1 Introduction 2 Imaginary Unit 3 Imaginary Numbers

Home , Imaginary unit

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 usual arithmetic rules. Practical rule 1. Expressions containing j can be manipulated like expressions containing numbers and/or variables with the added bonus that powers of j can be simpliﬁed.

3 Imaginary numbers

An imaginary number is a product of j, the imaginary unit, with a real number, 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 kinds of roots we need. For example, just as for positive numbers we can ﬁnd two square roots of any 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 q is a non-negative integer, r = 0, 1, 2 or 3. Hence 1 when r = 0  j when r = 1 j4q+r = −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 1 + j, −1 + j, −2 − j3, 3 − j . 2 In general, let a and b be real numbers. Then the complex number z = a+jb is said to be in standard (or cartesian) form. The real part of z is a, the imaginary part of z is b. Re(z) = a, Im(z) = b.

2 5.1 Argand diagram 4 quadrants, show a number in each complex 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 are often said to be purely real. The imaginary numbers are those complex numbers whose real parts are 0. We normally write j3 rather than 0 + j3. A complex number whose real part is zero is also said to be purely imaginary. 0 + j0 is both purely real and purely imaginary and is normally written as 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. Simply expand and collect terms as appropriate for each operation. Only the division is somewhat tricky. Notice the way to remove the imaginary unit from the denominator. Let z1 = 1 + j3, z2 = 2 − j. Introduce the following operations ﬁrst using z1 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 • Division z a + jb 1 = 1 1 z2 a2 + jb2 (a + jb )(a − jb ) = 1 1 2 2 (a2 + jb2)(a2 − jb2)

(a1 + jb1)(a2 − jb2) = 2 2 a2 − (jb2) (a1 + jb1)(a2 − jb2) = 2 2 a2 + b2 (a1a2 + b1b2) + j(b1a2 − a1b2) = 2 2 a2 + b2 (a1a2 + b1b2) (b1a2 − a1b2) = 2 2 + j 2 2 a2 + b2 a2 + b2 • z−1. This is the same as 1/z.

6.1 Solving quadratic equations Any quadratic equation has two roots, possibly complex. The familiar formula applies. Example. Solve the equation z2 − 2z + 5 = 0. Solution: √ √ √ √ 2 ± 4 − 4 × 5 2 ± 4 − 20 2 ± −16 2 ± j 4 z = = = = = 1 ± j2. 2 2 2 2 6.2 Complex conjugate If z = a + jb then the complex conjugate of z is denoted by z∗ orz ¯ and is obtained 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 parts and 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 prescribed properties. Example. Find a complex number z such that 1 z + z∗ = 3 + j2. 2 Solution. Let the standard form of z be z = x+jy. Then the above becomes 1 x + jy + (x − jy) = 3 + j2, 2 we now expand the expression and collect the terms as appropriate to get 1 1 x + x + j(y − y) = 3 + j2, 2 2 which simpliﬁes to 3 1 x + j( y) = 3 + j2. 2 2 The real and imaginary parts of the number to the left of the equality must be equal to the corresponding parts on the right. Hence, 3 1 x = 3, y = 2, 2 2 which gives x = 2, y = 4. Hence the answer is z = 2 + j4. We can verify this by substituting it into the equation above: 1 1 z + z∗ = 2 + j4 + (2 − j4) = 2 + j4 + 1 − j2 = 3 + j2, as required. 2 2 Note: There is no “less then” relation betwen complex numbers!

6.4 Modulus and argument 1) Show on the picture 2) deﬁne modulus, mention r > 0. Note that |z| = zz∗. 3) deﬁne argument, 4) discuss on z = 1 + j 5) 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

4 Example: z = −3 + j4. |z| = 5, tan(θ) = − 3 . 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, √ 3 3 3 z = + j . 2 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 particular, ez1+z2 = ez1 ez2 , (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 a complex 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.

jθ1 jθ2 If z1 = r1e , z2 = r2e , then

jθ1 jθ2 jθ1 jθ2 j(θ1+θ2) z = z1z2 = r1e r2e = r1r2e e = (r1r2)e , i.e., to get the product we multiply the moduli and add the arguments. Similarly, for division,

jθ1 jθ2 jθ1 −jθ2 j(θ1−θ2) z1/z2 = r1e /(r2e ) = r1r2e e = (r1r2)e , i.e., we divide the moduli and subtract the arguments. Geometrical representation of multiplication by j and other complex numbers.

6.7.2 Addition in exponential form Example. Find the real and imaginary parts of z = ejπ/4 + 2ejπ/3 Now, 1 1 ejπ/4 = cos(π/4) + j sin(π/4) = √ + j √ 2 2 √ 1 3 ejπ/3 = cos(π/3) + j sin(π/3) = + j 2 2 Hence, √ ! 1 1 1 3 z = √ + j √ + 2 + j 2 2 2 2 1 1 √ = (1 + √ ) + j(√ + 3) 2 2 = 1.7071 + j2.4392 The exponent is extremely useful. We apply it below to derive the De Moivre’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 the exponent we get,

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

Hence the expressions on the right-hand sides of the last two equations are equal, as required. De Moivre’s formula works for any n including negative. However, if n is not 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,

ejθ + e−jθ ejθ − e−jθ cos θ = , sin θ = . 2 2 We use these relations to deﬁne these functions for complex arguments. So, for any complex number z we deﬁne

ejz + e−jz cos z = , 2 ejz − e−jz sin z = . 2

6.7.5 Powers of sin(θ) and cos(θ) in terms of sin and cos of multiple angles We can use the above results to rewrite cosn θ and sinn(θ) in a way which is much easier to integrate. For example,

8 Ex. 1

ejθ + e−jθ 3 cos3 θ = 2 1 = ej3θ + 3ej2θe−jθ + 3ejθe−j2θ + e−j3θ 23 1 = ej3θ + e−j3θ + 3ejθ + 3e−jθ 8 1 = (ej3θ + e−j3θ) + 3(ejθ + e−jθ) 8 1 = (2 cos(3θ) + 3 × 2 cos θ) 8 1 3 = cos(3θ) + cos(θ) 4 4 Ex. 2 (Exercise) 1 5 5 sin5 θ = sin 5θ − sin 3θ + sin θ 16 16 8 Note. For real powers of sin θ the result is in terms of cosines. (see Example 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 is z1 z2 needed. In particular, if z1 and z2 are such that e = e we cannot conclude without 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, √ 3 |z| = 42 + 32 = 5, arg(z) = tan−1( ) = 0.6435, 4 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 ( n r)ejθ/n = rejθ. √ Hence n rejθ/n is an nth root of z. Similarly to the logarithm we actually have more roots. Dissimilarly, there are exactly n diﬀerent nth roots. These are 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, √ √ 1/2 1 j(0.6435+2kπ) j(0.3217+kπ) z = 5e 2 = 5e .

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,

1/3 1/3 1 j(0+2kπ) j 2kπ 1 = 1 e 3 = e 3 , k = 0, 1, 2.

Putting k = 0, 1, 2 we get

ej0, ej2π/3, ej4π/3, which can be written also as √ √ 1 3 1 3 1, − + j , − − j . 2 2 2 2

8.0.10 Rational powers Let p and q be integers and

z = rej(θ+2kπ).

Then p/q p/q p j(θ+2kπ) p/q j( θp + 2kπp ) z = r e q = r e q q √ n 1/n Note.√ In principle, z and z are equivalent. However, if z is real the n symbol 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: 1 cosh x = (ex + e−x) 2 1 sinh x = (ex − e−x) 2 *** draw graphics ... ***

11 From these two we can get others sinh x tanh x = cosh x cosh x 1 coth x = = sinh x tanh x 1 sech x = cosh x 1 cosech x = sinh x Practical rule 2. Any problem involving hyperbolic functions can be solved 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 the exponential function. The approved UMIST calculator can calculate them directly with the help of the “hyp” button.

• hyp, sin x gives sinh x. • hyp, shift, sin y gives sinh−1 y.

9.2 Similarities between hyperbolic and trigonometric functions Although shapes of these curves are very diﬀerent to the trigonometric functions, 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 complicated side of the equality and manipulates it to get the other side. In this case we take the right-hand side: cosh x cosh y + sinh x sinh y 1 1 = (ex + e−x)(ey + e−y) + (ex − e−x)(ey − e−y) 2 2 1 = (ex+y + e−x+y + ex−y + e−x−yex+y − e−x+y − ex−y + e−x−y) 4 2 = (ex+y + +e−x−y) 4 2 = (ex+y + +e−(x+y)) 4 = 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 and change the sign of every product or implied product of two sines.

Note. sin2 appears implicitly in tan2 and a few other similar cases where the 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 2 tan x tan 2x = . 1 − tan2 x We replace tan with tanh. We also change the sign before tan2 because it contains a sin2. So, 2 tanh x tanh 2x = . 1 + tanh2 x

9.3.2 Example 2. Show that cosh u − 11/2 u = tanh . cosh u + 1 2 2 u 2 u Now, cosh u = 2 cosh 2 − 1 and cosh u = 2 sinh 2 + 1. Hence cosh u − 1 = 2 sinh2 u cosh u + 1 = 2 cosh2 u.

Hence, !1/2 cosh u − 11/2 2 sinh2 u u = 2 = tanh . cosh u + 1 2 u 2 2 cosh 2 u Question: Why the negative square root, − tanh 2 can be discarded?

9.4 Reason for name Equation of hyperbola is x2 y2 − = 1 a2 b2 (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; then Draw a graph of the hyperbole x2 − y2=1 and show the hyperbolic functions meaning.

14 9.5 Solving equations involving hyperbolic functions Example. Find x, given that 2 cosh x − sinh x = 3. From deﬁnitions ex + e−x ex − e−x 2 − = 3 2 2 i.e. 1 3 ex + e−x = 3. 2 2 Multiply by 2ex and rearrange (ex)2 − 6ex + 3 = 0.

Hence, √ 6 ± 36 − 12 √ ex = = 3 ± 6. 2 Hence, √ x = ln(3 ± 6) 2 solutions.

9.6 Relation between trigonometric and hyperbolic functions Deﬁnitions 1 cosh z = (ez + e−z) 2 1 sinh z = (ez − e−z) 2 If z = jθ then 1 cosh(jθ) = (ejθ + e−jθ) = cos θ 2 1 sinh(jθ) = (ejθ − e−jθ) = j sin θ 2 Similarly, 1 cos(jθ) = (ej(jθ) + e−j(jθ)) 2 1 = (e−θ + eθ) 2 = cosh θ

15 and 1 sin(jθ) = (ej(jθ) − e−j(jθ)) 2j 1 = (e−θ − eθ) 2j 1 = sinh θ j = 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

π Find real and imaginary parts of cos( 4 +2j). Use cos(A+B) = cos A cos B − sin A sin B, and the the above relations. We have, π π π cos( + 2j) = cos( ) cos(2j) − sin( ) sin(2j) 4 4 4 1 1 = √ cosh 2 − √ (j sinh 2) 2 2 cosh 2 sinh 2 = √ − j √ 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. −1 1 N.B. 1 sinh x NOT same as 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.

Also cosh−1(cosh y) = cosh−1 x = y. Note: The ﬁrst equation above is always true but the second one is true only 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 1 x = cosh y = (ey + e−y), 2 so 2x − ey − e−y = 0. Multiply by −ey and rearrange

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

Hence, √ 2x ± 4x2 − 4 ey = √ 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 √ 1 ln(x − x2 − 1) = ln √ 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 sinh y ey − e−y x = tanh y = = cosh 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) 1 + x 1 + x e2y = − = . x − 1 1 − x So, 1 + x 2y = ln . 1 − x

18 So, 1 1 + x y = tanh−1 x = ln 2 1 − x Eg. Show that a2 − 1 tanh−1 = ln a a2 + 1 Put x = (a2 − 1)/(a2 + 1). Then

2 a −1 ! 2 2 1 1 + 2 1 a + 1 + (a − 1) a +1 = 2 a2−1 2 a2 + 1 − (a2 − 1) 1 − a2+1 1 = ln a2 2 = ln a

9.9 Derivatives of hyperbolics

d x x d −x −x N.B. Recall d x e = e and d x e = e . 1 x −x 1. cosh x. If y = cosh x = 2 (e + e ), so d y 1 = (ex − e−x) = sinh x, d x 2 i.e., d cosh x = sinh x. d x 1 x −x 2. sinh x. If y = sinh x = 2 (e − e ), so d y 1 = (ex + e−x) = cosh x, d x 2 i.e., d sinh x = cosh x. d x sinh x 1. tanh x. If y = tanh x = cosh x , so d y cosh x(cosh x) − sinh x(sinh x) = . d x cosh2 x But cosh2 x − sinh2 x = 1. So, d 1 tanh x = = sech2 x. d x cosh2 x

19 Similarly: d coth x = − cosech2 x d x d sech x = − sech x tanh x d x d cosech x = − cosech x coth x. d x Eg. Calculate derivative of 1 y = . 1 + sech x Now, let z = 1 + sech x. Then y = 1/z. So, using function of a function (chain) rule,

d y d y d z = = (−1/z2)(0 − sech x tanh x) d x d z d x we get d y = −(1 + sech x)−2(0 − sech x tanh x) d 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. Then sinh y = x. Now, the derivative of the left-hand side of this equation should be equal to the derivative of the right-hand side. We diﬀerentiate and denote by y0 the derivative of y with respect to x: d (sinh y) = y0 cosh y, d x d x = 1. d x So, y0 cosh y = 1.

20 Hence, 1 y0 = . cosh y This is a solution but the right-hand side depends on y while we require result in terms of x. To simplify it we notice that

cosh2 y = sinh2 y + 1 = x2 + 1, and so q √ cosh y = ± sinh2 y + 1 = ± x2 + 1. Hence 1 y0 = ±√ . 1 + x2 But y is an increasing function of x. Hence y0 > 0, so we can discard the minus sign and deduce that the only solution for y0 is 1 0 = √ . x2 + 1

Alternatively, put y = sinh−1 x, so sinh y = x. Then d x = cosh y. d y Now, cosh2 y = sinh2 y + 1 = x2 + 1, so √ cosh y = ± x2 + 1. Hence, d y 1 = √ . d x 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 = sinh y. d y

21 Now, sinh2 y = cosh2 y − 1 = x2 − 1, so √ sinh y = ± x2 − 1. Also, d y 1 1 = = ±√ . d x d x 2 d y x − 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 tanh y = y0 sech2 y = y0(1 − tanh2 y), d x and d x = 1. d x Hence, y0(1 − tanh2 y) = 1, i.e., 1 y0 = , 1 − tanh2 y Hence, d y 1 = . d x 1 − x2 Put y = tanh−1 x. So x = tanh y. Then d x = sech2 y = 1 − tanh2 y = 1 − x2. d y So, d y 1 = . d x 1 − x2

22 9.11 Integrals 9.11.1 Example 1 Evaluate Z 1 1 I = √ d x. 2 0 x + 1 Using above −1 1 I = sinh x 0 = sinh−1 1 − 0 = 0.8814.

9.11.2 Example 2 Evaluate Z √ 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, s s √ 9 3 2 4 + 9x2 = 4 1 + x2 = 2 1 + x 4 2 So, Z √ Z q 2 3 2 I = 4 + 9x d x = 2 1 + 2 x d x 2 Z √ 3 2 = 2 × 1 + u2 d u where u = x, d x = d u, 3 2 3 4 Z p = 1 + sinh2 v cosh v d v where u = sinh v, d u = cosh v d v, 3 4 Z p = cosh2 v cosh v d v 3 4 Z = cosh2 v d v 3 2 Z 1 = (cosh(2v) + 1) d v because cosh2 v = (cosh(2v) + 1) 3 2 2 1 = v + sinh(2v) + C. 3 3

23 Now substituting v with u and u with x as appropriate, we get the result in terms of the original variable x: 2 1 I = v + sinh(2v) + C 3 3 2 2 = v + sinh v cosh v + C 3 3 2 2 √ = sinh−1 u + u 1 + u2 + C 3 3 2 q = sinh−1( 3 x) + x 1 + ( 3 x)2 + C. 3 2 2

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 example in Example sheet 2. Evaluate: Z Z I1 = cos(bx) d x I2 = sin(bx) d x

Solution. We note that Z Z jbx I1 + jI2 = (cos(bx) + j sin(bx)) d x = e d x.

Hence, 1 −jb −j I + jI = ejbx = ejbx = ejbx + C, 1 2 jb b2 b where C = C1 + jC2 is a constant. Now I1 is equal to the real part of the right-hand side (RHS) while I2 is equal to the imaginary part of the RHS. By straightforward manipulation we get −j −j 1 −1 ejbx = (cos bx + j sin(bx)) = sin(bx) − j cos bx. b b b b So, 1 1 I = sin(bx) + C I = − cos bx + C . 1 b 1 2 b 2 9.12 Concluding remarks • Complex numbers can be manipulated according to the usual arithmetic rules. We only lose the possibility to say which of two complex numbers is greater since there is no “less than” relation between complex numbers.

24 • There are simple relations between the complex exponent, ez, the ordinary trigonometric functions and the hyperbolic functions.

• The elementary functions (such as ex, ln x, sin x) have the familiar properties when applied to complex arguments. The few notable exceptions are 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.