Using the geunetricali.ntopremtionof r, r,fad tto frduct i(l-t). fnus, using(3'10)and The complexnumber1-i hasargument-n/4and modulus "8. (3.11),its product with i has argument*n/4 and unchangedmodulus u5. lte complex number with modulus.f and argument*n/4 is I * i and so i ( l - t ) : l +t , as is easilyverifiedby direct multiplication.< The division of two complex numbers is similar to their multiplication but requiresthe notion of the complex conjugate (seethe following subsection)and so discussionis postponeduntil subsection3.2.5. 3.2.4 Complex conjugate If z has the convenientform x * iy then the complex conjugate,denoted by z-, may be found simply by changing the sign of the imaginary part, i.e.tf z: x*iy then z' : x-i!. More generally,we may define the complex conjugate of z as the (complex) number having the same magnitude as z that when multiplied by z leavesa real result, i.e. there is no imaginary component in the product. COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS Figure3.6 The complexconjugateas a mirror imagein the real axis' In the case where z can be written in the form x + iy it is easily verified, by direct multiplication of the components,that the product zz' givesa real result: zz' : (x* iyXx - iy) : x2 - ixy * ixy - i2y2 : f a y2 : 1212' complex conjugation corresponds to a reflection of z in the real axis of the Argand diagram, as may be seenin figure 3.6. a*2i+31b. of z = a + Zi + 3Ib. cogugateoJ > Finrl the complexco4jugate The complexnumberis written in the standardform z : a't i(2 * 3b\; then,replacingi by -i, we obtain z ': a - i ( 2 +3 b ) . < In some cases,however,it may not be simple to rearrangethe expressionfor z into the standard form x * iy. Nevertheless, given two complex numbers, z1 and 22, it is straightforward to show that the complex conjugate of their sum (or difference)is equal to the sum (or difference)of their complex conjugates,i.e. (q * z). : zi f zi. Similarly, it may be shown that the complex conjugateof the product (or quotient) of z1 and z2 is equal to the product (or quotient) of their complex conjugates,i.e. (zrz)' : ziz) and (q/z)' : ri/ti. Using these results,it can be deduced that, no matter how complicated the expression, its complex conjugate may always be flound by replacing every i by -i. To apply this rule, however, we must always ensure that all complex parts are first written out in full. so that no i's are hidden. 3.2 MANIPULATION OF COMPLEX NUMBERS 5i of thecomptexrusnberz - alsv+2a1, "rtq! " -.1 Although we do not discusscomplexpowersuntil section3.5,the simplerule givenabove still enablesus to find the complexconjugateofz. In this casew itself containsreal and imaginarycomponentsand so must be written out in full, i.e. >fira * Wt* t-iaut z : w3t+2i\: (x + 5i)3.u+2i*. Now we can replace each i by -i to obtain z-:G-5i )(3)-2N ). It can be shown that the product zz' is real, as required. < The following properties of the complex conjugate are easily proved and others may be derived from them. If z : x * iy then (3.r2) (z')' : z, (3.1 3) (3.14) z * z ': 2 F . e z : 2 x , z - z :z l l m z : z l y , z /x2-y2\ ./ 2xy \ r:\FT})+t\;rr7) (3.r5) The derivation ofthis last relation relieson the resultsofthe following subsection. 3.2.5 Division The division of two complex numbers zy &nd z2 bears some similarity to their multiplication. Writing the quotient in component form we obtain zt 22 xr * iYr xz * iyz (3.r6) In order to separate the real and imaginary components of the quotient, we multiply both numerator and denominator by the complex conjugate of the denominator. By definition, this process will leave the denominator as a real quantity.Equation(3.16)gives zt _ (\ + iyt)(x2 - iy) _ (xrxz + yty2)I i(x2y1- xflz) ":1"ra;rtYxr-t141 xlx2 + ytV l .x)V t '| + fi W - xtv2 ']+ fi Hence we have separatedthe quotient into real and imaginary components,as required. In the specialcasewhere zz: zi. so that xz : xr and y2 : -yr, the general result reducesto (3.15). 9l COMPLEXNUMBERSAND HYPERBOLICFT]NCTIONS |Express z in theform x * iY,when 3-2i ',:- -t+4i Multiplying numerator and denominator by the complex conjugate of the denominator we obtain : -11 - 10' (3-2i )(-1-4i ) --n (-t +4r.11t -4t) 11 l0 :----1.< l7 t7 In analogy to (3.10) and (3.11),which describethe multiplication of two complex numbers,the following relations apply to division: l zt I lr l l ztl (3.r7) l z2l /zr \ : arg zt - arg 22 arsl-l - \zz/ (3.1 8) The proof of theserelationsis left until subsection3.3.1. 3.3 Polar representationof complex numbers Although consideringa complex number as the sum of a real and an imaginary proveseasierto manipulate. part is often useful,sometimesthepolar representation which is definedby function, This makes use of the complex exponential z2 ez:expz=1*z+r.+ 23 t+ (3.re) Strictly speakingit is the function expz that is defined by (3.19).The number e is the value of exp(l), i.e. it is just a number. However, it may be shown that e' and expz are equivalent when z is real and rational and mathematiciansthen define their equivalencefor irrational and complex z. For the purposesof this book we will not concern ourselvesfurther with this mathematical nicety but, rather. assumethat (3.19)is valid for all z' We also note that, using (3.19)'by multiplying together the appropriate serieswe may show that (seechapter 24) ez te'z _ ez tl r z , which is analogousto the familiar result for exponentialsof real numbers. ( 3.20) 3.3 POLAR REPRESENTATIONOF COMPLEX NUMBERS lmz Figure 3.7 The polar representation of a complex number. From (3.19), it immediately follows that for z : i0,0 teal, (3.2r) e i o :rt*- f.- ' o. :'-f* f- +;( o - $ .$)- (3.22) and hence that eio:cos? f isin0, (3.23) where the last equality follows from the series expansions of the sine and costne functions (seesubsection4.6.3).This last relationshipis called Euler's equation.lt also follows from (3.23)that e i n a : c o s n ? *i s i n n 0 for all n. From Euler's equation (3.23)and figure 3.7 we deducethat r e i e : r ( c o s l *i s i n 0 ) : x *i ! . Thus a complex number may be representedin the polar form z : reio (3.24) Referring again to figure 3.7, we can identify r with lzl and 0 with arg z' The simplicity of the representationof the modulus and argument is one of the main reasonsfor using the polar representation.The angle 0 lies conventionallyin the range-n <0 <n, but, sincerotation by 0 is the sameas rotation by 2nn*0' where n is any integer, teiq - feilq+2nrl. 93 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS rg2si(oiozl Figure 3.8 The multiplication of two complex numbers. In this case rr and 12 are both greater than unity. The algebra of the polar representation is diflerent from that of the real and imaginary component representation,though, of course,the resultsare identical. Someoperationsprove much easierin the polar representation,others much more complicated.The best representationfor a particular problem must be determined by the manipulation required. 3.3.1 Muhiplication and division in polat form Multiplication and division in polar form are particularly simple.The product of zt : rgiot and z2: r2eiozis given bY Z1z2 : 71gi0tYrai9z : Yr1'rsilqi0z). (3.2s) The relations lzrzzl: lzllz2l and arg(42) : arg zr { arg z2 follow immediately' An exampleof the multiplication of two complex numbers is shown in figure 3'8' Division is equally simple in polar form; the quotient of zl and z2 is given by zt ttgiot rl ila.-a- (3.26\ --'-!ei (at-42). ;:;"^ The relations lq/zzl -- lzlllzzl and arg(z1f z) 94 : arg zt - atg22 are again 3.4 DE MOIVRE'STHEOREM Imz Figure 3.9 The division of two complexnumbers.As in the previousfigure, ry atrd12are both greaterthan unity. immediately apparent. The division of two complex numbers in polar form is shown in figure 3.9. 3.4 de Moivre's theorem We now derive an extremelyimportant theorem. Since (et0)' : ei'o, we have (c os0* i si n0)' : cosn0* i si nn0, (3.21) where the identity eing: cosnl * isinno follows from the seriesdefinition of ei"0 (see(3.21)).This result is calledile Moiure's theoremand is often used in the manipulation of complex numbers. The theorem is valid for all n whether real, imaginary or complex. There are numerous applications of de Moivre's theorem but this section examinesjust three: proofs of trigonometric identities; finding the nth roots of unity; and solving polynomial equationswith complex roots. 3.4.1 Trigonometric identities The use of de Moivre's theorem in finding trigonometric identities is best illustrated by example. We consider the expression of a multiple-angle function in terms of a polynomial in the single-anglefunction, and its converse. 95 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS >Exgew atut cnl3,Alft wntl of pow*s Usingde Moivre'stheorem, c os 30+ js in39: ( co s 0+ i s i n g ) 3 g) + i(3sin0cos20 g - 3cos0sin2 : (cos3 -sin3 g;. (3.28) i'e. We can equatethe real and imaginarycoefficientsseparately, cos39: cos3d - 3 cosgsin20 :4cos3d-3cosd (3.2e) and 0 - sins0 sin39: 3 sin0cos2 :3sin0-4sin30. < This method can clearly be applied to finding power expansionsof cos,4dand sinnd for any positive integer n. The converseprocessusesthe following propertiesol z : eiq, u * ;:2cosno, (3.30) z" - ! (3.31) :2i stnnl . zn These equalities follow from simple applications of de Moivre's theorem, l.e. ," * I V : (cosd* i si n0)'* (cos0+ i si n0)-' : cosnd * i sinn0 * cos(-no)* i sin(-no) : cosn' * i sinn0 * cosn0 - i sinn0 :2 cosn9 and "l zn : (cos0 * isin0)'r- (cos0 * isin0)-' : cosn0* i sinn0- cosn0* isinn0 :2i stnn?. In the particular casewhere n: l, I _ eio+ e,io :2cos0, z I : ,i0 - t-io :2i si n9. z 96 (3.32) (3.33) 3.4 DE MOIVRE'STHEOREM w30 Nt M ryO. > Fird an ayessianfer cod 0 ir te'rasof {w30 Using(3.32), cos3o : l \' | / F \' *r ) t= ( s*1,+l+ 1) Now using(3.30)and 2" ,/ z b\ l\ r/. r \z-+ " )* (3.32),we find l\ 3/ s\' * ;) cos30: fcos30+ lcos0.< This result happensto be a simple rearrangementoi (3.29),but casesinvolving larger valuesol n arebetter handled using this direct method than by rearranging polynomial expansionsof multiple-anglefunctions. 3.4.2 Finding the nth roots of unrty The equation z2 : I has the familiar solutions z : !1. However, now that we have introduced the concept of complex numbers we can solve the general equation zn : l. Recalling the fundamental theorem of algebra' we know that the equation has n solutions.In order to proceedwe rewrite the equation as zn : e2iko, where k is any integer. Now taking the nth root of each side of the equation we find z : e2ikn/n' Hence, the solutions of zn : I are 2r.2.....n-- I, e2in/n, . . . , e2i\n-llo/n , correspondingto the values0,1,2,...,n- I for k-Larger integervaluesof ft do not give new solutions, since the roots already listed are simply cyclically repeated f or k : n, nll, nl2, et c . >Find trv aoh$btrsta thee4uationz1 : l. By applying the above method we find z : e2th/3 : l, zz: s2in/3o4: sain/t'16''note that, as expected, Hence the three solutions arezr:3i the next solution, for which & : 3, gives 24 : s6ia/3: | : zt so that there are only three separate solutions. { 97 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS Figure3.10 The solutionsof zr : l' Not surprisingly,given that lz3l : lzl3 from (3.10),all the roots of unity have unit modulus, i.e. they all lie on a circle in the Argand diagram of unit radius. The three roots are shown in figure 3.10. The cube roots of unity are often written 1, co and az.The propertiescoj:1 and 1 + a * a2 : 0 are easily proved. 3'4.3 Solving polynomial equations A third application of de Moivre's theorem is to the solution of polynomial equations.Complex equationsin the form of a polynomial relationshipmust first be solved for z in a similar fashion to the method lor finding the roots of real polynomial equations.Then the complex roots of z may be found. > Solue tlw eEration tt - 7s I 4/ - 6zt + ?22 - 8z * 8 = 0. We first factoriseto give ( 2 3- 2 ) ( 2 2 + 4)e - ll :0. Hencezr :2 or z2: -4 or z : l.The solutionsto the quadraticequationare z : l2i; to find the complexcuberoots,we first write the equationin the form z3:2:2e2'kn, where /r is any integer. If we now take the cube root, we get z :21/3e2i k a/3 98 3.5 COMPLEX LOGARITHMS AND COMPLEX POWERS To avoid the duplicationof solutions,we usethe fact that -T <arg z < z and find - - 1t/3 / r A\ 22 : 2 t / 3 e 2 n:i / 3z r / ' | - i + +i I ' -n< ); , z t : 2 1 / 3 e 2 n i /-3 2 t / t t - ; - +t ,,, I I -/ \- I The complexnumbers21,22 Lfld23,togetherwith zo:2i, z5 : -)i and z6: I are the solutionsto the original polynomialequation. --As ofalgebra,we find thatthe total number e"pectedfroi the fundamentaliheorem of.o.pir* .oots (six,in this case)is equalto the largestpower of z in the polynomial.< A useful fesult is that the roots of a polynomial with real coemcientsoccur in conjugate pairs (i.e. if zr is a root, then zi is a seconddistinct root, unlessz1 is real). This may be proved as follows. Let the polynomial equation of which z ts a root be Q r zn+ a n - : z n - l + " '+ a l z + a o : 0 ' Taking the complex conjugateof this equation, aiQ ' ) " * a l r Q ') '- t + " '+ a\2" + ai:g' But the an are real, and so z- satisfies an/ . ) n* a n - r l z .l n - t + " ' + a 1 z *| a s : Q ' and is also a root of the original equation. 3.5 Complex logarithms and complex powers The concept of a complex exponentialhas already been introduced in section3.3, where it was assumedthat the definition of an exponentialas a serieswas valid for complex numbers as well as for real numbers. Similarly we can define the logarithm of a complex number and we can use complex numbers as exponents. Let us denote the natural logarithm of a complex number z by w : Ln z, where the notation Ln will be explainedshortly. Thus, w must satisfy z : e *, Using (3.20),we seethat z 1z 2: g| | r gw z : g\| r +w z , and taking logarithms of both sideswe find Ln(ztz): wt * wz : Ln zr 4- Ln 22, (3'34) which shows that the familiar rule for the logarithm of the product of two real numbersalso holds for complexnumbers. COMPLEXNI,]MBERSAND HYPERBOLICFIJNCTIONS We may use (3.34)to investigatefurther the propertiesol Ln z. We have already noted that the argumentof a complexnumberis multivalued,i.e.argz : 0 +2nn' where n is any integer.Thus, in polar form, the complex number z should strictly be written as 2 : Ygil9*2nn) Taking the logarithm of both sides,and using (3.34),we find (3.35) Ln z : lnr * i(0 4-2nn), where lnr is the natural logarithm of the real positive quantity r and so is written normally. Thus from (3.35)we seethat Lnz is itself multivalued.To avoid this multivaluedbehaviourit is conventionalto defineanotherfunction lnz, the principal ualueof Ln z, which is obtained from Ln z by restricting the argument of z t o lie in t he r ange- n < 0 < n . > EoaluateLn (-i). By rewriting -i as a complex exponential,we find n/z+znntl : ien/2 Ln (-i) : Ln leir -i n/2,3i r/2, wh e r e n is a n y in te g e r . He nce Ln(-i ): principal value of Ln(-,), is given by ln(-i) : -inl2. < + 2nn), " w e note that l n(-i )' the lf z and I are both complex numbers then the zth power of r is defined by f : e'Lnt. Since Ln t is multivalued, so too is this definition. >Simplify the expression z : i-2i Firstly we take the logarithm of both sides of the equation to give Lnz: -2i Lni . Now inverting the processwe find eL".:Z:g2iLni, We can write i: where n is any integer,and hence ei\n/2+2,n), Lni :Ln l e' Grz* z* 11 :i (nl 2' t2nn). We can now simpliiy z to give i(n1)+2m\ ;-)i-.L ' 'o'**t' _ which, perhaps surprisingly,is a real quantity rather than a complex one. < Complex powers and the logarithms of complex numbers are discussed further in chapter 24. 100 3.6 APPLICATIONSTO DIFFERENTIATION AND INTEGRATION 3.6 Applicationsto differentiationand integration We can usethe exponentialform of a complexnumbertogetherwith de Moivre's functions. of trigonometric theorem(seesection3.4)to simplifythedifferentiation >Fitd the tteriuatioewith respectto x of et'ooafx. We could differentiate this function straightforwardly using the product rule (see subsection 2.1.2). However, an alternative method in this case is to use a complex exponential. Let us consider the complex number z: e1"(cos4x+jsin4x): etxe4ix- el314i\', where we have used de Moivre's theorem to rewrite the trigonometric functions as a complex exponential. This complex number has e3'cos4x as its real part. Now, differentiating z with respect to x we obtain dz :( J*4 i;stl +41':(3+4i )e3'(cos4x*i si n4x), -ax where we have again used de Moivre's theorem. Equating real parts we then find (3.36) d (el' cos4x) : e3*(3cos4x - 4sin4x). -4 X By equating the imaginary parts of (3.36), we also obtain, as a bonus, +'; : e3'(4 cos 4x * 3 sin 4x)' < {d x' 1"'in In a similar way the complex exponential can be used to evaluate integrals containing trigonometric and exponential functions. > Eualuatethe integratt - t f cotbxdx. Let us consider the integrand as the real part of the complex number eo'(cosbx * i sin Dx) : eaaeib\ - etu+iblx ' where we use de Moivre's theorem to rewrite the trigonometric functions as a complex exponential. Integrating we find r ^(a+ib)Y lp r o *i rt,dx:i I +c a+tD (a - iblga+'ot' l a - i bl l a -l i bl :+(aei b'_ a2+bz' (3.37) i bei b,)+c, where the constant of integration c is in general complex. Denoting this constant by c : ct * ic2 and equating real parts in (3.37) we obtain eo' : [,", a2 [email protected]+bsi nbx)*cr' .,' o sb xdx which agrees with result (2.37) found using integration by parts. Equating imaginary parts in (3.37) we obtain, as a bonus, t: J: I I pax d' s t n b x d x : p j *@ s i n b x - b c o s b x ) *c 2 . 101 < COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS 3.7 Hyperbolic functions The hyperbolicfunctionsare the complexanaloguesof the trigonometricfunctions. The analogy may not be immediately apparent and their definitions may appear at first to be somewhat arbitrary. However, careful examination of their properties revealsthe purpose of the definitions.For instance,their close relationship with the trigonometric functions, both in their identities and in their calculus' means that many of the familiar propertiesof trigonometricfunctionscan also be applied to the hyperbolic functions. Further, hyperbolic functions occur regularly, and so giving them specialnamesis a notational convenience. 3.7.1 Definitions The two fundamental hyperbolic functions are coshx and sinh x, which, as their namessuggest,are the hyperbolic equivalentsof cosx and sin x. They are defined by the following relations: coshx: l { e" + e-" ), -1. srnhx:;l e' -e (3.38) (3.3e) Note that coshx is an even function and sinhx is an odd function. By analogy with the trigonometric functions, the remaining hyperbolic functions are e '- e * sinhx , (3.40) t a n nr : e a +e - ^ coshx ,t2 (3.41) --_--:secn.I-: e- f e-" cosnx l2 (3.42) cosechx:-: ..' e'- - e-' srnnx 1 e t *e - * ' .I-: (3.43) Cuin tanhx e\-e x A11the hyperbolic functions above have beendefinedin terms ofthe real variable x. However, this was simply so that they may be plotted (seefigures 3'11-3'13); the definitions are equally valid for any complex number z. 3.7'2 H yperbolic-trigonometric analogie s In the previous subsectionswe have alluded to the analogy betweentrigonometric and hyperbolic functions.Here, we discussthe closerelationshipbetweenthe two groups of functions. Recalling (3.32)and (3.33)we find cosi x: ){e' + e-' ), sinjx : \i1e' - e-'). t02 3.7 HYPERBOLICFUNCTIONS Figure 3.ll Graphs of coshx and sech-x. \ cosechr sinh -r \a \ -L cosechr \ \4 Figure 3.I 2 Graphs of sinh x and cosechr. Hence, by the definitions given in the previous subsection, coshx:cosix, (3.44) i sinhx : sin ix, (3.45) cosx:coshix, (3.46) i sinx : sinhix. (3.47) These useful equations make the relationship between hyperbolicand trigono103 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS Figure3'13 Graphsof tanhx and cothx' metric functions transparent.The similarity in their calculusis discussedfurther in subsection3.7.6. 3.7.3 ldentities of hyperbolic functions The analogiesbetweentrigonometric functions and hyperbolic functions having been established,we should not be surprisedthat all the trigonometric identities also hold for hyperbolic functions, with the following modification. wherever sin2x occurs it must be replaced by - sinh2x, and vice versa. Note that this : sin2x/ cos2x replacementis necessaryeven if the sin2x is hidden, e.g. tan2x : - tanh2x' and so must be replacedby (- sinh2x/ cosh2x) Using the rulesstatedabovecos2x is replacedby cosh2x, and sin2x by - sinh2x, and so the identity becomes cosh2x-sinh2x: l. usingthe definitionsofcoshx and sinhx; see substitution, This can be verifiedby direct (3.38)and (3.39).< Some other identitiesthat can be proved in a similar way are sech2x:1-tanh2x, cosech2x:coth2x-1, sinh 2x : 2 sinh x coshx, cosh2x : cosh2x * sinh2x. (3.48) (3.4e) (3.s0) (3.51) 3.7 HYPERBOLICFUNCTIONS 3.7.4 Solving hyperbolic equations When we are presented with a hyperbolic equation to solve, we may proceed by analogy with the solution of trigonometric equations.However, it is almost always easierto expressthe equation directly in terms of exponentials. the hypobotk epatiott coshx - 5 siahx - 5 = 0. Substituting the definitions of the hyperbolic functions we obtain 1,(e' : .- +e') - ltt -e-')-5 0. Rearranging,and then multiplying throughby -e', givesin turn - 2 e 'I 3 e - '- 5 : 0 and 2 e 2 '+5 e '- 3 : 0 . Now we can factoriseand solve: ( 2 e '- l \ ( e '+ 3 ) : 0 . of the -3. Hencex: -ln2 orx: ln(-3). The interpretation Thus e* :l/2 or d: in section3'5. < logarithmof a negativenumber has beendiscussed 3.7.5 Inverces of hyperbolic functions Just like trigonometric functions, hyperbolic functions have inverses.lt y : coshx then x: cosh-ly, which servesas a definitionofthe inverse.By using the fundamental definitions of hyperbolic functions, we can find closed-form expressionsfor their inverses.This is best illustrated by example. >Fitd a cloxd-farm atpessiotfor tte bvercehyperbotk First we write x as a function of y, i.e. .l:sinh-rx + x:sinh.y. Now,sincecoshy: l1e,+e-t1and sinhy : llet -e-t\, er : coshy* sinhy r----------:- : V 1 +s i n h 'y *s i n h y e:Jta*a*, and hence y :tn(Jt + r] + r). < In a similar fashion it can be shown that - t + x;. cosh-lx : tn1..,6z J : sint-r x' COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS Figure3.14 Graphsofcosh-'x and sech-'x First we write x as a function of y, i.e' + tt:tanh-rx x:tanhY Now, using the definition of tanhy and rearranging, we find e' - e-! " Thus. it follows + l ) e- y : (x * ( l - x ) eY. eY+e- ! that ,.. 1 *x + . e':\l I -X tt+" 1- , V I -x . y:rn\/ lT+' '. , Y t -I | tanhrx:=l n(; t /l-1-Y\ -'l '< \t-x/ Graphs of the inversehyperbolic functions are given in figures 3'14-3'16' 3.7.6 Calculus of hyperboliefunctions Just as the identitiesof hyperbolic functions closelyfollow those of their trigonometric counterparts,so their calculusis similar. The derivativesof the two basic 106 3.7 HYPERBOLICFUNCTIONS Figure 3.15 Graphs of sinh ' .x and cosech 'x. F ig u r e 3.16 Graphs of tanh 1x and coth-rx. hyperbolic functions are given by d {c o s nx ) : s l n nx , ^ : cosh.x. f trtnn') (3.s2) r1 51) They may be deducedby consideringthe definitions (3.38),(3.39)as follows. t07 COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS >Ver{v tlte * - *iehx. Using the definition of coshx. c o s h x : ; ( e '+e - " ) , and differentiatingdirectly,we find 4 tcosh : I {e*- e-*} dx ") -- sinhx. < Clearly the integrals of the fundamental hyperbolic functions are also defined by these relations.The derivativesof the remaining hyperbolic functions can be derived by product differentiation and are presentedbelow only for completeness. f ftAunnl: sech2x, : x tanhx, ft{r""t x) -sech (3.54) {"or."t x) : -cosechxcothx, (3.56) [email protected]: -cosech2x. (3.55) (3.s7) The inverse hyperbolic functions also have derivatives, which are given by the following: d t .-rX\ , lc osh '- l : 4X \ A/ d / ,-rx\ - - ls lnn '- , : dx \ d , 4X d , 4X a,/ t ,-rX \ lla nn--l : \ A ,/ r ,-r X \ tco tn'-l : \ O/ I (3.58) I- ' lx. - a. I (3.5e) FTa' a a 2- x 2 ' -a x '- 4 " fot x2 < a2, (3.60) for x2 > a2. (3.61) These may be derived from the logarithmic form of the inverse (see subsection 3.7.5). 108 3.8 EXERCISES From the resultsof section3.7.5. )trt-\t *) : a, ltn(x+ r/* + t)) fr {rint,' : ; ; , ? '| T ] 1 /.' - @ l x (.6uTt+x\ | x + ftTT ::. I \ r-l \ { "tt+l / J x 2 +l 3.8 Exercises 3.1 T wo co m p le x n u m bers z andw are gi ven by z:3I4i Argand diagram, plot andw :2- i . On an (a\ z * w, (b) w - z, (c) wz, (d) z/w, (e\ z' w * w' z, (f\ w2, 1g1ln z, (h) (l * z -t w)t/2. 3.2 3.3 3.4 By considering the real and imaginary parts of the product sias'o prove the standard formulae for cos(d + d) and sin(0 * d). By writing n/12: (n/3\ - (n/4) and consideringsi"/r2,evaluate cot(n/12). Find the locus in the complex z-plane of points that satisfy the following equations. lal z -r: (b), / I +,r\ p is realandt is a realparameter p\T;J , wherec is complex, that varies in the range -.4 < t <... : a + bt + ct2, in which t is a real parameter and a, b, and c are complex numbers with b/c rcal. Evaluate (a) Re(exp2iz), (b) Im(cosh2z), (c) (-1 + \6i\1/2, (e) exp(i3),(f) Im(2i+r), (g) ,', (h) lnt(\6 +,)31. (d) | exp(,'/'z)1, 3.6 Find the equations in terms of x and y of the sets of points in the Argand diagram that satisfy the following: ( a ) Re z2 :lm zz; (b\ (lmz2)lzz : -i; ( c) a r e lz/( z- r ) l:n /2. Show that the locus of all points z:x* l z - i al : 3.8 ry in the complex plane that satisfy i l z -l i al ' )' > 0' is a circle of radius l21a/(l -,12)l centred on the point z : ial(l + )J)10 - L'z\|. Sketch the circlesfor a few typical valuesofI, including A < l,1 > 1 and,l : 1. The two sets of points z : a, z : b, z : c, and z : A, z : B, z : C arc the corners of two similar triangles in the Argand diagram. Express in terms of a ,b ,...,C COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS (a) the equalitiesofcorrespondingangles,and (b) the constantratio of correspondingsides, in the two triangles. as By noting that any complexquantity can be expressed 2 : lzlexp(iargz), deducethat a(B- C) + b(C- A) + c(.4- B) :0. For the real constanta find the loci of all points z : x I i! in the complexplane that satisfy ( /.-;^\'l ( a) Re{ ln{ ": i* } } : c , | \ z + t o/ ) ,0, ,,"{,n1=) | 3.10 \ z + u/ c>0, } : o, 0<k<n/2 ) Identify the two families of curves and verify that in case (b) all curves pass through the two points +id. The most general type of transformation between one Argand diagram, in the z-plane, and another, in the Z-plane, that gives one and only one value of Z for each value of z (and conversely) is known as the general bilinear transformation and takes the form aZ -l b :: 17af,' (a) Confirm that the transformation from the Z-plane to the z-plane is also a general bilinear transformation. (b) Recalling that the equation of a circle can be written in the form lz - z1 | l-l: ^+ ^' t, show that the general bilinear transformation transforms circles into circles (or straight lines). What is the condition that zt, zz and ,1must satisly if the transformed circle is to be a straight line? 3. 1I Sketch the parts of the Argand diagram in which 3. 12 ( a ) Re z2 < 0 , zt/2 <2; ( b ) 0 < a tg z' < n /2 ; qr. ( c) le xp z3 l a s l zl -0 What is the area of the region in which all three sets of conditions are satisfied? Denote the r4throots of unity by l, to,,,.?,, . .. , ai,-t. (a) Prove that n-1 ti r \-,,,'*n : o "' 1r:0 (r) f[a;i: (-1)'*1. r:0 (b) Express x2 + y2 + 22 - yz - zx - xy as the product of two factors, each linear in x, y and z, with coefficients dependent on the third roots of unity (and those of the x terms arbitrarilv taken as real). 3.8 EXERCISES 3. 13 Prove that x2n+r -a2n+r, where m is an integer > 1, can be written as x2n+t - a2n+t: ,"- ",_ll - zo".o, (#il l*' 3. 14 The complex position vectors of two parallel interacting equal fluid vortices moving with their axes of rotation always perpendicular to the z-plane are z1 and zz.The equations governing their motions are dzi i dti i -;:-:-- z2 -.1 -' Z 1- 22 AI 3. 15 . r] ut Deduce that (a) zr * zz, (b) lzr - z2l and (c) 1ztl2+ lzzl2are all constant in time, and hence describe the motion geometrically. Solve the equation z7 _ 426+6zs -624 +623 -1222 *82*4:0, (a) by examining the effect of setting z3 equal to 2, and then (b) by factorising and using the binomial expansion of (z + a\a' 3. 16 3. 17 Plot the sevenroots of the equation on an Argand plot, exemplifying that complex roots ofa polynomial equation always occur in conjugate pairs if the polynomial has real coefficients. The polynomial /(z) is defined by - 624+ l5z3 - 34221 362 - 48. l(r) : "s (a) Show that the equation f(z) :0 has roots of the form z : tri, where ) is real, and hence factorize /(z). (b) Show further that the cubic factor of f(z) can be written in the lorm (z + a\3 Ib, where a and b are real, and hence solve the equation f(z) : 0 completely. The binomial expansion of (1 f x)', discussed in chapter 1, can be written for a positive integer n as (t+x, . ":2_+,,^ c,x. where "C : nl/Irl(n - r)ll. (a) Use de Moivre's theorem to show that the sum S1 ( n ) :' 6 0 -"C z*uC t-... *(-l )" "C 2^, n-l <2m<n, has the value 2"/2cos(nn/4). (b) Derive a similar result for the sum Sz( n ) :"6 t -'C t*'C s-... nC 2,,a1, n-l * (-l )^ and verify it for the casesn : 6,7 and 8. 3. 18 By considering(l + exp i0)", prove that _l: ) "C cosrd : 2' co{'(0 /2)cos(n0/2), -,:o )-'C, -,:o where 'C, : nl/lrr.(n - r)ll. sin r0 : 2' cos'(0 / 2\ sin(nO/ 2), <2m+| <n, COMPLEX NUMBERS AND HYPERBOLICFUNCTIONS 3.l9 to provethat g + l. : d - 8cos2 8cosa cos49 Usede Moivre'stheoremwith n:4 and deducethat l/2 /^ .o.* : d\"/ 3.20 3.21 ^\ Expresssinag entirelyin termsof the trigonometricfunctio-nsof multiple angles and deducethat its averagevalueover a completecycleis f. Usede Moivre'stheoremto Drovethat tan)u: )-zL [' * rl ' l r ) - 1 0 r r +5 r tL lot, + l' wheret : tan 0. Deducethe valuesof tan(nn/ l0) for n : l, 2, 3, 4. Provethe following resultsinvolving hyperbolicfunctions. (a) That c o s h x - c o s h y : 2 s i n h ( +) r ''n ( ? ) \z/\ ( b ) ln a t,llY:sln h 'x. + ,!dx :0. u' +tt2 dx' 3.23 Determine the conditions under which the equation ccoshx*bsi nhx:c, c>0, has zero, one, or two real solutions for x. What is the solution if a2 : c2 + b2'! Use the definitions and properties ofhyperbolic functions to do the following: ( a ) So lveco sh x : si nhx* 2sechx. (b) Show that the real solution x of tanhx : cosech x can be written in the form x : ln(u+ Ju\.Find an explicit value for u. (c) Evaluate tanhx when x is the real solution ofcosh2x:2coshx. 3.25 Express sinha x in terms of hyperbolic cosines of multiples of x, and hence find the real solutions of 2 cosh4x - 8 cosh 2x + 5 : 0. 3.26 In the theory of special relativity, the relationship between the position and time coordinates of an event, as measured in two frames of referencethat have parallel x-axes, can be expressedin terms of hyperbolic functions. If the coordinates are x and t in one frame and x' and t'in the other, then the relationship take the form x':xcoshd-cl si nh{, cr' : -x sinh d + ct cosh d. Express x and ct in terms ofx', ct' and Q and show that x2 - (ct)2 : 1x')2- (ct')2. t12 3.9T{INTSAND ANSWERS 3.27 A closed barrel has as its curved surface the surface obtained by rotating about the -x-axisthe part of the curve y: al 2- cosh(x/a)l lying in the range -b <,t < b, where b < acosh-r2. Show that the total surlace area,A, of the barrel is given by A : ral9a - 8a exp(-b / al -f aexp(-2b / a) - Zbl. 3.28 The principal value of the logarithmic function of a complex variable is defined to have its argument in the range -n < arg z < n. By writing z : tanw in terms oi exponentialsshow that l. , tan,z:=l nl _ zr /1+i z\ \t-rz/ l. Use this result to evaluate /- \ tanrllJ::1 \'/ ..\ I 3.9 Hints and answers 3 .1 3 .3 3 .5 3.7 3 .9 3.11 3.13 3 .1 5 3 .1 7 l.l9 3.21 (f) 3-4t; (c)10+5i ; (d) 2/5+l 1i /5:(e)4; ( a ) 5 + 3 t;( b ) - l- 5 ,; ( g ) ln 5 + ifta n t( 4 /3\ + 2nn1;$) +(2.521+ 0.595,). tJsesin zl4 : co sn /4 - l /nD , si nn13: l /2 and cosn13: nE /2. co tn /1 2 : 2 + 1 5 . ( a ) e xp ( - 2 y) co s2 .t; (b) (si n2ysi nh2x)/2; (c) .uDexp(zi/3) or l D exp(4tti /3); ( d ) e xp ( l/J2 ) o r e x p(-l l .,D ); (e) 0.540-0.841t; (f) 8si n(l n2): 5.11; (g) exp(-n/2 - 2nn): (h) ln 8 * i(6n + | /2)n. Starting from l-x* iy - ia :,l.Jx * iy * ial, show that the coefficientsof -x and y are equal, and write the equation in the lorm -x2+ (l - u)2 : 12. ( a ) Cir cle se n clo si ngz: -i a, w i th,l : expc > l . (b) The condition is that arg[(z ia)/(z'tia)]: k. This can be rearrangedto give a(z * z') : (a2- zl2)tank. which becomesin x,y coordinates the equation of a circle with centre (-a cot k, 0) and radius d coseck. All three conditions are satisfiedin 3n12 < 0 < 7n/4, lzl < 4; area : 2n. - d2''+r as a product of factors Denoting expl2ni/(2m't l)] by O, express-x2"'+1 like (x - aO') and then combine those containing O' and Q2D'+r-/.[Jse the fact that o2"'+l : 1. T h e r o o ts a r e 2 t/3 exp(2nni /3)torn:0, 1,2; l +31/a;l +3t/4i . Co n sid e r( l + tf' . ( b) S zh\:2'/z si n(nr/4).5r(6) : -8, S z(7): -8, S 2(8): 0 tlse th e b in o m ia l e x pansi onof {cos0 f i si n{}}4. Show that cos50 : l6cs - 20c3+ 5c, where c : cos0, and correspondingly for sin 50. Ljsecos-20 : | + an2 0. The four required values are (s+ v20)'2. [(s- .l,T)/slt/r,(5- !t0)'2, [(5+ \,00)/5]r/r, 3.23 3 .2 5 Reality of the root(s) requiresc2 + b2 > a2 and a + b > 0. With theseconditions, there are two roots if a2 > b2, but only one if b2 > a2. F o r a 2 : c2 + b 2 ,x: j l n[(c - h)/(a + b)1. Re d u ceth e e q u a tionto l 6si nhax: I, yi el di ngx - *0.481. 113 C] O M PLEXNT J M B E R S A N D H Y P E R B O L I CF I ] N C T I O N S 3. 21 Sh o w th a t /5 : ( co shr/a)d-r, curved surface area : na2[8 sinh(b/ a) sinh(2bI a)) - 2rab fla t e n d s a r e a : 2 r il4 4cosh(b/a)+ coshr(b/d)].