semester 1 2017{18 - university of...
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University of Sheffield
School of Mathematics and Statistics
MAS140: Mathematics (Chemical)
MAS152: Civil Engineering Mathematics
MAS152: Essential Mathematical Skills & Techniques
MAS156: Mathematics (Electrical and Aerospace)
MAS161: General Engineering Mathematics
Semester 1 2017–18
Outline Syllabus
• Functions of a real variable. The concept of a function; odd, even and periodic
functions; continuity. Binomial theorem.
• Elementary functions. Circular functions and their inverses. Polynomials.
Exponential, logarithmic and hyperbolic functions.
• Differentiation. Basic rules of differentiation: maxima, minima and curve sketching.
• Partial differentiation. First and second derivatives, geometrical interpretation.
• Series. Taylor and Maclaurin series, L’Hopital’s rule.
• Complex numbers. basic manipulation, Argand diagram, de Moivre’s theorem,
Euler’s relation.
• Vectors. Vector algebra, dot and cross products, differentiation.
Module Materials
These notes supplement the video lectures. All course materials, including examples sheets
(with worked solutions), are available on the course webpage,
http://engmaths.group.shef.ac.uk/mas140/
http://engmaths.group.shef.ac.uk/mas151/
http://engmaths.group.shef.ac.uk/mas152/
http://engmaths.group.shef.ac.uk/mas156/
http://engmaths.group.shef.ac.uk/mas161/
which can also be accessed through MOLE.
1
1 Complex Numbers
The equation
x2 − 1 = 0
has the solutions x = ±1 and the equation
x2 − 6x+ 5 = 0
⇒ (x− 5)(x− 1) = 0
has the roots 5 and 1. However, the equation
x2 + 1 = 0
is not satisfied by any (real) number x (indeed, x2 ≥ 0 for all x ∈ R).
We define the quantity i by i =√−1 so that i2 = −1. The last equation now has the two
roots i and −i. (Do not be tempted to think that one of them is positive and one negative.
The quantities i and −i are neither positive nor negative!)
A quantity z of the form
z = x+ iy,
where x and y are real numbers, is called a complex number. They occur naturally when
solving quadratic equations. Thus if
25z2 − 20z + 13 = 0
then (using the standard formula)
z =20±
√202 − 4× 25× 13
2× 25
=2±√−9
5
=2
5± i3
5
For any real numbers x and y, z = x+ iy is a complex number. In particular, choosing y to
be zero shows that all real numbers are complex. In the same way, all integers are rational.
C ⊃ R ⊃ Q ⊃ Z
We do not distinguish between any of the following:
x+ iy, x+ yi, iy + x, yi+ x.
The symbol j is often used for√−1, especially in electrical engineering.
In what follows, x and y will always be real numbers and z a complex number.
2
If z = x+ iy then
x is called the real part of z
y is called the imaginary part of z
and we write
x = Re(z) and y = Im(z).
Note. The imaginary part of a complex number is real .
Example
If z = 4− 3i then Re(z) = 4 and Im(z) = −3.
(<(z) and =(z) are also used.)
A complex number of the form iy is called pure-imaginary . Its real part is zero.
The complex conjugate of the complex number z = x+ iy is denoted by z and is defined by
z = x− iy.
Example z = 4− 3i⇒ z = 4 + 3i.
Taking the complex conjugate twice gives the original number. i.e. if z1 = z, then z1 = z.
If z = z then Im(z) = 0 and z is a real number.
1.1 Algebraic Operations
Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers.
• Equality
z1 = z2 ⇔ x1 = x2 and y1 = y2
• Additionz1 + z2 = (x1 + iy1) + (x2 + iy2)
= (x1 + x2) + i(y1 + y2)
• Subtractionz1 − z2 = (x1 + iy1)− (x2 + iy2)
= (x1 − x2) + i(y1 − y2)
• Multiplication
z1z2 = (x1 + iy1)(x2 + iy2)
= (x1x2 − y1y2) + i(x1y2 + x2y1)
• Divisionz1z2
=x1 + iy1x2 + iy2
=(x1 + iy1)
(x2 + iy2)
(x2 − iy2)(x2 − iy2)
=(x1x2 + y1y2)
(x22 + y22)+ i
(−x1y2 + y1x2)
(x22 + y22)
3
Example: Find the real and imaginary parts of z =3− 4i
5 + 2i
Solution3− 4i
5 + 2i=
(3− 4i)
(5 + 2i)
(5− 2i)
(5− 2i)
=(3× 5− 4× 2) + i(−3× 2− 4× 5)
(25 + 4)
=7− 26i
29
The real part is Re(z) = 7/29 and the imaginary part is Im(z) = −26/29.
For any complex numbers z1, z2 and z3 we have
z1 + z2 = z2 + z1 (commutative)
z1z2 = z2z1
z1 + (z2 + z3) = (z1 + z2) + z3 (associative)
z1(z2z3) = (z1z2)z3
Note
• z + z = 2Re(z) and z − z = 2iIm(z)
• z1 + z2 = z1 + z2, z1 − z2 = z1 − z2
• z1z2 = (z1)(z2), z1/z2 = z1/z2
1.2 Geometrical Representation: The Argand Diagram
The complex number z = x+iy can be represented by the point
P with coordinates (x, y) in a plane. In this plane, known as
the Argand Diagram , the x-axis is called the real axis and the
y-axis is the imaginary axis.
θ
y
x P
O
1.2.1 Modulus and Argument
Instead of using Cartesian coordinates (i.e. x and y) to specify the position of P in the Argand
diagram, we may use polar coordinates r and θ. Then
x = r cos θ and y = r sin θ
4
and so
z = x+ iy = r(cos θ + i sin θ).
This is called the polar form of the complex number z. The non-negative real number r is
the modulus of z and θ is the argument of z.
We use the notation
|z| = r and arg z = θ.
Clearly
r =√x2 + y2, sin θ = y/r and cos θ = x/r.
The angle θ is measured positive in the anti-clockwise direction from the real axis. Values of θ
which differ by 2π (or 4π etc) correspond to the same direction in the Argand diagram. The
unique angle θ such that −π < θ ≤ π is called the principal value of the argument and we
shall denote it by Argz.
Example: Find the modulus and principal argument of
(a) 1 + i√
3 (b) −1 + i
Solution
(a) Let z = 1 + i√
3, then
|z| =√
12 + (√
3)2 = 2 and θ is such that
cos θ =1
2, sin θ =
√3
2and − π < θ ≤ π.
Hence the principal value of the argument is π/3.
(b) | − 1 + i| =√
(−1)2 + (1)2 =√
2 and θ is such that
cos θ = − 1√2, sin θ =
1√2
and − π < θ ≤ π.
Hence the principal value of the argument is 3π/4.
Note
(a) |z| = |z|
(b) z = x+ iy ⇒ zz = |z|2 = x2 + y2
(c) Re(z) ≤ |z| and Im(z) ≤ |z|
It follows from (b) that1
z=
z
|z|2.
5
In particular, if |z| = 1 then 1/z = z.
If the complex number z has modulus r and argument θ, then this is sometimes written as
z = 〈r, θ〉. Also
z = 〈r, θ〉 ⇒ z = 〈r,−θ〉
1.2.2 Addition and the Argand Diagram
If P represents z1 and Q represents z2 then R will
represent their sum z1 + z2. Note that OPRQ is a
parallelogram. Also the triangle inequality gives
OP + PR ≥ OR
and hence
OP +OQ ≥ OR.
This implies that
|z1 + z2| ≤ |z1|+ |z2|.
y
xP
QR
O
1.3 Multiplication of Complex Numbers in Polar Form
Letz1 = r1(cos θ1 + i sin θ1)
and z2 = r2(cos θ2 + i sin θ2)
thenz1z2 = r1(cos θ1 + i sin θ1)r2(cos θ2 + i sin θ2)
= r1r2[(cos θ1 cos θ2 − sin θ1 sin θ2)+
i(sin θ1 cos θ2 + cos θ1 sin θ2)]
= r1r2[(cos(θ1 + θ2) + i sin(θ1 + θ2)]
Or
〈r1, θ1〉〈r2, θ2〉 = 〈r1r2, θ1 + θ2〉
Alsoz1z2
=r1(cos θ1 + i sin θ1)
r2(cos θ2 + i sin θ2)
=r1r2
(cos θ1 + i sin θ1)(cos θ2 − i sin θ2)
=r1r2
[(cos θ1 cos θ2 + sin θ1 sin θ2)+
i(sin θ1 cos θ2 − cos θ1 sin θ2)]
=r1r2
[(cos(θ1 − θ2) + i sin(θ1 − θ2)]
6
Or〈r1, θ1〉〈r2, θ2〉
=
⟨r1r2, θ1 − θ2
⟩Hence we have the following results for multiplication and division:
(a) |z1z2| = |z1||z2|
(b) |z1/z2| = |z1|/|z2|
(c) arg(z1z2) = arg z1 + arg z2
(d) arg(z1/z2) = arg z1 − arg z2
Note that it is not possible to use Arg in results (c) and (d).
1.4 Loci
If the complex number z satisfies the equation
|z| = 1
then its distance from the origin is 1. So z lies on the circle of radius 1 centred on the origin.
This circle is often referred to as the unit circle .
Alternatively,
|z| = 1 ⇒√x2 + y2 = 1
⇒ x2 + y2 = 1
which is the equation of a circle, radius 1 and centre at the origin.
If |z| satisfies
|z − z0| = r
(where z0 is a fixed complex number and r is a fixed positive number) then z lies on the circle
of radius r with centre at z0.
To see this, let
z = x+ iy and z0 = x0 + iy0
then
z − z0 = (x− x0) + i(y − y0)
and so
|z − z0| = r ⇒√
(x− x0)2 + (y − y0)2 = r
⇒ (x− x0)2 + (y − y0)2 = r2
7
which is the equation of a circle, centre (x0, y0) and radius r.
Example: If z satisfies the equation
|z − i| = |z + i| (?)
what is the locus of z in the Argand diagram?
Solution 1: Geometry
The given equation implies that the distance between the point P representing z and the
point i is the same as that between P and −i. It follows that P lies on the perpendicular
bisector of the points i and −i.So P lies on the real axis.
i
- i
- 2 2
Solution 2: Analysis
Set z = x+ iy so that the equation (?) becomes
|x+ i(y − 1)| = |x+ i(y + 1)|⇒√x2 + (y − 1)2 =
√x2 + (y + 1)2
⇒ x2 + (y − 1)2 = x2 + (y + 1)2
⇒ x2 + y2 − 2y + 1 = x2 + y2 + 2y + 1
⇒ 4y = 0
⇒ y = 0
Thus the imaginary part of z is zero and so z is a real number and hence lies on the real axis.
Example: If |z + 1| =√
2|z − 1|, what is the locus of z?
Solution
Let z = x+ iy, then
|z + 1| = |(x+ 1) + iy| =√
(x+ 1)2 + y2
and |z − 1| = |(x− 1) + iy| =√
(x− 1)2 + y2
8
Thus
(x+ 1)2 + y2 = 2[(x− 1)2 + y2]
x2 + 2x+ 1 + y2 = 2[x2 − 2x+ 1 + y2]
= 2x2 − 4x+ 2 + 2y2
x2 + y2 − 6x+ 1 = 0
(x− 3)2 − 9 + y2 + 1 = 0
⇒ (x− 3)2 + y2 = 8
Thus the point representing z in the Argand diagram lies on a circle with radius√
8 and
centred on (3, 0). This result is known as the circle of Apollonius.
Example
Show that multiplying any complex number by i has the same effect as an anticlockwise
rotation of π/2.
Solution
If z has modulus r and argumentθ then
z = 〈r, θ〉 and i = 〈1, π/2〉
and so
iz = 〈r, θ〉〈1, π/2〉= 〈r × 1, θ + π/2〉= 〈r, θ + π/2〉
So iz has the same modulus as z but the argument is increased by π/2.
z
iz
9
1.5 De Moivre’s Theorem
Let z1 have modulus r1 and argument θ1. Also let z2 have modulus r2 and argument θ2. Then
z1z2 = 〈r1, θ1〉〈r2, θ2〉= 〈r1r2, θ1 + θ2〉
If θ1 = θ2 = θ and r1 = r2 = r then we see that
〈r, θ〉2 = 〈r2, 2θ〉
and so
(cos θ + i sin θ)2 = cos 2θ + i sin 2θ
Also
〈r, θ〉3 = 〈r2, 2θ〉〈r, θ〉= 〈r3, 3θ〉
and
(cos θ + i sin θ)n = cosnθ + i sinnθ
for any natural number n.
The generalization of this result is de Moivre’s theorem which states that
(cos θ + i sin θ)q = cos qθ + i sin qθ
for any rational number q.
More accurately, the r.h.s. gives one of the values of the l.h.s. If q = m/n (where m and n are
integers, n positive) then the l.h.s. has n possible values whereas the r.h.s. has just one value.
1.6 Euler’s Relation
If x is any real number, then
ex = 1 + x+x2
2!+x3
3!+x4
4!+ . . .
sinx = x− x3
3!+x5
5!− x7
7!+ . . .
cosx = 1− x2
2!+x4
4!− x6
6!+ . . .
10
These expansions are now taken to apply to all complex numbers. Thus
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!+ . . .
= 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!+ . . .
= 1− θ2
2!+θ4
4!+ . . .+ i
(θ − θ3
3!+θ5
5!+ . . .
)
= cos θ + i sin θ
which gives Euler’s relation,
eiθ = cos θ + i sin θ
Note
(a) e−iθ = cos θ − i sin θ,
(b) eiπ/2 = i,
(c) eiπ = −1,
(d) ei3π/2 = −i,
(e) ei2pπ = 1 for any integer p.
From the results
eiθ = cos θ + i sin θ and e−iθ = cos θ − i sin θ
it follows that
cos θ =1
2
(eiθ + e−iθ
)and
sin θ =1
2i
(eiθ − e−iθ
)Also, de Moivre’s theorem can now be written as(
eiθ)q
= eiqθ
for any rational number q.
Example: Find the real and imaginary parts of
z =(√
3 + i)6
(i− 1)3.
Solution
Let z1 = (√
3 + i) and z2 = (i− 1), and write z1 and z2 in polar form:
|z1| = |√
3 + i| =√
(√
3)2 + 12 =√
4 = 2
11
and if Argz1 = θ then
cos θ =
√3
2, sin θ =
1
2, − π < θ ≤ π ⇒ θ =
π
6
Thus z1 = (√
3 + i) = 2eiπ/6.
Also
|z2| = |i− 1| =√
(−1)2 + 12 =√
2
and if Argz2 = φ then
cosφ = − 1√2, sinφ =
1√2, − π < φ ≤ π ⇒ φ =
3π
4
Thus z2 = (i− 1) =√
2ei3π/4.
The numerator of z is therefore
(√
3 + i)6 =(2eiπ/6
)6= 26e6iπ/6 = 26e−iπ = −26
Also, the denominator of z is
(i− 1)−3 =(√
2ei3π/4)−3
= 2−3/2e−9iπ/4 = 2−3/2e−iπ/4
= 2−3/2[cos(π
4
)− i sin
(π4
)]= 2−3/2
[1√2− i 1√
2
]= 2−2(1− i)
Therefore
z =(√
3 + i)6
(i− 1)3= −26 × 2−2(1− i) = 16(i− 1) =⇒ Re(z) = −16, Im(z) = 16.
1.7 Applications of De Moivre’s Theorem
Example: Show that each of
z0 = 〈1, 0〉, z1 = 〈1, 2π/3〉, z2 = 〈1, 4π/3〉
satisfies the equation z3 = 1. Plot all 3 numbers in the Argand diagram.
Solution
Firstly z0 = 1⇒ z30 = 1.
Also, since z = 〈r, θ〉 = r(cos θ + i sin θ)
⇒ z3 = 〈r3, 3θ〉 = r3(cos 3θ + i sin 3θ)
it follows that
z31 = 〈13, 3× 2π/3〉 = 〈1, 2π〉 = 1
and z32 = 〈13, 3× 4π/3〉 = 〈1, 4π〉 = 1
Since all of the z’s have modulus 1, they lie on the unit circle. They are called the
cube roots of unity.
12
2π/3
2π/3
..
.z
z
z
0
1
2
In the last example we verified that three given numbers were the cube roots of unity. We
now show how to find them directly. The method used will allow us to find all the roots of
any complex number.
Example: Find the cube roots of unity, i.e. the three values of 11/3.
Solution
First write 1 in the most general polar form. Since |1| = 1 and Arg(1) = 0 it follows that
arg(1) = 2pπ for any integer p and that
1 = 1[cos(2pπ) + i sin(2pπ)] is the most general polar form of the (complex) number 1. Now
we require z = 11/3 and so, by de Moivre’s theorem,
z = {1[cos(2pπ) + i sin(2pπ)]}1/3
= cos
(2pπ
3
)+ i sin
(2pπ
3
)for any integer p. Setting p equal to 0, 1 and 2 in turn gives the required 3 values. It is usual
to denote the corresponding z values by the subscript p.
p = 0 ⇒ z0 = cos 0 + i sin 0 = 1
p = 1 ⇒ z1 = cos2π
3+ i sin
2π
3= −1
2+ i
√3
2
p = 2 ⇒ z2 = cos4π
3+ i sin
4π
3= −1
2− i√
3
2
These are the 3 numbers in the previous example.
Note
(a) Any other value of p will give one of z0, z1, z2. Thus there are exactly 3 distinct
values of z for which z3 = 1.
(b) z1 and z2 form a complex conjugate pair. This illustrates a general result, viz. that if
z is a root of a polynomial with real coefficients, the z is also a root. (Of course, if z is
real then this says nothing.)
13
1.7.1 The nth Roots of Unity
If n is any positive integer, we say that z is an nth root of unity if zn = 1. To find z, proceed
as before:
1 = 1[cos(2pπ) + i sin(2pπ)] (p any integer)
z = 11/n = [cos(2pπ) + i sin(2pπ)]1/n
= cos
(2pπ
n
)+ i sin
(2pπ
n
)Taking p = 0, 1, 2, . . . , (n− 1) will give the n roots. (Other values of n will just repeat the
values.) They all have modulus 1 and so lie on the unit circle.
..z
z
0
1
.
.
z2
zn -1
θ
θ
θ
The angle θ is 2π/n. The roots are equally spaced around the unit circle.
Example: Find all values z for which
z3 = 1− i√
3
Solution: First write 1− i√
3 in the most general polar form. Since
|1− i√
3| =√
12 + (−√
3)2 =√
4 = 2
and, for the principal argument θ,
cos θ =1
2, sin θ = −
√3
2, − π < θ ≤ π ⇒ θ = −π
3
Thus
z3 = 2[cos(−π
3+ 2pπ
)+ i sin
(−π
3+ 2pπ
)]where p is any integer. De Moivre’s theorem now gives
z = 21/3
[cos
(−π
9+
2pπ
3
)+ i sin
(−π
9+
2pπ
3
)]
14
Now set p equal to 0,1,2 in turn to give the three values for z:
z0 = 21/3[cos(−π
9
)+ i sin
(−π
9
)]z1 = 21/3
[cos
(5π
9
)+ i sin
(5π
9
)]
z2 = 21/3
[cos
(11π
9
)+ i sin
(11π
9
)]z0, z1, z2 are equally spaced around the circle |z| = 21/3.
1.7.2 Deriving multi-angle trigonometric formulae using de Moivre’s theorem
De Moivre’s theorem (and the Binomial expansion) can be used to easily find expressions for
multi-angle trigonometric functions. For any positive integer n, de Moivre’s theorem states
that
(cos θ + i sin θ)n = cosnθ + i sinnθ.
Equating real and imaginary parts, we see that
cosnθ = Re [(cos θ + i sin θ)n] , sinnθ = Im [(cos θ + i sin θ)n] .
Example: Expand cos 2θ and sin 2θ as powers of cos θ and sin θ.
Solution: Since cos 2θ + i sin 2θ = (cos θ + i sin θ)2, we see that cos 2θ and sin 2θ are the
real and imaginary parts of (cos θ + i sin θ)2. Thus we expand (cos θ + i sin θ)2 by the
binomial theorem and equate real and imaginary parts:
(cos θ + i sin θ)2 = cos2 θ + 2i cos θ sin θ + (i sin θ)2
= cos2 θ − sin2 θ + 2i sin θ cos θ.
Hencecos 2θ = cos2 θ − sin2 θ
sin 2θ = 2 sin θ cos θ.
Using cos2 θ + sin2 θ = 1, we can also write cos 2θ as
cos 2θ = 2 cos2 θ − 1
or cos 2θ = 1− 2 sin2 θ.
Example: Expand cos 3θ as a sum of powers of cos θ.
Solution: From de Moivre’s theorem with n = 3 we have
cos 3θ + i sin 3θ = (cos θ + i sin θ)3
15
and so cos 3θ is the real part of (cos θ + i sin θ)3. Again we use the Binomial Theorem to
expand this expression:
(cos θ + i sin θ)3 = (cos θ)3 + 3(cos θ)2(i sin θ) + 3 cos θ(i sin θ)2 + (i sin θ)3
= cos3 θ + 3i sin θ cos2 θ − 3 sin2 θ cos θ − i sin3 θ.
Hence cos 3θ = cos3 θ − 3 sin2 θ cos θ. Since cos2 θ + sin2 θ = 1, we can write cos 3θ as
cos 3θ = cos3 θ − 3 cos θ(1− cos2 θ)
= 4 cos3 θ − 3 cos θ.
Exercise: Show thatsin 3θ = 3 sin θ − 4 sin3 θ
cos 4θ = 8 cos4 θ − 8 cos2 θ + 1
sin 4θ = 4 cos θ(sin θ − 2 sin3 θ).
1.7.3 Finding the mth roots of a complex number
De Moivre’s theorem holds when n is rational (and not just when n is an integer). However, in
this case, cosnθ + i sinnθ is just one of several values that (cos θ + i sin θ)n may have.
Suppose that n = 1/m (where m is an integer), then
(cos θ + i sin θ)1/m = cosθ
m+ i sin
θ
m.
This gives one of the mth roots of the complex number z = cos θ + i sin θ. However, there are
a total of m distinct mth roots of z. We can obtain them all by writing z in the form
z = cos θ + i sin θ = cos(θ + 2πp) + i sin(θ + 2πp)
where p is any integer. De Moivre’s theorem now gives
z1/m = [cos(θ + 2πp) + i sin(θ + 2πp)]1/m
= cos
(θ + 2πp
m
)+ i sin
(θ + 2πp
m
).
Taking p = 0, 1, 2, . . . ,m− 1 gives all the different roots. (Other values of p could be used,
but they will just repeat previous roots.) We can extend this method to any complex number
by writing it in the form
z = r(cos θ + i sin θ), r > 0
(modulus-argument form) or
z = r{cos(θ + 2πp) + i sin(θ + 2πp)}, r > 0
where again p is any integer. Applying de Moivre’s theorem now gives
z1/m = r1/m [cos(θ + 2πp) + i sin(θ + 2πp)]1/m
= r1/m[cos
(θ + 2πp
m
)+ i sin
(θ + 2πp
m
)].
16
Alternatively we may use the exponential notation and then
z = reiθ = rei(θ+2πp),
so that
z1/m = r1/m exp
[i
(θ + 2πp
m
)]= r1/m
[cos
(θ + 2πp
m
)+ i sin
(θ + 2πp
m
)],
where p = 0, 1, 2, . . . ,m− 1.
Example: Find the two square roots of z = 4i.
Solution: In modulus-argument (polar) form,
z = 4[cos(π
2+ 2πp
)+ i sin
(π2
+ 2πp)]
for any integer p. Therefore, using de Moivre’s theorem,
z1/2 = 41/2[cos(π
4+ πp
)+ i sin
(π4
+ πp)], p = 0, 1.
p = 0 ⇒ z1/2 = 2(
cosπ
4+ i sin
π
4
)= 2
(1√2
+i√2
)=√
2 + i√
2
p = 1 ⇒ z1/2 = 2
(cos
5π
4+ i sin
5π
4
)= 2
(− 1√
2− i√
2
)= −√
2− i√
2
Alternatively, we may use the exponential notation:
z = 4ei(π/2+2πp) ⇒ z1/2 = 41/2ei(π/4+πp) = 2ei(π/4+πp), p = 0, 1
and now substitute in the values for p to get
p = 0 ⇒ z1/2 = 2eiπ/4 = 2(
cosπ
4+ i sin
π
4
)=√
2 + i√
2
p = 1 ⇒ z1/2 = 2ei5π/4 = 2
(cos
5π
4+ i sin
5π
4
)= −√
2− i√
2
Example: Find the cube roots of z =√
3 + i and show them on an Argand diagram.
Solution: In polar form
z = 2[cos(π
6+ 2πp
)+ i sin
(π6
+ 2πp)]
⇒ z1/3 = 21/3
[cos
(π
18+
2πp
3
)+ i sin
(π
18+
2πp
3
)], p = 0, 1, 2
p = 0 ⇒ z0 = z1/3 = 21/3(
cosπ
18+ i sin
π
18
)p = 1 ⇒ z1 = z1/3 = 21/3
(cos
13π
18+ i sin
13π
18
)p = 2 ⇒ z2 = z1/3 = 21/3
(cos
25π
18+ i sin
25π
18
)17
Argand Diagram to show the roots
on the circle |z| = 21/3.
z
z
z
0
1
2
Example: Find the four distinct fourth roots of z = −1.
Solution: In exponential form,
z = ei(π+2πp)
⇒ z1/4 = ei(2p+1)π/4
for any integer p. Therefore
p = 0 ⇒ z0 = z1/4 = eiπ/4 = cosπ
4+ i sin
π
4=
1 + i√2
p = 1 ⇒ z1 = z1/4 = ei3π/4 = cos3π
4+ i sin
3π
4=−1 + i√
2
p = 2 ⇒ z2 = z1/4 = ei5π/4 = cos5π
4+ i sin
5π
4=−1− i√
2
p = 3 ⇒ z3 = z1/4 = ei7π/4 = cos7π
4+ i sin
7π
4=
1− i√2
To summarise, there are three steps to carry out to find the mth roots of a complex number:
1. Write the complex number z in general polar form
z = r[cos(θ + 2πp) + i sin(θ + 2πp)], r > 0
or z = rei(θ+2πp)
2. Use de Moivre’s theorem
z1/m = r1/m[cos
(θ + 2πp
m
)+ i sin
(θ + 2πp
m
)]or z1/m = r1/mei(θ+2πp)/m
3. Let p = 0, 1, 2, . . . ,m− 1 to get the m roots.
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1.7.4 Expressing Powers of Trigonometric Functions as Multiple Angle
Functions
It is sometimes useful to be able to express cosn θ and sinn θ, where n is a positive integer, in
terms of cosines and sines of multiples of θ. This is necessary when such expressions have to
be integrated. To achieve this we use the Binomial expansion and the identities
cos θ =1
2(eiθ + e−iθ), sin θ =
1
2i(eiθ − e−iθ)
.
Example: Express cos2 θ and sin2 θ in terms of cos 2θ.
Solution:
cos2 θ = (cos θ)2 =
[1
2(eiθ + e−iθ)
]2=
1
4
[(eiθ)2 + 2eiθ.e−iθ + (e−iθ)2
]=
1
4(e2iθ + 2 + e−2iθ).
But e2iθ + e−2iθ = 2 cos 2θ, and so
cos2 θ =1
2(1 + cos 2θ).
We can then use the relation sin2 θ = 1− cos2 θ to deduce that
sin2 θ =1
2(1− cos 2θ).
Example: Express cos3 θ in multiple angles.
Solution:
cos3 θ = (cos θ)3 =
[1
2(eiθ + e−iθ)
]3
=1
8(e3iθ + 3e2iθ.e−iθ + 3eiθ.e−2iθ + e−3iθ)
=1
8(e3iθ + 3eiθ + 3e−iθ + e−3iθ)
=1
8
[(e3iθ + e−3iθ) + 3(eiθ + e−iθ)
]=
1
8[2 cos 3θ + 6 cos θ]
=1
4(cos 3θ + 3 cos θ).
19
Exercise: Find
1. the fifth roots of −6√
3 + 6i
2. the sixth roots of −1
3. the fifth roots of i
4. the cube roots of 1 + i.
Except for (2), leave the answer in polar form.
Answers
1. 1·64(cosφ+ i sinφ), where φ = π6, 17π
30, 29π
30, 41π
30, 53π
30.
2. ±i,√32± i
2,−√32± i
2.
3. cosφ+ i sinφ, where φ = π10, π2, 9π10, 13π
10, 17π
10.
4. 1·12(cosφ+ i sinφ), where φ = π12, 3π
4, 17π
12.
Exercise: Show that
1. sin3 θ = 14(3 sin θ − sin 3θ)
2. cos4 θ = 18(cos 4θ + 4 cos 2θ + 3)
3. sin4 θ = 18(cos 4θ − 4 cos 2θ + 3)
In all of these examples, the expansion of a power of cosine gives a series involving only
cosines of multiples of the angle. This is also true for an even power of sine, but an odd
power of sine gives a series with sines of multiple angles. This reflects the fact that both
cosn θ and sin2n θ are even functions (i.e. their value is unchanged if θ is replaced by −θ) but
sin2n+1 θ is an odd function (its value changes sign if θ is replaced by −θ).
1.8 Circular and Hyperbolic Functions of Complex Numbers
We have already shown that
cos θ =1
2
(eiθ + e−iθ
)sin θ =
1
2i
(eiθ − e−iθ
)
20
Also the definitions of the basic hyperbolic functions are
coshx =1
2
(ex + e−x
)sinhx =
1
2
(ex − e−x
)If these last two definitions are also taken to apply to complex numbers, then we have
cosh iθ = cos θ
sinh iθ = i sin θ
and also
cos iθ = cosh θ
sin iθ = i sinh θ
The results are true for any complex number θ (not just real and pure-imaginary).
All of the standard trigonometric identities are valid for all complex numbers. For
example, from
sin(θ + φ) = sin θ cosφ+ cos θ sinφ
it follows that
sin(x+ iy) = sinx cos(iy) + cos x sin(iy)
= sin x cosh y + i cosx sinh y
So, if x and y are real, then
Re(sin(x+ iy)) = sinx cosh y
Im(sin(x+ iy)) = cosx sinh y
Example: Find a z for which sin z = 2.
Solution: Let z = x+ iy, then we require
Re(sin(x+ iy)) = sinx cosh y = 2
Im(sin(x+ iy)) = cosx sinh y = 0
and these are satisfied by x = π/2 = 1.57079 and y = cosh−1 2 = 1.31695.
21