3

Complex numbers andhyperbolic functions

This chapter is concerned with the representation and manipulation of complexnumbers. Complex numbers pervade this book, underscoring their wide appli-cation in the mathematics of the physical sciences. The application of complexnumbers to the description of physical systems is left until later chapters andonly the basic tools are presented here.

3.1 The need for complex numbers

Although complex numbers occur in many branches of mathematics, they arisemost directly out of solving polynomial equations. We examine a specific quadratic

equation as an example.Consider the quadratic equation

z"-42*5:0.

Equation (3.1) has two solutions, zy a.nd 22, such that

(z - zr)Q - z) :0.

Using the lamiliar formula for the roots of a quadratic

solutions z1 and 22, written in brief as 2r.2, are

(3.1 )

(3.2)

equation, (1.4), the

4+ J;$ -411" '

(3.3)

Both solutions contain the square root of a negative number. However, it is nottrue to say that there are no solutions to the quadratic equation. Thefundamentaltheorem of algebra states that a quadratic equation will always have two solutionsand these are in fact given by (3.3). The second term on the RHS of (3.3) iscalled an imaginary term since it contains the square root of a negative number;

83

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

Figure 3.1 The function f (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

IQ): z2 -42 * 5 is shown in f igure 3.1. I t wi l l be seen that the plot does not

intersect the z-axis, corresponding to 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 representation of 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* iy,

where i is used to denote the square root of - 1 . The real part x and the imaginarypart y arc usually denoted by Rez andlmz respectively. We note at this point

that some physical scientists, engineers in particular, use j instead of i. However,

for consistency, we will use i throughout this book.In our particular example, lZ :2J-t - 2i, and hence the two solutions of

(3.1 ) are

zrz:2tz j :zt i .

Thus, here x:2 and,y : *1.For compactness a complex number is sometimes written in the form

z : (x,y),

where the components of z may be thought ofas coordinates in an xy-plot. Such

a plot is called an Argand diagram and is a common representation of 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 generalised readily to

polynomials whose highest power (degree) is greater than2, e.g. cubic equations

(degree 3), quartic equations (degree 4) and so on. For a general polynomial /(z),

of degree n, the fundamental theorem of algebra states that the equation f(z) : O

will have exactly n solutions. We will examine cases of higher-degree equations

in subsection 3.4.3.

The remainder of this chapter deals with: the algebra and manipulation of

complex numbers; their polar representation, which has advantages in many

circumstances ; complex exponentials and logarithms; the use of complex numbers

in finding the roots of polynomial equations; and hyperbolic functions.

3.2 Manipulation of complex numbers

This section considers basic complex number manipulation. Some analogy may

be drawn with vector manipulation (see chapter 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 components and the imaginary components are added

separately and in a like manner to the familiar addition of real numbers:

4 + 22 :(xr * iyr) * (xz * iy) : (x1+x2) + i(y I yz),

85

lrn z

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

4+22

Figure 3.3 The addition of two complex numbers.

or in component notation

z1 | z2 - (xr"yr) * 6z,y) : (xr * xz,yr * yz).

The Argand representation of 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.

z1*22:22+zt,

zr*kz*4):Qrlzz) lzt .

Thus it is immaterial in what order complex numbers are added.

r sum the compl* nambas 1 + 2,, 3 - 4i, -2 + ,.

Summing the real terms we obtain

l+3-2:2.

and summing the imaginary terms we obtain

2i-4i+i : - i .

(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.

Hence

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

l t l : \ ,Fr t . (3.4)

Hence the modulus of the complex number is the distance of the corresponding

point irom the origin in the Argand diagram, as may be seen in figure 3.4.

The argument of the complex number z is denoted by arg z and is defined as

a,sz-*1l- ' ( i ) (3.5)

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 cases give the same value for the ratio of y to x.

>Fuil tlw mdubs aul of the canplex rupiber z :2 -ii.

Using (3.4), the modulus is given by

Ej: JY * 1-3Y : '81.

Using (3.5), the argument is given by

ary z : tan-t (-)) .

The two angles whose tangents equal -1.5 are -0.9828 rad and 2.1588 rad. Since x : 2 andy : -3, z clearly lies in the fourth quadrant; therefore atg z : -0.9828 is the appropriateanswer. <

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

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).

(3.7)

associative,

(3.8)

(3.e)

(3.10)

(3.1 I )

lz1z2l : l5 - l4tl : .rFT elry : "m

1a1: J3aY:1V'tzr :JerY+(4Y:", tm,

lzllz2l: "ryi"m

: Jm : Etzzl. <

(3.6)

the complex wnbers zr * 3 * 2i otd z2: -l - 4i

Bv direct multiolication we find

422:(3+2i)(- l -4 i ): -3-2i- l2 i -8 i2:5-14i .<

The multiplication of complex numbers is both commutative and

i.e.

ztz2: z2zl ,

(z1z)23: ztQzzi .

The product of two complex numbers also has the simple properties

laz2l : lz1l lz2l ,

arg(zP) : arg zt I arg 22.

These relations are derived in subsection 3.3.1.

>Vertfy tlnt (3.14) hoills , Wiluct of 4 :3 + A uid zz: *l -

From (3.7)

We also find

and hence

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 see immediately

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 corresponding rotations of n12 or -nf2 respectively. This geometrical

interpretation of multiplication is shown in figure 3.5.

>Using the geunetrical i.ntopremtion of r, r,fad tto frduct i(l-t).

The complex number 1-i has argument -n/4and modulus "8.

fnus, using (3'10) and(3.11), its product with i has argument *n/4 and unchanged modulus u5. lte complexnumber with modulus .f and argument *n/4 is I * i and so

i( l - t ) : l+t ,

as is easily verified by direct multiplication. <

The division of two complex numbers is similar to their multiplication but

requires the notion of the complex conjugate (see the following subsection) and

so discussion is postponed until subsection 3.2.5.

3.2.4 C omplex conjugate

If z has the convenient form 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 leaves a real result, i.e. there is no imaginary component in the product.

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

Figure 3.6 The complex conjugate as a mirror image in 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' gives a real result:

zz' : (x* iyXx - iy) : x2 - ixy * ixy - i2 y2 : f a y2 : 1212'

complex conjugation corresponds to a reflection of z in the real axis of the

Argand diagram, as may be seen in figure 3.6.

cogugate of z = a*2i+31b.> Finrl the complex co4jugate oJ z = a + Zi + 3Ib.

The complex number is written in the standard form

z : a't i(2 * 3b\;

then, replacing i by -i, we obtain

z' :a- i (2+3b).<

In some cases, however, it may not be simple to rearrange the expression for

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 conjugate of 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

>fira * Wt* t-iaut of the comptex rusnber z - alsv+2a1, "rtq! " -.1 5i

Although we do not discuss complex powers until section 3.5, the simple rule given abovestill enables us to find the complex conjugate ofz.

In this case w itself contains real and imaginary components and so must be writtenout in full, i.e.

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

(z') ' : z,

z*z ' :2F.ez:2x,

z-z:z l lmz:zly,

z /x2-y2\ . / 2xy \r : \FT})+t \ ; r r7)

The derivation ofthis last relation relies on the results ofthe 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 xr * iYr22 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

quanti ty. Equation (3.16) gives

zt _ (\ + iyt)(x2 - iy) _ (xrxz + yty2) I i(x2y1 - xflz)

" :1"ra;r tYxr- t141 W

xlx2 + ytVl .x)Vt - xtv2

' |+ f i ' ]+ f iHence we have separated the quotient into real

required.In the special case where zz: zi . so that xz

result reduces to (3.15).

(3.r2)

(3.1 3)

(3.14)

(3. r 5)

and imaginary components, as

: xr and y2 : -yr, the general

9l

COMPLEX NUMBERS AND HYPERBOLIC FT]NCTIONS

|Express z in the form x * iY, when

3 -2i, : -' - t+4i

Multiplying numeratorwe obtain

and denominator by the complex conjugate of the denominator

(3-2i)(-1-4i):

(- t +4r.11- t -4t)11 l0:----1.<l7 t7

-11 - 10'--n

In analogy to (3.10) and (3.11), which describe the mult ipl icat ion of two

complex numbers, the following relations apply to division:

lz t l

lz2l

: arg zt - arg 22

The proof of these relat ions is left unti l subsection 3.3.1.

3.3 Polar representation of complex numbers

Although considering a complex number as the sum of a real and an imaginary

part is often useful, sometimesthe polar representation proves easier to manipulate.

This makes use of the complex exponential function, which is defined by

lz t Il r l

(3. r 7)

(3.1 8)/zr \ars l - l

- \zz/

(3. r e)

Strictly speaking it 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 mathematicians then

define their equivalence for irrational and complex z. For the purposes of this

book we will not concern ourselves further with this mathematical nicety but,

rather. assume that (3.19) is val id for al l z ' We also note that, using (3.19)' by

multiplying together the appropriate series we may show that (see chapter 24)

ezte'z _ ezt l rz, (3.20)

which is analogous to the familiar result for exponentials of real numbers.

z2 23ez:expz=1*z+r.+ t+

lmz

3.3 POLAR REPRESENTATION OF COMPLEX NUMBERS

Figure 3.7 The polar representation of a complex number.

From (3.19), it immediately follows that for z : i0,0 teal,

eio:cos? f is in0,

(3.2r)

(3.22)

(3.23)

where the last equality follows from the series expansions of the sine and costne

functions (see subsection 4.6.3). This last relationship is called Euler's equation. lt

also follows from (3.23) that

eina:cosn? * is inn0

for all n. From Euler's equation (3.23) and figure 3.7 we deduce that

reie:r(cosl * is in0)

:x* i ! .

Thus a complex number may be represented in 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 representation of the modulus and argument is one of the main

reasons for using the polar representation. The angle 0 lies conventionally in the

range -n <0 <n, but, since rotat ion by 0 is the same as rotat ion by 2nn*0'

where n is any integer,

teiq - feilq+2nrl.

eio:r t*- f . - 'o.

: ' - f* f - +;(o-$.$- )and hence that

93

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

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 results are identical.

Some operations prove much easier in the polar representation, others much more

complicated. The best representation for a particular problem must be dete rmined

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: r2eioz is given bY

Z1z2 : 71gi0t Yrai9z

: Yr1'rsilqi0z). (3.2s)

The relations lzrzzl: lzllz2l and arg(42) : arg zr { arg z2 follow immediately'

An example of 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 t tg iot r l i la.-a-

; : ; "^ -- ' - !e i (at-42). (3.26\

The relations lq/zzl -- lzlllzzl and arg(z1f z) : arg zt - atg22 are again

rg2si(oiozl

94

Imz

3.4 DE MOIVRE'S THEOREM

Figure 3.9 The division of two complex numbers. As in the previous figure,

ry atrd 12 are both greater than 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 extremely important theorem. Since (et0)' : ei'o, we have

(cos0 * is in0) ' : cosn0 * is inn0, (3.21)

where the identity eing : cosnl * isinno follows from the series definition of

ei"0 (see (3.21)). This result is called ile Moiure's theorem and 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

examines just three: proofs of trigonometric identities; finding the nth roots of

unity; and solving polynomial equations with complex roots.

3.4. 1 Trigonometric identities

The use of de Moivre's theorem in finding trigonometric identities is best illus-

trated by example. We consider the expression of a multiple-angle function in

terms of a polynomial in the single-angle function, and its converse.

95

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

atut cnl3,A lft wntl of pow*s>Exgew

Using de Moivre's theorem,

cos30 + js in39: (cos0 + is ing)3: (cos3 g - 3cos0sin2 g) + i(3sin0cos20 -sin3 g;.

We can equate the real and imaginary coefficients separately, i'e.

(3.28)

cos39 : cos3 d - 3 cos gsin2 0:4cos3d-3cosd

sin39 : 3 sin0cos2 0 - sins 0:3sin0-4sin30. <

(3.2e)

This method can clearly be applied to finding power expansions of cos,4d and

sinnd for any positive integer n.

The converse process uses the following properties ol z : eiq,

and

u *; :2cosno,

z" - ! :2 istnnl .zn

These equalities follow from simple applications of de Moivre's theorem, l.e.

I, " * V

: (cosd * is in0) ' * (cos0 + is in0)- '

: cos nd * i sin n0 * cos(-no) * i sin(-no)

: cosn' * i sin n0 * cos n0 - i sin n0

:2cosn9

and

: (cos 0 * isin 0)'r - (cos 0 * isin 0)-'

: cosn0 * i sin n0 - cosn0 * isinn0

:2istnn?.

In the particular case where n: l,

" lzn

I

zI

z

:2cos0,

:2is in9.

(3.30)

(3.31)

(3.32)

(3.33)

_ eio + e,io

: , i0 - t-io

96

3.4 DE MOIVRE'S THEOREM

{w30 Nt> Fird an ayessian fer cod 0 ir te'ras of w30 M ryO.

Using (3.32),

cos3 o :

Now using (3.30) and (3.32), we find

cos30: fcos30+ lcos0. <

This result happens to be a simple rearrangement oi (3.29), but cases involving

larger values ol n are better handled using this direct method than by rearranging

polynomial expansions of multiple-angle functions.

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 proceed we 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-l lo/n ,

corresponding to the values 0,1,2,.. . ,n- I for k-Larger integer values of f t do

not give new solutions, since the roots already listed are simply cyclically repeated

fork:n,nl l ,n l2,etc.

>Find trv aoh$btrs ta the e4uation z1 : l.

By applying the above method we find

z : e2th/3

Hence the three solutions are zr:3i : l, zz: s2in/3o4: sain/t'16'' note that, as expected,

the next solution, for which & : 3, gives 24 : s6ia/3 : | : zt so that there are only three

separate solutions. {

| / l \ '

F\ '*r)t=(s*1,+l+1)b\ z 2", /

r / . l \ 3/ l \r \z-+

")* s\ ' * ; )

97

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

Figure 3.10 The solutions of 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 properties coj: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 equations in the form of a polynomial relationship must 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 factorise to give

(23 - 2)(22 + 4)e - l l :0.

Hence zr :2 or z2: -4 or z : l.The solutions to the quadratic equation are z : l2i;to find the complex cube roots, we first write the equation in the form

z3:2:2e2'kn,

where /r is any integer. If we now take the cube root, we get

z :21/3e2ika/3

98

3.5 COMPLEX LOGARITHMS AND COMPLEX POWERS

To avoid the duplication of solutions, we use the fact that -T <arg z < z and find

- - 1t /3

/ r A\22: 2t /3e2ni /3 : zr / ' | - i + +i I '

) ; -n<, , , I I ,

zt:21/3e 2ni /3 - 2t / t t - ; - +t I\ - - /

The complex numbers 21, 22 Lfld 23, together with zo:2i, z5 : -)i and z6: I are the

solutions to the original polynomial equation.--As e"pected froi the fundamentaliheorem ofalgebra, we find thatthe total number

of.o.pir* .oots (six, in this case) is equal to the largest power of z in the polynomial. <

A useful fesult is that the roots of a polynomial with real coemcients occur in

conjugate pairs (i.e. if zr is a root, then zi is a second distinct root, unless z1 is

real). This may be proved as follows. Let the polynomial equation of which z ts

a root be

Qrzn + an-:zn- l + " '+ alz + ao:0 '

Taking the complex conjugate of this equation,

aiQ')" * a l rQ') ' - t + " '+ a\2" + ai :g '

But the an are real, and so z- satisfies

an/. )n * an-r lz. ln- t + " ' + a1z* | as : Q'

and is also a root of the original equation.

3.5 Complex logarithms and complex powers

The concept of a complex exponential has already been introduced in section 3.3,

where it was assumed that the definition of an exponential as a series was 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 explained shortly. Thus, w must satisfy

z:e*,

Using (3.20), we see that

z1z2: g| | rgwz : g\ | r+wz,

and taking logarithms of both sides we 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

numbers also holds for complex numbers.

COMPLEX NI,]MBERS AND HYPERBOLIC FIJNCTIONS

We may use (3.34) to investigate further the properties ol Ln z. We have already

noted that the argument of a complex number is mult ivalued, 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

Ln z : lnr * i(0 4- 2nn), (3.35)

where lnr is the natural logarithm of the real positive quantity r and so is

written normally. Thus from (3.35) we see that Lnz is itself multivalued. To avoid

this mult ivalued behaviour i t is conventional to define another function lnz, the

principal ualue of Ln z, which is obtained from Ln z by restricting the argument

of z to l ie in the range -n < 0 < n.

> Eoaluate Ln (-i).

By rewriting -i as a complex exponential, we find

Ln (-i) : Ln leir n/z+znntl : ien/2 + 2nn),

where n is any integer. Hence Ln(- i ) : - in/2,3ir /2, " we note that ln(- i ) ' the

principal value of Ln(-,), is given by ln(-i) : -inl2. <

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: -2iLni .

Now inverting the process we find

eL". :Z:g2iLni ,

We can write i: ei\n/2+2,n), where n is any integer, and hence

Lni:Ln le 'Grz*z*11: i (n l2 ' t2nn).

We can now simpliiy z to give

; - ) i - .L i (n1)+2m\'

_ '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 APPLICATIONS TO DIFFERENTIATION AND INTEGRATION

3.6 Applications to differentiation and integration

We can use the exponential form of a complex number together with de Moivre's

theorem (see section 3.4) to simplify the differentiation of trigonometric functions.

>Fitd the tteriuatioe with respect to x of et'ooafx.

We could differentiate this function straightforwardly using the product rule (see subsec-

tion 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 com-plex exponential. This complex number has e3'cos4x as its real part. Now, differentiating

z with respect to x we obtain

dz

- : (J*4i ;st l+41' : (3+4i)e3'(cos4x*is in4x), (3.36)

axwhere we have again used de Moivre's theorem. Equating real parts we then find

d

- (el' cos4x) : e3*(3 cos4x - 4sin4x).

4XBy equating the imaginary parts of (3.36), we also obtain, as a bonus,

{ 1"- 'in

+'; : e3'(4 cos 4x * 3 sin 4x)' <dx'

In a similar way the complex exponential can be used to evaluate integrals

containing trigonometric and exponential functions.

> Eualuate the integrat t - t f cotbx dx.

Let us consider the integrand as the real part of the complex number

eo'(cos bx * i sin Dx) : eaa eib\ - etu+iblx '

where we use de Moivre's theorem to rewrite the trigonometric functions as a complex

exponential. Integrating we findr ^(a+ib)Ylpro* i r t ,dx: i -

+cI a+tD

(a - iblga+'ot'

la - ib l la - l ib l

:+(aeib '_ ibeib,)+c,a2+bz'

where the constant of integration c is in general complex. Denoting thisc : ct * ic2 and equating real parts in (3.37) we obtain

t : [ , " , eo'

. , 'osbxdx :

a2 +62@cosbx+bsinbx)*cr '

which agrees with result (2.37) found using integration by parts. Equating imaginary parts

in (3.37) we obtain, as a bonus,I pax

J: I

d 'stnbxdx: pj*@sinbx-bcosbx)*c2. <

(3.37)

constant by

101

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

3.7 Hyperbolic functions

The hyperbolic functions are the complex analogues of the trigonometric functions.

The analogy may not be immediately apparent and their definitions may appear

at first to be somewhat arbitrary. However, careful examination of their properties

reveals the 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 properties of trigonometric functions can also be applied

to the hyperbolic functions. Further, hyperbolic functions occur regularly, and so

giving them special names is a notational convenience.

3.7.1 Definitions

The two fundamental hyperbolic functions are cosh x and sinh x, which, as their

names suggest, are the hyperbolic equivalents of cos x and sin x. They are defined

by the fol lowing relat ions:

coshx: l {e"+e-") ,srnhx:; le ' -e -1.

Note that coshx is an even function and sinhx is an odd function. By analogy

with the trigonometric functions, the remaining hyperbolic functions are

, s inhx e ' -e *tann r :

coshx ea+e-^

, t2secn .I- : --_--:-cosnx e- f e-"

l2cosechx:- : - . . '

srnnx e' - e- '

' 1 et*e-*Cuin .I- :

tanhx e\-e x

(3.38)

(3.3e)

(3.40)

(3.41)

(3.42)

(3.43)

A11 the hyperbolic functions above have been defined in terms ofthe real variable

x. However, this was simply so that they may be plotted (see figures 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 subsections we have alluded to the analogy between trigonometric

and hyperbolic functions. Here, we discuss the close relationship between the two

groups of functions.Recalling (3.32) and (3.33) we find

cosix: ) {e '+e- ' ) ,sin jx : \i1e' - e-').

t02

3.7 HYPERBOLIC FUNCTIONS

Figure 3.ll Graphs of coshx and sech-x.

\ cosech r

\a\ -L

cosech r \\4

Figure 3. I 2 Graphs of sinh x and cosechr.

Hence, by the definitions given in the previous subsection,

coshx:cosix,

i sinh x : sin ix,

cosx:coshix,

i sin x : sinh ix.

These useful equations make the relationship between

sinh -r

(3.44)

(3.45)

(3.46)

(3.47)

hyperbolic and trigono-

103

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

Figure 3'13 Graphs of tanhx and cothx'

metric functions transparent. The similarity in their calculus is discussed further

in subsection 3.7.6.

3.7.3 ldentities of hyperbolic functions

The analogies between trigonometric functions and hyperbolic functions having

been established, we should not be surprised that all the trigonometric identities

also hold for hyperbolic functions, with the following modification. wherever

sin2 x occurs it must be replaced by - sinh2 x, and vice versa. Note that this

replacement is necessary even if the sin2 x is hidden, e.g. tan2 x : sin2 x/ cos2 x

and so must be replaced by (- sinh2 x/ cosh2 x) : - tanh2 x'

Using the rules stated above cos2 x is replaced by cosh2 x, and sin2 x by - sinh2 x, and so

the identity becomescosh2x-sinh2x: l .

This can be verified by direct substitution, using the definitions ofcoshx and sinhx; see(3.38) and (3.39). <

Some other identities that can be proved in a similar way are

sech2x:1-tanh2x,

cosech2x:coth2x-1,

sinh 2x : 2 sinh x cosh x,

cosh 2x : cosh2 x * sinh2 x.

(3.48)(3.4e)(3.s0)(3.51)

3.7 HYPERBOLIC FUNCTIONS

3.7.4 S olving 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 easier to express the 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') - l t t -e- ' ) -5 : 0.. -Rearranging, and then multiplying through by -e', gives in turn

-2e' I3e- ' -5:0

and2e2'+5e'-3:0.

Now we can factorise and solve:

(2e' - l \ (e '+ 3) :0.

Thus e* : l /2 or d: -3. Hence x: - ln2 orx: ln(-3). The interpretat ion of thelogarithm of a negative number has been discussed in section 3'5. <

3.7.5 Inverces of hyperbolic functions

Just like trigonometric functions, hyperbolic functions have inverses. lt y :

coshx then x: cosh-ly, which serves as a definit ion ofthe inverse. By using

the fundamental definitions of hyperbolic functions, we can find closed-form

expressions for their inverses. This is best illustrated by example.

>Fitd a cloxd-farm atpessiotfor tte bverce hyperbotk J : sint-r x'

First we write x as a function of y, i.e.

. l :s inh-rx + x:s inh.y.

Now, since coshy: l1e, +e-t1and sinhy : llet -e-t\,

er : coshy * sinhyr----------:-

: V1+sinh'y*sinhy

e:Jta*a*,

y :tn(Jt + r] + r). <

In a similar fashion it can be shown that

cosh-l x : tn1..,6z - t + x;.

and hence

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

Figure 3.14 Graphs ofcosh-'x and sech-'x

First we write x as a function of y, i.e'

t t : tanh-rx + x: tanhY

Now, using the definition of tanhy and rearranging, we find

e' - e-!" eY+e-!

Thus. it follows that

, . . 1*xI -X

+ (x * l )e-y : ( l -x)eY.

. t t+"+ e' : \ l 1- ,

V I -x

. lT+'y:rn\ / ' . ,Y t - I

| / l -1-Y\tanhrx:=ln(;

- ' l '<t \ t -x/

Graphs of the inverse hyperbolic functions are given in figures 3'14-3'16'

3.7.6 Calculus of hyperbolie functions

Just as the identities of hyperbolic functions closely follow those of their trigono-

metric counterparts, so their calculus is similar. The derivatives of the two basic

106

3.7 HYPERBOLIC FUNCTIONS

Figure 3.15 Graphs of sinh ' .x and cosech 'x.

Figure 3.16 Graphs of tanh 1x and coth-rx.

hyperbolic functions are given by

d

^

{cosn x) : s lnn x,

f trtnn ') : cosh.x.

(3.s2)

r1 51)

They may be deduced by considering the definitions (3.38), (3.39) as follows.

t07

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

>Ver{v tlte * - *iehx.

Using the definition of cosh x.

coshx:;(e '+e-") ,

and differentiating directly, we find

4 tcosh ")

: I {e* - e-* }dx

-- sinh x. <

Clearly the integrals of the fundamental hyperbolic functions are also definedby these relations. The derivatives of the remaining hyperbolic functions can bederived by product differentiation and are presented below only for complete-ness.

The inverse hyperbolic functions also have derivatives, which are given by thefollowing:

ftAunnl: sech2x,

ft{r""t x) : -sech x tanhx,

f {"or."t x) : -cosech xcothx,

ft@otni: -cosech2x.

II - '

lx. - a.I

FTa'a

a2 - x2'-a

x ' -4"

fot x2 < a2,

for x2 > a2.

(3.54)

(3.55)

(3.56)

(3.s7)

(3.58)

(3.5e)

(3.60)

(3.61)

d t . - rX\, lcosh'- l :

4X \ A/

d / , - rx\-- ls lnn'- , :dx \ a, /

d t , - rX\, l lann-- l :

4X \ A, /d r , - r X\, tcotn ' - l :

4X \ O/

These may be derived from thetion 3.7.5).

logarithmic form of the inverse (see subsec-

108

3.8 EXERCISES

From the results of section 3.7.5.) t r t - \ t

fr {rint, ' *) : a, ltn (x + r/* + t))| / . x \: ; ; ,? 'T]1 ' -@l

| ( .6uTt+x\r - l

x + ftTT \ "tt+l

/I: : . {

Jx2+l

3.1

3.8 Exercises

Two complex numbers z andw are given by z:3I4i andw:2- i . On anArgand diagram, plot

(a\ z * w, (b) w - z, (c) wz, (d) z/w,(e\ z' w * w' z, (f\ w2 , 1g1 ln z, (h) (l * z -t w)t/2 .

By considering the real and imaginary parts of the product sias'o prove thestandard formulae for cos(d + d) and sin(0 * d).By writing n/12: (n/3\ - (n/4) and considering si"/r2, evaluate cot(n/12).Find the locus in the complex z-plane of points that satisfy the following equa-tions.

/ I +,r \lal z -r: p\T;J , where c is complex, p is real and t is a real parameter

that varies in the range -.4 < t <...(b), : 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,(d) | exp(,'/'z)1, (e) exp(i3), (f) Im(2i+r), (g) ,', (h) lnt(\6 +,)31.

Find the equations in terms of x and y of the sets of points in the Arganddiagram that satisfy the following:

(a) Rez2: lmzz;(b\ (lmz2)lzz : -i;(c) arelz/(z-r) l :n/2.

Show that the locus of all points z:x* ry in the complex plane that satisfy

lz - ia l : i lz - l ia l ' ) ' > 0 '

is a circle of radius l21a/(l -,12)l centred on the point z : ial(l + )J)10 - L'z\|.Sketch the circles for a few typical values ofI, 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 arcthe corners of two similar triangles in the Argand diagram. Express in terms ofa,b, . . . ,C

3.2

3.33.4

3.6

3.8

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

(a) the equalities ofcorresponding angles, and(b) the constant ratio of corresponding sides,

in the two triangles.By noting that any complex quantity can be expressed as

2 : lzlexp(iargz),

deduce thata(B - C) + b(C - A) + c(.4 - B) :0.

For the real constant a find the loci of all points z : x I i! in the complex planethat satisfy

( / . - ;^ \ ' l(a) Re{ ln{" : i * } } :c,

| \z+to/ )

,0, , ," {,n 1=) } : o,| \z+u/ )

Identify the two families of curves and verify that in case (b) all curves pass

through the two points +id.3.10 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 foreach value of z (and conversely) is known as the general bilinear transformationand takes the form

aZ - lb: :

17af, '

(a) Confirm that the transformation from the Z-plane to the z-plane is also ageneral bilinear transformation.

(b) Recalling that the equation of a circle can be written in the form

c>0,

0<k<n/2

3.1 I

3.12

^+ t ,

show that the general bilinear transformation transforms circles into circles(or straight lines). What is the condition that zt, zz and ,1 must satisly if thetransformed circle is to be a straight line?

Sketch the parts of the Argand diagram in which

(a) Rez2 <0, zt /2 <2;(b) 0< atgz'<n/2;(c) lexpz3l -0

as lz l - qr .

What is the area of the region in which all three sets of conditions are satisfied?Denote the r4th roots of unity by l, to,,, .?,, . .. , ai,-t.

(a) Prove that

lz - z1 |l - l :

^ '

n-1

t i r \ - , , , ' : o" ' 1-

*n

r:0

(r) f [a; i : ( -1) '*1.r:0

(b) Express x2 + y2 + 22 - yz - zx - xy as the product of two factors, each linearin x, y and z, with coefficients dependent on the third roots of unity (andthose of the x terms arbitrarilv taken as real).

3.8 EXERCISES

3.13

3.14

3.15

Prove that x2n+r -a2n+r, where m is an integer > 1, can be written as

x2n+t - a2n+t: ," - ",_ll l*' - zo".o, (#il . r]

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 iAI Z1-22

dti i- ; : - : - - - 'ut z2 - .1

Deduce that (a) zr * zz, (b) lzr - z2l and (c) 1ztl2 + lzzl2 are 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'

Plot the seven roots of the equation on an Argand plot, exemplifying that complexroots ofa polynomial equation always occur in conjugate pairs if the polynomial

has real coefficients.The polynomial /(z) is defined by

l(r) : "s

- 624 + l5z3 - 3422 1 362 - 48.

(a) Show that the equation f(z) :0 has roots of the form z : tri, where ) isreal, 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) : 0completely.

The binomial expansion of (1 f x)', discussed in chapter 1, can be written for apositive 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): '60 -"Cz*uCt-. . . *(- l ) " "C2^, n- l <2m<n,

has the value 2"/2cos(nn/4).(b) Derive a similar result for the sum

Sz(n):"6t - 'Ct* 'Cs-. . . * ( - l )^ nC2,,a1, n- l <2m+| <n,

and verify it for the cases n : 6,7 and 8.

By considering (l + exp i0)", prove that

_l:) "C cosrd : 2' co{'(0 /2)cos(n0 /2),-,:o

)-'C, sin r0 : 2' cos' (0 / 2\ sin(nO / 2),-,:o

where 'C, : nl/lrr.(n - r)ll.

3.16

3.17

3.18

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS

3. l9 Use de Moivre's theorem with n:4 to prove that

cos49 : 8cosa d - 8cos2 g + l .

and deduce that/^

^\

l /2

.o.* : [ ' * r l ' ld\" /

Express sina g entirely in terms of the trigonometric functio-ns of multiple anglesand deduce that its average value over a complete cycle is f.Use de Moivre's theorem to Drove that

r)-10rr+5rtan)u :

tL lot , + l '

where t : tan 0. Deduce the values of tan(nn / l0) for n : l, 2, 3, 4.Prove the following results involving hyperbolic functions.

(a) That

coshx-coshy:2sinh (+) r ' 'n (?)\z/ \

(b) lnat , l lY:s lnh 'x.

u'+ t t2 +, ! :0.dx' dx

Determine the conditions under which the equation

ccoshx*bsinhx: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) Solve coshx : s inhx* 2sech x.(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.

Express sinha x in terms of hyperbolic cosines of multiples of x, and hence findthe real solutions of

2 cosh4x - 8 cosh 2x + 5 : 0.

3.20

3.21

) -zL

3.23

3.25

3.26 In the theory of special relativity, the relationship between the position and timecoordinates of an event, as measured in two frames of reference that have parallelx-axes, can be expressed in terms of hyperbolic functions. If the coordinates arex and t in one frame and x' and t'in the other, then the relationship take theform

x' :xcoshd-cls inh{,

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

3.9 T{INTS AND ANSWERS

3.27 A closed barrel has as its curved surface the surface obtained by rotating aboutthe -x-axis the part of the curve

y: al2- cosh(x/a) l

lying in the range -b <,t < b, where b < acosh-r2. Show that the total surlacearea, A, of the barrel is given by

A : ral9a - 8a exp(-b / al -f aexp(-2b / a) - Zbl.

The principal value of the logarithmic function of a complex variable is definedto have its argument in the range -n < arg z < n. By writing z : tanw in termsoi exponentials show that

3.28

Use this result to evaluate

3.9 Hints and answers

3.1 (a)5+3t;(b) - l -5, ; (c)10+5i ; (d) 2/5+l1i /5:(e)4; ( f ) 3-4t ;(g) ln 5 + i f tan t (4/3\ + 2nn1; $) +(2.521 + 0.595,) .

3.3 tJse sinzl4 : cosn/4 - l /nD, s inn13: l /2 and cosn13 : nE/2.cotn/12: 2 + 15.

3.5 (a) exp(-2y)cos2.t ; (b) (s in2ysinh2x)/2; (c) .uD exp(zi /3) or lDexp(4t t i /3) ;(d) exp( l /J2) or exp(- l l . ,D); (e) 0.540 -0.841t; ( f ) 8sin( ln2): 5.11;(g) exp(-n/2 - 2nn): (h) ln 8 * i(6n + | /2)n.

3.7 Starting from l-x * iy - ia :,l.Jx * iy * ial, show that the coefficients of -x and yare equal, and write the equation in the lorm -x2 + (l - u)2 : 12.

3.9 (a) Circ les enclosing z: - ia, wi th, l : expc > l .(b) The condition is that arg[(z ia)/(z'tia)]: k. This can be rearranged to give

a(z * z') : (a2 - zl2)tank. which becomes in x,y coordinates the equationof a circle with centre (-a cot k, 0) and radius d cosec k.

3.1 1 All three conditions are satisfied in 3n12 < 0 < 7n/4, lzl < 4; area : 2n.3.13 Denoting expl2ni/(2m't l)] by O, express -x2"'+1 - d2''+r as a product of factors

like (x - aO') and then combine those containing O' and Q2D'+r-/. [Jse the factthat o2"'+l : 1.

3.15 The roots are2t/3exp(2nni /3) tor n:0, 1,2; l+31/a; l+3t /4 i .

3.17 Consider ( l + t f ' . (b) Szh\:2 ' /z s in(nr/4). 5r(6) : -8, Sz(7) : -8, S2(8) : 0l . l9 t lse the binomial expansion of {cos0 f is in{}}4.3.21 Show that cos50 : l6cs - 20c3 + 5c, where c : cos0, and correspondingly for

sin 50. Ljse cos-2 0 : | + an2 0. The four required values are

[(s - .l,T)/slt/r, (5 - !t0)'2, [(5 + \,00)/5]r/r, (s + v20)'2.3.23 Reality of the root(s) requires c2 + b2 > a2 and a + b > 0. With these conditions,

there are two roots if a2 > b2, but only one if b2 > a2.For a2 : c2 + b2, x: j ln[(c - h) / (a + b)1.

3.25 Reduce the equat ion to l6sinhax: I , y ie ld ing x - *0.481.

, l . /1+iz\tan,z:=lnl_ l .

zr \ t - rz/

/ - \ . . \tanr l lJ : :1 I

\ ' /

113

C]OMPLEX NTJMBERS AND HYPERBOLIC FI]NCTIONS

3.21 Show that /5: (coshr/a)d-r ,curved surface area : na2[8 sinh(b / a) sinh(2b I a)) - 2rabf lat ends area: 2r i l4 4cosh(b/a) + coshr(b/d)] .