1 Introduction

We introduce the integer numbers because otherwise the equation

a + x = b

cannot always be solved.We introduce the rational numbers because otherwise the equation

ax = b

cannot always be solved.We introduce the irrational numbers because otherwise equations like

x2 = 2

cannot always be solved.We introduce the complex numbers because otherwise equations like

x2 = −1

cannot always be solved.

2 Imaginary unit

We postulate that there is a number j such that

j2 = −1.

This number is called the imaginary unit. We assume that j obeys the usualarithmetic rules.

Practical rule 1. Expressions containing j can be manipulated like ex-pressions containing numbers and/or variables with the added bonus thatpowers of j can be simplified.

3 Imaginary numbers

An imaginary number is a product of j, the imaginary unit, with a realnumber, e.g.

j2, j3.1, jπ, j(−1).

Imaginary numbers can be multiplied, as in

j2× j5 = j × j(2× 5) = j210 = (−1)10 = −10

1

The square of an imaginary number is always negative:

(ja)2 = j2a2 = −a2 < 0

j and −j are both square roots of −1 because

j2 = 1, (−j)2 = (−1)2j2 = 1(−1) = −1.

We shall see later that the imaginary unit is sufficient to express all kindsof roots we need.

For example, just as for positive numbers we can find two square roots ofany negative number as in

√−2 =

√−1√

2 = j√

2 or − j√

2

4 Powers of j

j2 = −1 (by definition)

j3 = j2j = −1j = −j

j4 = j2j2 = 1

j0 = 1

j1 = j

j5 = j4j = j

(Omitted) Any positive integer n can be written as n = 4q + r, where qis a non-negative integer, r = 0, 1, 2 or 3. Hence

j4q+r =

1 when r = 0

j when r = 1

−1 when r = 2

−j when r = 3

5 Complex numbers

The sum of a real and an imaginary number is called a complex number, e.g.

1 + j, −1 + j, −2− j3,√

3− j1

2.

In general, let a and b be real numbers. Then the complex number z = a+jbis said to be in standard (or cartesian) form. The real part of z is a, theimaginary part of z is b.

Re(z) = a, Im(z) = b.

2

5.1 Argand diagram

4 quadrants, show a number in eachcomplex plane

5.2 Notational conventions

The terminology is natural but if you are baffled by questions like:

“Is the complex number z real?” or

“Is the complex number z purely imaginary?”,

you may find the following explanations useful.The real numbers are those complex numbers whose imaginary part is

0. We normally write 2 rather than 2 + j0. For emphasis such numbers areoften said to be purely real.

The imaginary numbers are those complex numbers whose real parts are0. We normally write j3 rather than 0 + j3. A complex number whose realpart is zero is also said to be purely imaginary.

0 + j0 is both purely real and purely imaginary and is normally writtenas 0.

We also write −j rather than j(−1), etc.

6 Arithmetic with complex numbers

Let z1 = a1 + jb1, z2 = a2 + jb2.We apply Practical rule 1 to arrive at appropriate definitions for the basic

arithmetic operations. There is no need to remember the formulas. Simplyexpand and collect terms as appropriate for each operation. Only the divisionis somewhat tricky. Notice the way to remove the imaginary unit from thedenominator.

Let z1 = 1+ j3, z2 = 2− j. Introduce the following operations first usingz1 and z2 and then using general letters.

• z1 + z2 = (a1 + a2) + j(b1 + b2)

• z1 − z2 = (a1 − a2) + j(b1 − b2)

• z1z2 = (a1a2 − b1b2) + j(a1b2 + b1a2)

3

• Divisionz1

z2

=a1 + jb1

a2 + jb2

=(a1 + jb1)(a2 − jb2)

(a2 + jb2)(a2 − jb2)

=(a1 + jb1)(a2 − jb2)

a22 − (jb2)2

=(a1 + jb1)(a2 − jb2)

a22 + b2

2

=(a1a2 + b1b2) + j(b1a2 − a1b2)

a22 + b2

2

=(a1a2 + b1b2)

a22 + b2

2

+ j(b1a2 − a1b2)

a22 + b2

2

• z−1. This is the same as 1/z.

6.1 Solving quadratic equations

Any quadratic equation has two roots, possibly complex. The familiar for-mula applies.

Example. Solve the equation

z2 − 2z + 5 = 0.

Solution:

z =2±

√4− 4× 5

2=

2±√

4− 20

2=

2±√−16

2=

2± j√

4

2= 1± j2.

6.2 Complex conjugate

If z = a + jb then the complex conjugate of z is denoted by z∗ or z̄ and isobtained by reversing the sign of the imaginary part of z:

z∗ = a− jb

Show it on Argand diagram!Important properties:

• z + z∗ = 2Re(z);

• z − z∗ = j(2Im(z));

• zz∗ = (a + jb)(a− jb) = a2 + b2 = Re(z)2 + Im(z)2, i.e. it is real;

• (z∗)∗ = z.

4

6.3 Equality of complex numbers

Two complex numbers are equal if and only if they have the same real partsand the same imaginary parts. Symbolically:

z1 = z2 if and only if Re(z1) = Re(z2) and Im(z1) = Im(z2).

This rule gives us a simple device to find complex numbers with prescribedproperties.

Example. Find a complex number z such that

z +1

2z∗ = 3 + j2.

Solution. Let the standard form of z be z = x+ jy. Then the above becomes

x + jy +1

2(x− jy) = 3 + j2,

we now expand the expression and collect the terms as appropriate to get

x +1

2x + j(y − 1

2y) = 3 + j2,

which simplifies to3

2x + j(

1

2y) = 3 + j2.

The real and imaginary parts of the number to the left of the equality mustbe equal to the corresponding parts on the right. Hence,

3

2x = 3,

1

2y = 2,

which givesx = 2, y = 4.

Hence the answer is z = 2 + j4. We can verify this by substituting it intothe equation above:

z +1

2z∗ = 2 + j4 +

1

2(2− j4) = 2 + j4 + 1− j2 = 3 + j2, as required.

Note: There is no “less then” relation betwen complex numbers!

6.4 Modulus and argument

1) Show on the picture2) define modulus, mention r > 0. Note that |z| = zz∗.3) define argument,4) discuss on z = 1 + j5) mention principal value, (−π, π];6) advice to use radians rather than degrees.

5

6.5 Polar form of a complex number

Say what it is, restriction on the parameters.Mention that it is unique only if the argument is restricted to an interval

like (−π, π].

6.5.1 Finding polar form from standard form

Example: z = −3+ j4. |z| = 5, tan(θ) = −43. So θ = 2.2143 or θ = −0.9272.

We choose the first one because it gives the correct quadrant. So

z = 5(cos 2.2143 + j sin 2.2143)

Discuss particular cases: z = 1, j,−1,−j.

6.5.2 Finding standard form from polar form.

Example. Given that |z| = 3 and arg(z) = π/3 write z in standard form.Solution: We have a = 3 cos(π/3) = 3/2, b = 3 sin(π/3) = 3

√3/2. So,

z =3

2+ j

3√

3

2.

6.6 Exponent of a complex number

Let b be real. We define ejb by

ejb = cos b + j sin b.

Let z = a + jb, where a and b are real. We define ez by

ez = ea(cos b + j sin b) = eaejb.

and we see that the modulus of ez is equal to ea while arg(z) = b.The function ez has the same properties as its real counterpart, in par-

ticular,ez1+z2 = ez1ez2 , (ez)n = enz.

The importance of the exponent cannot be overstated.One magnificient identity which follows from the definition above is

ejπ = −1.

It gives a simple relation between some fundamental mathematical constants.Other special cases with modulus one:

ejπ/4, ejπ/2, e−jπ/2, ej3π/2, e2πj, etc.

6

6.7 Exponential form of a complex number

The complex exponent provides another way to write the polar form of acomplex number. Namely,

z = r(cos θ + sin θ) = reiθ,

where r is the modulus of z, θ its argument.Example.

z = 1 + j =√

2ejπ/4.

6.7.1 Multiplication and division in exponential form.

If z1 = r1ejθ1 , z2 = r2e

jθ2 , then

z = z1z2 = r1ejθ1r2e

jθ2 = r1r2ejθ1ejθ2 = (r1r2)e

j(θ1+θ2),

i.e., to get the product we multiply the moduli and add the arguments.Similarly, for division,

z1/z2 = r1ejθ1/(r2e

jθ2) = r1r2ejθ1e−jθ2 = (r1r2)e

j(θ1−θ2),

i.e., we divide the moduli and subtract the arguments.Geometrical representation of multiplication by j and other complex num-

bers.

6.7.2 Addition in exponential form

Example. Find the real and imaginary parts of

z = ejπ/4 + 2ejπ/3

Now,

ejπ/4 = cos(π/4) + j sin(π/4) =1√2

+ j1√2

ejπ/3 = cos(π/3) + j sin(π/3) =1

2+ j

√3

2Hence,

z =1√2

+ j1√2

+ 2

(1

2+ j

√3

2

)= (1 +

1√2) + j(

1√2

+√

3)

= 1.7071 + j2.4392

The exponent is extremely useful. We apply it below to derive the DeMoivre’s formula.

7

6.7.3 De Moivre’s formula

This is de Moivre’s formula:

(cos(θ) + j sin(θ))n = cos(nθ) + j sin(nθ).

Indeed, let z = cos(θ) + j sin(θ). Then

zn = (cos(θ) + j sin(θ))n.

On the other hand, using the exponential form and the properties of theexponent we get,

zn = (ejθ)n = ejnθ = cos(nθ) + j sin(nθ).

Hence the expressions on the right-hand sides of the last two equations areequal, as required.

De Moivre’s formula works for any n including negative. However, if n isnot an integer care is needed (see the section about roots).

6.7.4 Cosine and sine in terms of exponentials

Direct calculation shows that the following relations hold for real θ,

cos θ =ejθ + e−jθ

2, sin θ =

ejθ − e−jθ

2.

We use these relations to define these functions for complex arguments. So,for any complex number z we define

cos z =ejz + e−jz

2,

sin z =ejz − e−jz

2.

6.7.5 Powers of sin(θ) and cos(θ) in terms of sin and cos of multipleangles

We can use the above results to rewrite cosn θ and sinn(θ) in a way which ismuch easier to integrate. For example,

8

Ex. 1

cos3 θ =

(ejθ + e−jθ

2

)3

=1

23

(ej3θ + 3ej2θe−jθ + 3ejθe−j2θ + e−j3θ

)=

1

8

(ej3θ + e−j3θ + 3ejθ + 3e−jθ

)=

1

8

((ej3θ + e−j3θ) + 3(ejθ + e−jθ)

)=

1

8(2 cos(3θ) + 3× 2 cos θ)

=1

4cos(3θ) +

3

4cos(θ)

Ex. 2 (Exercise)

sin5 θ =1

16sin 5θ − 5

16sin 3θ +

5

8sin θ

Note. For real powers of sin θ the result is in terms of cosines. (seeExample sheet).

7 Logarithm of a complex number

By definition a number z is said to be a logarithm of w = cejθ if

ez = w = cejθ,

where c is the modulus of w, θ its argument.By direct substitution (and using the properties of the exponent) we can

see that the number ln c + jθ has this property,

eln c+jθ = eln cejθ = cejθ = w.

However, we can see that for any integer k we have

eln c+j(θ+2πk) = eln cejθej2πk = cejθ = w,

since ej2πk = 1. Hence,

ln w = ln c + j(θ + 2πk),

where k can be any integer.

9

So, we may be in trouble when a single number for the logarithm isneeded. In particular, if z1 and z2 are such that ez1 = ez2 we cannot concludewithout additional information that z1 = z2. We can only say that

z1 − z2 = j2πk,

for some integer k.

7.0.6 Example 1. Find ln(−3).

We write −3 in exponential form: −3 = 3ejπ). So,

ln(−3) = ln(3ejπ) = ln 3 + j(π + 2kπ).

7.0.7 Example 2. Find ln z where z = 4 + j3.

Now,

|z| =√

42 + 32 = 5, arg(z) = tan−1(3

4) = 0.6435,

So,ln(4 + j3) = ln 5 + j(0.6435 + 2kπ), k = 0,±1,±2, . . . .

8 Roots of complex numbers

1) Consider square and cube roots.2) Let n be a natural number and

z = rejθ.

Obviously, (( n√

r)ejθ/n)n

= rejθ.

Hence n√

rejθ/n is an nth root of z. Similarly to the logarithm we actuallyhave more roots. Dissimilarly, there are exactly n different nth roots. Theseare given by,

n√

rej(θ+2πk)/n, k = 0, 1, . . . , n− 1.

8.0.8 Example: Square root of z = 4 + j3

Using above,z = 4 + j3 = 5ej(0.6435+2kπ).

So,z1/2 =

√5e

12j(0.6435+2kπ) =

√5ej(0.3217+kπ).

10

Now we put k = 0 and k = 1 to get two different roots of z

k = 0√

5ej0.3217 = 2.1214 + j0.7070

k = 1√

5ej0.3217+π = −2.1214− j0.7070

Other values of k give one of the two roots above.

8.0.9 Example: Cube root of 1

We have mod (1) = 1, arg(1) = 0, so 1 = ej0. Hence,

11/3 = 11/3e13j(0+2kπ) = ej 2kπ

3 , k = 0, 1, 2.

Putting k = 0, 1, 2 we get

ej0, ej2π/3, ej4π/3,

which can be written also as

1, −1

2+ j

√3

2, −1

2− j

√3

2.

8.0.10 Rational powers

Let p and q be integers and

z = rej(θ+2kπ).

Thenzp/q = rp/qe

pqj(θ+2kπ) = rp/qej( θp

q+ 2kπp

q)

Note. In principle, n√

z and z1/n are equivalent. However, if z is real thesymbol n

√z tends to mean the “arithmetic root” or the positive root. For

example, in a formula√

π usually means the positive square root.

9 Hyperbolic Functions

Definitions:

cosh x =1

2(ex + e−x)

sinh x =1

2(ex − e−x)

*** draw graphics ... ***

11

From these two we can get others

tanh x =sinh x

cosh x

coth x =cosh x

sinh x=

1

tanh x

sech x =1

cosh x

cosech x =1

sinh x

Practical rule 2. Any problem involving hyperbolic functions can besolved by replacing them with exponentials according to the definitions above.

9.1 Calculating hyperbolic functions

The hyperbolic functions can be calculated using their definitions and theexponential function.

The approved UMIST calculator can calculate them directly with thehelp of the “hyp” button.

• hyp, sin x gives sinh x.

• hyp, shift, sin y gives sinh−1 y.

9.2 Similarities between hyperbolic and trigonometricfunctions

Although shapes of these curves are very different to the trigonometric func-tions, mathematically there are many similarities.

From the definitions of cosh and sinh we get

cosh x + sinh x = ex,

cosh x− sinh x = e−x.

Multilplying the two equations we get

cosh2 x− sinh2 x = (cosh x + sinh x)(cosh x− sinh x)

= exe−x

= 1.

Dividing both sides by cosh2 x we get

1− tanh2 x = sech2 x.

12

Similarly,coth2 x− 1 = cosech2 x.

For this kind of problem one normally starts with the more complicatedside of the equality and manipulates it to get the other side. In this case wetake the right-hand side:

cosh x cosh y + sinh x sinh y

=1

2(ex + e−x)(ey + e−y) +

1

2(ex − e−x)(ey − e−y)

=1

4(ex+y + e−x+y + ex−y + e−x−yex+y − e−x+y − ex−y + e−x−y)

=2

4(ex+y + +e−x−y)

=2

4(ex+y + +e−(x+y))

= cosh(x + y).

Similarly (exercise)

sinh(x + y) = sinh x cosh y + cosh x sinh y.

Special cases.

cosh 2x = cosh2 x + sinh2 x

sinh 2x = 2 sinh x cosh x.

Also, since cosh2 x− sinh2 x = 1 we have

cosh 2x = 1 + 2 sinh2 x = 2 cosh2 x− 1.

9.3 Osborne’s Rule.

Osborne’s Rule. Take a trigonometric identity, replace each ordinary (cir-cular) trigonometric function with the corresponding hyperbolic function andchange the sign of every product or implied product of two sines.

Note. sin2 appears implicitly in tan2 and a few other similar cases wherethe sign should be changed as well.

13

9.3.1 Example 1.

Problem: Write down a formula for tanh 2x in terms of tanh x.Solution: We start with the trig formula

tan 2x =2 tan x

1− tan2 x.

We replace tan with tanh. We also change the sign before tan2 because itcontains a sin2. So,

tanh 2x =2 tanh x

1 + tanh2 x.

9.3.2 Example 2.

Show that (cosh u− 1

cosh u + 1

)1/2

= tanhu

2.

Now, cosh u = 2 cosh2 u2− 1 and cosh u = 2 sinh2 u

2+ 1. Hence

cosh u− 1 = 2 sinh2 u

cosh u + 1 = 2 cosh2 u.

Hence, (cosh u− 1

cosh u + 1

)1/2

=

(2 sinh2 u

2

2 cosh2 u2

)1/2

= tanhu

2.

Question: Why the negative square root, − tanh u2

can be discarded?

9.4 Reason for name

Equation of hyperbola isx2

a2− y2

b2= 1

(show a graph!)In parametric form it is

x = a cosh u

y = b sinh u

Draw a graph of a circle and show the trig functions meaning; thenDraw a graph of the hyperbole x2 − y2=1 and show the hyperbolic func-

tions meaning.

14

9.5 Solving equations involving hyperbolic functions

Example. Find x, given that

2 cosh x− sinh x = 3.

From definitions

2

(ex + e−x

2

)−(

ex − e−x

2

)= 3

i.e.1

2ex +

3

2e−x = 3.

Multiply by 2ex and rearrange

(ex)2 − 6ex + 3 = 0.

Hence,

ex =6±

√36− 12

2= 3±

√6.

Hence,x = ln(3±

√6) 2 solutions.

9.6 Relation between trigonometric and hyperbolic func-tions

Definitions

cosh z =1

2(ez + e−z)

sinh z =1

2(ez − e−z)

If z = jθ then

cosh(jθ) =1

2(ejθ + e−jθ) = cos θ

sinh(jθ) =1

2(ejθ − e−jθ) = j sin θ

Similarly,

cos(jθ) =1

2(ej(jθ) + e−j(jθ))

=1

2(e−θ + eθ)

= cosh θ

15

and

sin(jθ) =1

2j(ej(jθ) − e−j(jθ))

=1

2j(e−θ − eθ)

=1

jsinh θ

= j sinh θ

cosh(jθ) = cos θ

cos(jθ) = cosh θ

sinh(jθ) = j sin θ

sin(jθ) = j sinh θ

(Always seems surprising no minuses here.)

9.6.1 Example

Find real and imaginary parts of cos(π4+2j). Use cos(A+B) = cos A cos B−

sin A sin B, and the the above relations. We have,

cos(π

4+ 2j) = cos(

π

4) cos(2j)− sin(

π

4) sin(2j)

=1√2

cosh 2− 1√2(j sinh 2)

=cosh 2√

2− j

sinh 2√2

= 2.6603− j2.5646

9.7 Inverse hyperbolic functions

Similar to sin−1 x, etc.If y = sinh−1 x this means

x = sinh y,

so y is the value whose sinh is equal to x.y = cosh−1 x means x = cosh y.y = tanh−1 x means x = tanh y.y = coth−1 x means x = coth y.N.B. 1 sinh−1 x NOT same as 1

sinh x.

N.B. 2 sinh−1 x also called arcsinh x.

16

9.7.1 Sketches

Graphs are same as for sinh, cosh but with axes switched.N.B. cosh−1 x only valid for x ≥ 1 — it has two values.(For x < 1 need complex values —see later.)

9.7.2 Reason for name ‘inverse’

If y = cosh−1 x then x = cosh y. So

cosh(cosh−1 x) = cosh(y) = x.

Alsocosh−1(cosh y) = cosh−1 x = y.

Note: The first equation above is always true but the second one is trueonly if appropriate choice of cosh−1 is made.

Similarly for sinh, sinh−1 etc.

9.8 Relation of inverse hyperbolics to logs (Omitted)

Let y = cosh−1 x. Then

x = cosh y =1

2(ey + e−y),

so2x− ey − e−y = 0.

Multiply by −ey and rearrange

(ey)2 − 2xey + 1 = 0, quadratic for ey.

Hence,

ey =2x±

√4x2 − 4

2

= x±√

x2 − 1.

Take ln of both sides,

y = ln(x±√

x2 − 1), 2 solutions.

17

Now

x−√

x2 − 1 =(x−

√x2 − 1(x +

√x2 − 1))

x +√

x2 − 1

=x− (x2 − 1)

x +√

x2 − 1

=1

x +√

x2 − 1,

so

ln(x−√

x2 − 1) = ln

(1

x +√

x2 − 1

)= − ln(x +

√x2 − 1)

So the two solutions can also be written as

y = cosh−1 x = ± ln(x +√

x2 − 1)

2. Similarly (left as exercise)

sinh−1 x = ln(x +√

x2 + 1), any x

(only one solution here.)3. Let y = tanh−1 x. Then

x = tanh y =sinh y

cosh y=

ey − e−y

ey + e−y

Multiply top and bottom by ey,

=e2y − 1

e2y + 1.

Hence,x(e2y + 1) = e2y − 1

Hence,

e2y(x− 1) = −1− x = −(1 + x)

e2y = −1 + x

x− 1=

1 + x

1− x.

So,

2y = ln

(1 + x

1− x

).

18

So,

y = tanh−1 x =1

2ln

(1 + x

1− x

)Eg. Show that

tanh−1

(a2 − 1

a2 + 1

)= ln a

Put x = (a2 − 1)/(a2 + 1). Then

1

2

(1 + a2−1

a2+1

1− a2−1a2+1

)=

1

2

(a2 + 1 + (a2 − 1)

a2 + 1− (a2 − 1)

)=

1

2ln a2

= ln a

9.9 Derivatives of hyperbolics

N.B. Recall dd x

ex = ex and dd x

e−x = e−x.1. cosh x. If y = cosh x = 1

2(ex + e−x), so

d y

d x=

1

2(ex − e−x) = sinh x,

i.e.,d

d xcosh x = sinh x.

2. sinh x. If y = sinh x = 12(ex − e−x), so

d y

d x=

1

2(ex + e−x) = cosh x,

i.e.,d

d xsinh x = cosh x.

1. tanh x. If y = tanh x = sinh xcosh x

, so

d y

d x=

cosh x(cosh x)− sinh x(sinh x)

cosh2 x.

Butcosh2 x− sinh2 x = 1.

So,d

d xtanh x =

1

cosh2 x= sech2 x.

19

Similarly:

d

d xcoth x = − cosech2 x

d

d xsech x = − sech x tanh x

d

d xcosech x = − cosech x coth x.

Eg. Calculate derivative of

y =1

1 + sech x.

Now, let z = 1 + sech x. Then y = 1/z. So, using function of a function(chain) rule,

d y

d x=

d y

d z

d z

d x= (−1/z2)(0− sech x tanh x)

we get

d y

d x= −(1 + sech x)−2(0− sech x tanh x)

=sech x tanh x

(1 + sech x)2

9.10 Derivatives of inverse hyperbolic functions

9.10.1 sinh−1 x

Put y = sinh−1 x. Thensinh y = x.

Now, the derivative of the left-hand side of this equation should be equal tothe derivative of the right-hand side. We differentiate and denote by y′ thederivative of y with respect to x:

d

d x(sinh y) = y′ cosh y,

d

d xx = 1.

So,y′ cosh y = 1.

20

Hence,

y′ =1

cosh y.

This is a solution but the right-hand side depends on y while we requireresult in terms of x. To simplify it we notice that

cosh2 y = sinh2 y + 1 = x2 + 1,

and so

cosh y = ±√

sinh2 y + 1 = ±√

x2 + 1.

Hence

y′ = ± 1√1 + x2

.

But y is an increasing function of x. Hence y′ > 0, so we can discard theminus sign and deduce that the only solution for y′ is

′ =1√

x2 + 1.

Alternatively, put y = sinh−1 x, so sinh y = x. Then

d x

d y= cosh y.

Now,cosh2 y = sinh2 y + 1 = x2 + 1,

socosh y = ±

√x2 + 1.

Hence,d y

d x=

1√x2 + 1

.

Only positive sign here, see sketch.***put sketch here!

9.10.2 cosh−1 x

Put y = cosh−1 x. So cosh y = x. Then

d x

d y= sinh y.

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Now,sinh2 y = cosh2 y − 1 = x2 − 1,

sosinh y = ±

√x2 − 1.

Also,d y

d x=

1d xd y

= ± 1√x2 − 1

.

Both signs possible here—see sketch.***put sketch here!

9.10.3 tanh−1

Put y = tanh−1 x. So x = tanh y. Then

d

d xtanh y = y′ sech2 y = y′(1− tanh2 y),

andd

d xx = 1.

Hence,y′(1− tanh2 y) = 1,

i.e.,

y′ =1

1− tanh2 y,

Hence,d y

d x=

1

1− x2.

Put y = tanh−1 x. So x = tanh y. Then

d x

d y= sech2 y = 1− tanh2 y = 1− x2.

So,d y

d x=

1

1− x2.

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9.11 Integrals

9.11.1 Example 1

Evaluate

I =

∫ 1

0

1√x2 + 1

d x.

Using above

I =[sinh−1 x

]10

= sinh−1 1− 0

= 0.8814.

9.11.2 Example 2

Evaluate

I =

∫ √4 + 9x2 d x

For this sort of integrals we can use the formula

cosh2 x− sinh2 y = 1,

by a change of variables like x = cosh u or x = sinh u.Some preparatory work is needed however. We first manipulate 4 + 9x2

to look like 1 + something2,

√4 + 9x2 =

√4

(1 +

9

4x2

)= 2

√1 +

(3

2x

)2

So,

I =

∫ √4 + 9x2 d x = 2

∫ √1 +

(32x)2

d x

= 2× 2

3

∫ √1 + u2 d u where u =

3

2x, d x =

2

3d u,

=4

3

∫ √1 + sinh2 v cosh v d v where u = sinh v, d u = cosh v d v,

=4

3

∫ √cosh2 v cosh v d v

=4

3

∫cosh2 v d v

=2

3

∫(cosh(2v) + 1) d v because cosh2 v =

1

2(cosh(2v) + 1)

=2

3v +

1

3sinh(2v) + C.

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Now substituting v with u and u with x as appropriate, we get the result interms of the original variable x:

I =2

3v +

1

3sinh(2v) + C

=2

3v +

2

3sinh v cosh v + C

=2

3sinh−1 u +

2

3u√

1 + u2 + C

=2

3sinh−1(3

2x) + x

√1 + (3

2x)2 + C.

9.11.3 Example 3

This illustrates how integrals involving trigonometric functions can be evalu-ated using complex exponent. You are invited to try a more useful examplein Example sheet 2.

Evaluate:

I1 =

∫cos(bx) d x I2 =

∫sin(bx) d x

Solution. We note that

I1 + jI2 =

∫(cos(bx) + j sin(bx)) d x =

∫ejbx d x.

Hence,

I1 + jI2 =1

jbejbx =

−jb

b2ejbx =

−j

bejbx + C,

where C = C1 + jC2 is a constant. Now I1 is equal to the real part of theright-hand side (RHS) while I2 is equal to the imaginary part of the RHS.By straightforward manipulation we get

−j

bejbx =

−j

b(cos bx + j sin(bx)) =

1

bsin(bx)− j

−1

bcos bx.

So,

I1 =1

bsin(bx) + C1 I2 = −1

bcos bx + C2.

9.12 Concluding remarks

• Complex numbers can be manipulated according to the usual arith-metic rules. We only lose the possibility to say which of two complexnumbers is greater since there is no “less than” relation between com-plex numbers.

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• There are simple relations between the complex exponent, ez, the or-dinary trigonometric functions and the hyperbolic functions.

• The elementary functions (such as ex, ln x, sin x) have the familiar prop-erties when applied to complex arguments. The few notable exceptionsare listed below.

• ez can be negative when z is complex.

• sin z and cos z are no longer restricted to the interval [−1, 1].

• Roots and logarithms exist for all numbers but are not unique.

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