Download 2.3 Complex Numbers - Defect Dynamics Group

Mathematics for Materials and Earth Sciences. Hilary Term. J. H. Woodhouse
2.3
Complex Numbers
It is well known that the equation x2 + 1 = 0 has no solutions, because the square of any number is always
positive. The basic new idea in this section is to introduce an abstract entity, the so called imaginary number
‘i’ that acts in much the same way as an ordinary number, but has the special property that i2 = 1. We
introduce a generalization of the concept of number to expressions of the form z = x + iy, where x and y are
ordinary numbers. The numbers z = x + iy are called complex numbers. x is called the real part of z, and y is
called the imaginary part of z. We write x = Re z, y = Im z. If the imaginary part vanishes the number is said
to be real – these are the ordinary numbers – and if the real part vanishes the number is said to be imaginary.
The remarkable developments that arise from this simple, if abstract, idea permeate the whole of mathematics. All the familiar properties of ordinary real numbers have their counterparts in the domain of
complex numbers. The generalization unifies and in many ways makes simpler many mathematical ideas.
For example, it turns out that in the complex domain a polynomial equation of degree n possesses precisely
n solutions; in the real domain the situation is more complicated, as polynomial equations have variable
numbers of solutions, up to a maximum of n, depending upon their coefficients (e.g. a quadratic can have
0 or 2 solutions, a cubic can have 1 or 3 solutions, a quartic can have 0, 2 or 4 solutions, etc.). A further
example of the unification and simplification of mathematics that occurs in the complex domain, as we shall
see, is that trigonometric functions and exponential functions, when viewed in the complex domain, are
found to be aspects the same underlying functions.
By virtue of their mathematical properties, complex numbers have many practical applications. For example, they are essential (or at least make the task much simpler) for analyzing wave motion and for studying
oscillatory systems, such as oscillating electrical circuits. They are central to the theory of Fourier analysis,
which is a technique for separating signals into components at different frequencies. The most fundamental practical application of complex numbers is in quantum mechanics, as at the heart of this theory is the
quantum wave function, which is a complex function of the dynamical variables of the system. Thus physical
reality itself is described in terms of complex numbers.
2.3.1
Arithmetic using complex numbers
We define addition and multiplication according to the usual rules of arithmetic. If z1 and z2 are two complex numbers, z1 = x1 + iy1 , z2 = x2 + iy2 , then their sum and product can be expressed as other complex
numbers:
z1 + z2 = (x1 + iy1 ) + (x2 + iy2 )
= (x1 + x2 ) + i(y1 + y2 )
= (x1 + iy1 )(x2 + iy2 ) = x1 x2 + ix1 y2 + iy1 x2 + i2 y1 y2
z1 z2
= (x1 x2
y1 y2 ) + i(x1 y2 + y1 x2 ).
In the second line we have used the fact that i = 1. Therefore sums and products of complex numbers
are themselves complex numbers, i.e they are expressible in the form x + iy, where x and y are ordinary
numbers. What about division? We can write
2
z1
z2
x1 + iy1
(x1 + iy1 )(x2 iy2 )
(x1 x2 + y1 y2 ) + i( x1 y2 + y1 x2 )
=
=
x2 + iy2
(x2 + iy2 )(x2 iy2 )
x22 + y22
✓
◆
✓
◆
x1 x2 + y1 y2
x1 y2 + y1 x2
=
+i
2
2
x2 + y2
x22 + y22
=
In the first line the numerator and the denominator have been multiplied by x2 iy2 in order to convert the
denominator into a real number, enabling us to obtain the result in the form of a complex number, i.e. in the
form x + iy where x and y are real numbers. To make sure that this makes sense, we should check that if we
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multiply the result by z2 we get z1 . The rather unpleasant looking calculation is:
✓
◆
✓
◆
x1 x2 + y1 y2
x1 y2 + y1 x2
(x2 + iy2 )
+i
x22 + y22
x22 + y22
x2 (x1 x2 + y1 y2 ) y2 ( x1 y2 + y1 x2 )
x2 ( x1 y2 + y1 x2 ) + y2 (x1 x2 + y1 y2 )
=
+i
x22 + y22
x22 + y22
2
2
⇠2
x1 x22 + ⇠
x2⇠
y1⇠
y2 + x1 y22 ⇠
y2 ⇠
y1 x
x2⇠
x1⇠
y⇠
y2 ⇠
x1⇠
x⇠
2 + y1 x2 + ⇠
2 + y 1 y2
=
+i ⇠
2
2
2
2
x2 + y2
x2 + y2
x1 (x22 + y22 )
y1 (x22 + y22 )
=
+i
= x1 + iy1 X
x22 + y22
x22 + y22
1
x iy
= 2
, and it is easy to verify that
x + iy
x + y2
In particular, for the reciprocal of a complex number, we have:
(x + iy) ⇥
x iy
= 1.
x2 + y 2
The foregoing has set up the basic arithmetic operations for complex numbers, which work in exactly the
same way as for ordinary numbers, provided only that we remember the trick for converting complex numbers
in the denominator into real numbers (if x + iy occurs in the denominator, multiply top and bottom by
x iy), and provided we replace i2 by 1 whenever it appears.
1
Example 2.3.1. Show that = i3 = i 5 = i.
i
1
i
i
= 2 =
= i.
i
i
1
i3 = i2 i = i.
1
1
1
1
i2
i 5= 5 = 2 2 =
= =
=
i
i i i
( 1)( 1) i
i
i
i.
Example 2.3.2. Solve the following equation for z: (2 + 3i) z + 2
4i = 5 + 6i.
We can solve it just as we would an equation involving ordinary real numbers.
z=
(5 + 6i) (2
2 + 3i
Check:
4i)
(2 + 3i)( 36
+
13
=
11
13
9i + 20i 30i2
36 + 11i
=
=
22 + 32
13
(3 + 10i)(2 3i)
3 + 10i
6
=
=
2 + 3i
(2 + 3i)(2 3i)
72
13
i) =
33
13
+
108
13
Example 2.3.3. Solve the quadratic equation 4z 2
+
22
13
i=
39
13
+
130
13
36
13
+
11
13
i
i = 3 + 10 i X
8z + 13 = 0 for complex z.
Using the usual formula for the solution of a quadratic:
p
p
8 ± 64 208
8±
144
8 ± 12 i
z=
=
=
= 1 ± 32 i
8
8
8
Check:
4(1 +
4(1
3
2
3
2
i)2
8(1 +
i)2
8(1
3
2
3
2
12 i + 13 = 0 X
i) + 13 = 4 + 12 i
9
8
i) + 13 = 4
9
8 + 12 i + 13 = 0 X
12 i
1
as a sum of partial fractions.
4z 2 8z + 13
From Example 2.3.3, the quadratic form 4z 2 8z + 13 has factors (z
Example 2.3.4. Express
in the case of real numbers. Thus we have 4z 2
Check:
(2z
2
3 i)(2z
2 + 3 i) = 4z 2
8z + 13 = (2z
4z + 6iz
4z + 4
2
6i
⇢
z1 ), (z
3 i)(2z
z2 ), where z1 = 1 ±
2
3
2
i, just as
2 + 3 i).
6iz + ⇢
6i + 9 = 4z 2
8z + 13. X
By the usual method (multiplying up by the denominator and equating the coefficients of the various powers of
z) the partial fraction expansion can be found to be: (Exercise)
1
1
i
i
1
1
6
6
=
=
+
4z 2 8z + 13
(2z 2 3 i)(2z 2 + 3 i)
2z 2 3 i
2z 2 + 3 i
Check:
1
6
2z
2
i
3i
+
1
6
2z
i
=
2 + 3i
1
6
i( ⇢
2⇢
z + 2 3i + ⇢
2⇢
z 2 3i)
=
(2z 2 3 i)(2z 2 + 3 i)
(2z
21
2
1
3 i)(2z
2 + 3 i)
.
X
Many calculators and many computer software systems can carry out calculations using complex numbers.
Of course, the computer does not do this because it somehow knows a value for i. Rather, it treats a complex
number as a pair, [x, y], of real numbers representing the real and imaginary parts. It then implements rules
for adding, multiplying and dividing complex numbers, that can be represented:
[x1 , y1 ] + [x2 , y2 ] :!
[x1 + x2 , y1 + y2 ]
[x1 , y1 ] ⇥ [x2 , y2 ] :!
[x1 x2
h
[x1 , y1 ] ÷ [x2 , y2 ] :!
y1 y2 , x1 y2 + y1 x2 ]
i
x1 x2 +y1 y2
,
x22 +y22
x1 y2 +y1 x2
x22 +y22
Real numbers are represented by pairs having the second element of the pair equal to zero, and as a special
case of the above rules, they satisfy the usual rules for addition, multiplication and division. The number
i is represented in memory as the pair [0, 1] and then i ⇥ i :! [ 1, 0] (i.e. i2 = 1) by virtue of the
multiplication rule above. In this scheme there is no need to regard i as an abstract oddity; it is simply the
pair of numbers [0, 1] stored in computer memory.
The mathematical theory of complex numbers can be developed in this way also, in which case i is just a
shorthand for the pair [0, 1]. Thought of in this way, equations such as those solved in Examples 2.3.2 and
2.3.3 can be thought of as pairs of equations, for the determination of the pairs of numbers that represent the
solution, i.e. as two equations for two unknowns. It is rarely a good idea, however to try to solve equations
by explicitly seeking a solution z = x + iy, where x and y are ordinary numbers. It is usually much easier to
manipulate the equation for z in the same way that we would for equations involving ordinary numbers.
2.3.2
Geometrical properties of complex numbers in the Argand plane
The modulus and argument of a complex number and the complex conjugate.
The real and imaginary parts of a complex number can be used as
coordinates in the plane, which is called the Argand plane or the Argand
diagram or, simply, the complex plane. The distance of a given complex
number z = x + iy from the origin in the complex plane is called the
modulus of z, and is denoted by |z|, and the angle subtended at the
origin with the Re z axis, is called the argument of z, and is denoted by
arg z.
Figure 2.3.1: The modulus r = |z| and arp
gument ✓ = arg z are polar coordinates in
The modulus, |z| = x2 + y 2 , is a real, positive quantity, and for real the Argand plane. x = Re z = r cos ✓,
numbers the notation |z| agrees with the notation for the absolute value. y = Im z = r sin ✓.
If r = |z| and ✓ = arg z then z = r(cos ✓ + i sin ✓). This is known as the
modulus-argument form or the polar form of the complex number. Any
complex number can be written either in terms of its real and imaginary parts x and y, say: z = x + i y or in terms of its modulus and
argument, r and ✓, say: z = r(cos ✓ + i sin ✓). The real and imaginary
parts x and y are Cartesian coordinates in the Argand plane and r and
✓ are polar coordinates in the Argand plane.
The relations between (x, y) and (r, ✓) are:
x = r cos ✓, y = r sin ✓
that is:
Re z = |z| cos(arg z), Im z = |z| sin(arg z).
The equations x = r cos ✓, y = r sin ✓ (with r > 0) determine ✓ only
up to an integer multiple of 2⇡, which means that if ✓ is an argument
of z, then so is ✓ + 2⇡n for any (positive or negative) integer n. This is
because cos(✓ + 2⇡n) = cos ✓ and sin(✓ + 2⇡n) = sin ✓ for any integer
n. Therefore, arg z is intrinsically multivalued. In terms of the Argand
22
Figure 2.3.2: The argument has an infinite
number of values, obtained by adding 2⇡n
to the angle ✓, where n is any (positive
or negative) integer. In the Argand plane
this corresponds to adding (anticloclockise
or clockise) circuits of the origin.
diagram, this is because we can arrive at the line joining z to the origin by adding complete 2⇡ cycles to ✓
(see Fig. 2.3.2). Therefore, if we want to enumerate all the arguments of z, i.e. all the angles ✓ such that
z = |z|(cos ✓ + i sin ✓) we have:
8
>
tan 1 (y/x) + 2⇡n
>
>
>
>
< tan 1 (y/x) ± ⇡ + 2⇡n
arg z =
>
⇡/2 + 2⇡n
>
>
>
>
:
⇡/2 + 2⇡n
if Re z > 0
if Re z < 0
if Re z = 0 and Im z > 0
if Re z = 0 and Im z < 0
where n is any integer. The complications here are to ensure that arg z is in the correct quadrant, which means
that we need to add or subtract ⇡ (it does not matter which) if z lies in the second or third quadrants of the
Argand diagram; tan 1 y/x is taken to be a number between ⇡/2 and ⇡/2, such as would be obtained
using a calculator. The best way to get the quadrant right in applications is to make a sketch of the Argand
diagram. The principal argument or the principal value of the argument of z can be defined as that argument
of z that is in the interval ( ⇡, ⇡] (i.e. ⇡ < ✓ 6 ⇡); however, in most cases in which we need to consider
the argument of a complex number, we need to take it into account that the argument is multivalued. Some
treatments introduce a special notation Arg z for the multivalued variety of arg, and use arg z for the principal value. Others use the opposite convention. Here, we use the single notation, arg z, assuming that it
represents the multivalued version unless stated otherwise.
The complex conjugate of a complex number z = x + iy (as always, x and y being real) is defined to be
z ⇤ = x iy. In the Argand diagram, complex conjugation corresponds to reflection in the real axis, e.g.
2⇤ = 2, (2 + 3i)⇤ = 2 3i, (2 3i)⇤ = 2 + 3i, i⇤ = i. The following relations can easily be verified:
Exercise
z z ⇤ = z ⇤ z = |z|2
(z1 + z2 )⇤ = z1⇤ + z2⇤
(z1 z2 )⇤ = z1⇤ z2⇤
Re z = 12 (z + z ⇤ )
Im z =
1
2i (z
|z ⇤ | = |z|
arg z ⇤ =
z⇤)
(z ⇤ )⇤ = z
arg z
As a result of these rules, the complex conjugate of any expression involving complex numbers can be
2 3i
2 + 3i
evaluated by everywhere replacing i by i. E.g. if z = (5 + 6i)2
, then z ⇤ = (5 6i)2
. If an
1 + 4i
1 4i
expression involves complex numbers z1 , z2 , say, then its complex conjugate can be obtained by everywhere
2 3i
2 + 3i
replacing z1 and z2 by their complex conjugates; e.g. if z = z15
, then z ⇤ = z1⇤5
.
z2 + 4iz1
z2⇤ 4iz1⇤
Example 2.3.5. For any complex numbers z1 , z2 , show that |z1
|z1
z2 | 2
=
=
(z1⇤
z2⇤ )(z1
2
|z1 | + |z2 |
2
z2 ) = z1⇤ z1
z1⇤ z2
z2 |2 = |z1 |2 + |z2 |2
z2⇤ z1 + z2⇤ z2 = |z1 |2 + |z2 |2
2 Re(z1⇤ z2 )
2 Re(z1⇤ z2 )
z1⇤ z2
(z1⇤ z2 )⇤
Example 2.3.6. For any complex numbers z1 , z2 , show that |z1 z2 | = |z1 | |z2 | and arg(z1 z2 ) = arg z1 + arg z2 .
We express z1 and z2 in modulus-argument form z1 = r1 (cos ✓1 + i sin ✓1 ), z2 = r2 (cos ✓2 + i sin ✓2 ). Then:
z1 z2
=
r1 r2 (cos ✓1 + i sin ✓1 )(cos ✓2 + i sin ✓2 )
=
r1 r2 [(cos ✓1 cos ✓2
=
r1 r2 [cos(✓1 + ✓2 ) + i sin(✓1 + ✓2 )] .
sin ✓1 sin ✓2 ) + i(sin ✓1 cos ✓2 + cos ✓1 sin ✓2 )]
This expression represents a complex number in modulus-argument form, having modulus r1 r2 and argument
✓1 + ✓2 , and thus |z1 z2 | = r1 r2 and arg(z1 z2 ) = ✓1 + ✓2 , as required.
This is a very important property of complex numbers: when complex numbers are multiplied together,
their moduli get multiplied and their arguments get added.
23
Example 2.3.7. Show that Re(z1⇤ z2 ) = |z1 | |z2 | cos(arg z2
arg z1 ).
In modulus-argument form:
z1⇤ z2
=
r1 r2 (cos ✓1
=
r1 r2 [(cos ✓1 cos ✓2 + sin ✓1 sin ✓2 ) + i(sin ✓2 cos ✓1
Therefore Re (z1⇤ z2 )
i sin ✓1 )(cos ✓1 + i sin ✓2 )
=
r1 r2 (cos ✓1 cos ✓2 + sin ✓1 sin ✓2 )
=
r1 r2 cos(✓2
as required.
cos ✓2 sin ✓1 )]
✓1 ) = |z1 | |z2 | cos(arg z2
.
,
arg z1 )
Putting together the results in Examples 2.3.5 and 2.3.7 we have
|z1 z2 |2 = |z1 |2 + |z2 |2 2 |z1 | |z2 | cos(arg z2 arg z1 ),
which is equivalent to the cosine rule for the triangle in the Argand
plane having vertices 0, z1 , z2 (see figure). Notice that |z2 z1 | is the
distance between z1 and z2 in the complex plane.
Figure 2.3.3: If a = |z1 |, b = |z2 | and c =
|z1 z2 |, the cosine rule states that c2 = a2 +
b2 2ab cos , where is the angle between
the sides of length a and b. = ✓2 ✓1 .
Example 2.3.8.
|z 2| = 1.
Find all complex numbers z satisfying |z
1| = 1 and
If |z
1| = 1, the distance between the point 1 and the point z in
the complex plane is 1, which means that z lies on a circle of radius 1
centred at 1 (see figure). Similarly, since |z 2| = 1, z lies on a circle of
radius 1 centred at 2. The possible values of z lie at the intersections of
the two circles, which are at
p
3
3
z= ±
i. Exercise.
2
2
q
p
Check:
|z 1| = 12 ± 23 i = 14 + 34 = 1.
q
p
1
|z 2| =
± 23 i = 14 + 34 = 1. X
2
2.3.3
de Moivre’s Theorem
Figure 2.3.4: The required values z1 , z2 lie
on circles of radius 1 centred at 1 and 2 in
the Argand plane.
We showed in Example 2.3.6 that if z1 and z2 are any complex numbers,⇥ having polar form z1 = r1 (cos
⇤ ✓1 +
i sin ✓1 ) and z2 = r2 (cos ✓2 + i sin ✓2 ), then their product is z1 z2 = r1 r2 cos(✓1 + ✓2 ) + i sin(✓1 + ✓2 ) . Here
we apply this to obtain expressions for the powers of a complex number.
If z = r(cos ✓ + i sin ✓) then
⇥
⇤
z 2 =z ⇥ z = r (cos ✓ + i sin ✓) ⇥ r (cos ✓ + i sin ✓) = r2 cos(✓ + ✓) + i sin(✓ + ✓)
=r2 (cos 2✓ + i sin 2✓)
⇥
⇤ 3
3
2
2
3
z =z ⇥ z = r (cos 2✓ + i sin 2✓) ⇥ r (cos ✓ + i sin ✓) = r cos(2✓ + ✓) + i sin(2✓ + ✓) =r (cos 3✓ + i sin 3✓)
⇥
⇤
z 4 =z 3 ⇥ z = r3 (cos 3✓ + i sin 3✓) ⇥ r (cos ✓ + i sin ✓) = r4 cos(3✓ + ✓) + i sin(3✓ + ✓) =r4 (cos 4✓ + i sin 4✓)
⇥
⇤
z 5 =z 4 ⇥ z = r4 (cos 4✓ + i sin 4✓) ⇥ r (cos ✓ + i sin ✓) = r5 cos(4✓ + ✓) + i sin(4✓ + ✓) =r5 (cos 5✓ + i sin 5✓)
24
Evidently this can be continued indefinitely, giving the result:
h
in
r (cos ✓ + i sin ✓) = rn (cos n✓ + i sin n✓)
for positive integers n. This takes its simplest form, and does not essentially lose generality, in the special case that r = 1, in which case
we get the result known as de Moivre’s theorem:
de Moivre’s
Theorem
(cos ✓ + i sin ✓)n = cos n✓ + i sin n✓
Because cos ✓ + i sin ✓ = 1, the number w = cos ✓ + i sin ✓ represents a point on the unit circle in the Argand diagram. The powers
w2 , w3 , w4 , ... all have modulus 1 and therefore are represented by
points on the unit circle, and since their arguments are 2✓, 3✓, 4✓, ...,
points around the unit circle having angular spacing ✓ (see Fig 2.3.5).
Figure 2.3.5: The complex numbers wn =
(cos ✓ + i sin ✓)n are equally spaced around
the unit circle.
they correspond to equally spaced
Example 2.3.9. Use de Moivre’s theorem to obtain expressions for sin 3✓ and cos 3✓ in terms of sin ✓ and cos ✓.
We have cos 3✓ + i sin 3✓
=
(cos ✓ + i sin ✓)3
=
cos3 ✓ + 3 cos2 ✓(i sin ✓) + 3 cos ✓(i sin ✓)2 + (i sin ✓)3
=
cos3 ✓ + 3i cos2 ✓ sin ✓
Taking real and imaginary parts: cos 3✓
=
sin 3✓
=
cos3 ✓
3 cos ✓ sin2 ✓
i sin3 ✓
3 cos ✓ sin2 ✓
3 cos2 ✓ sin ✓
sin3 ✓
Here we have used the general formula (a + b) = a + 3a b + 3ab2 + b3 to expand (cos ✓ + i sin ✓)3 . Later in the
term we shall see this as a special case of the Binomial Theorem, which enables us to expand (a + b)n , for any n.
3
2.3.4
3
2
The Euler Relation
Up to now we have seen how to carry out arithmetic and algebra with complex numbers. Now we turn our
attention to determining the meaning of exp z, sin z, cos z for complex z. In order to do this we need to find
arithmetic forms for these functions, so that we can generalize them by simply inserting complex, rather
than real quantities. The necessary forms for these functions are as follows:
exp x = 1 +
sin x = x
cos x = 1
1
1
x + x2 +
1!
2!
1 3
1
x + x5
3!
5!
1 2
1
x + x4
2!
4!
1 3
1
x + x4 +
3!
4!
1 7
1
x + x9
7!
9!
1 6
1
x + x8
6!
8!
1 5
1
x + x6 + · · ·
5!
6!
1 11
x + ···
11!
1 10
x + ···
10!
These are called the power series for exp x, sin x and cos x, and we shall prove them in a couple of weeks
time (week 5), however, at this point we shall take them as given, so that we can use them to define the
corresponding functions for complex z. A homework problem asks you to verify that they give correct
results. When reviewing this section after week 5, such power series representations will be familiar.
We use the power series to define exp z (or ez which is another notation for exp z), sin z, cos z, for complex z
simply be replacing x by z.
exp z
= 1+
sin z
= z
cos z
= 1
1
1
z + z2 +
1!
2!
1 3
1
z + z5
3!
5!
1 2
1
z + z4
2!
4!
1 3
1
z + z4 +
3!
4!
1 7
1
z + z9
7!
9!
1 6
1
z + z8
6!
8!
1 5
1
z + z6 + · · ·
5!
6!
1 11
z + ···
11!
1 10
z + ···
10!
Consider first the case z = i✓ where ✓ is real. We have
25
(2.3.1)
(2.3.2)
(2.3.3)
1
1 2
1 3
1
1
1 6
1 7
1
1
1 10
= 1 + i✓
✓
i ✓ + ✓4 + i ✓5
✓
i ✓ + ✓8 + i ✓9
✓ + ···
1!
2!
3!
4!
5!
6!
7!
8!
9!
10!
✓
◆
✓
◆
1 2
1 4
1 6
1 8
1 3
1 5
1 7
1 9
=
1
✓ + ✓
✓ + ✓ + ··· + i ✓
✓ + ✓
✓ + ✓ + ···
2!
4!
6!
8!
3!
5!
7!
9!
= cos ✓ + i sin ✓.
ei✓
This remarkable result is known as the Euler Relation
Euler
Relation
ei✓ = cos ✓ + i sin ✓
Recalling de Moivre’s Theorem, (cos ✓ + i sin ✓)n = cos n✓ + i sin n✓, we can write
ei✓
n
= ein✓
and since this is a well known property of the exponential function, de Moivre’s theorem is automatic by
virtue of the Euler relation. A general complex number in modulus-argument form (r = |z|, ✓ = arg z), can
now be written
z
= r ei✓ .
The fact that, for two arbitrary complex numbers z1 , z2 , their product satisfies: |z1 z2 | = |z1 | |z2 | and
arg(z1 z2 ) = arg z1 + arg z2 (Example 2.3.6) is also automatic:
r1 ei✓1 r2 ei✓2 = r1 r2 ei(✓1 +✓2 )
With familiarity, we develop the habit that
when we see an expression ei✓ we automatically picture a point on the unit circle in the
Argand plane having argument ✓; e.g.:
ei⇡/4
= cos ⇡/4 + i sin ⇡/4
=
ei⇡/2
i⇡
e
4⇡i/3
e
e
⇡i/4
p1
2
+ i p12 ,
= cos ⇡/2 + i sin ⇡/2 = i,
= cos ⇡ + i sin ⇡ =
1,
= cos 4⇡/3 + i sin 4⇡/3
=
1
2
=
p1
2
p
3
2 ,
i p12 .
i
Notice that if the angle is incremented by an
integer multiple of 2⇡ it results in the same
complex number, e.g.:
e
⇡i/4
2⇡i
e
= e
=e
=
i p12
7⇡i/4
p1
2
4⇡i
= e
15⇡i/4
=e
28⇡i
=e
9⇡i/4
Figure 2.3.6: The unit circle and some illustrative points in the Argand
plane.
.
= 1.
The result ei⇡ = 1 was endowed with almost mystical significance by Euler, as it embodies four key
ideas in mathematics in a single relation: the unit negative number 1, the unit imaginary number i,
the number ⇡ and the number e!
26
2.3.5
The N th roots of a (real or complex) number
Example 2.3.10. Find all complex numbers z satisfying z 3 = 8.
In the domain of real numbers, we know that the only solution of the equation x3 = 8 is x = 81/3 = 2. In this
example we look for complex roots.
Let z have modulus r and argument ✓, i.e. z = r ei✓ , where r and ✓ are real. We need:
r ei✓
3
= 8,
i.e.
r3 e3i✓ = 8.
Taking the modulus, we find r3 = 8 and since, by definition, r is a real, positive number, the only solution for r
is r = 2. Now we need to find ✓ such that 8 e3i✓ = 8, i.e. e3i✓ = 1. We have seen above that e2⇡ni = 1 for any
integer n, and therefore we conclude that the equation will be satisfied if 3✓ = 2⇡n for any integer n. Thus we
find that z = 2 e2⇡ni/3 for any integer n. Are there, therefore, an infinite number of roots? Actually no. Let us
list the possibilities, sat for n = 4 to n = 4:
z 4 = 2 e 8⇡i/3 = 2 e4⇡i/3
z 3 = 2 e 2⇡i
=2
z 2 = 2 e 4⇡i/3 = 2 e2⇡i/3
z 1 = 2 e 2⇡i/3 = 2 e4⇡i/3
z0 = 2 e0
=2
z1 = 2 e2⇡i/3
= 2 e2⇡i/3
4⇡i/3
z2 = 2 e
= 2 e4⇡i/3
2⇡i
z3 = 2 e
=2
z4 = 2 e8⇡i/3
= 2 e2⇡i/3
The second equality on each line is obtained by adding integer multiples of 2⇡i to the exponents, as we know
that this has no effect on the value. We see that there are three distinct values that repeat. The situation is made
much clearer by sketching the roots in the Argand diagram. We see that there are three roots, equally spaced
around the circle of radius 2, and that as n increases zn steps around the circle by (1/3)-cycle, i.e. by an angle
2⇡/3, visiting each root in turn and wrapping around on itself. Thus there are three roots:
2 e2⇡i/3 = 2 (cos 120 + i sin 120 ),
z2 = 2 e4⇡i/3 = 2 (cos 240 + i sin 240 )
2 (⇣ cos 60 +⌘ i sin 60 )
= 2 (⇣ cos 60 ⌘ i sin 60 )
p
p
p
p
3
1
1
= 2
+i 2 = 1+i 3
= 2
i 23 = 1 i 3
2
2
p
p
The equation z 3 = 8 has three solutions, z = 2, z = 1 + i 3 and z = 1 i 3.
p 3
p
p 2
p
p
p
p
Check: ( 1 + i 3)p = ( 1 + p
i 3)( 1 + i 3) = ( 1 + i 3)(1 2i 3 3) = ( 1 + i 3)( 2 2i 3)
= 2( 1 + i 3)( 1 i 3) = 2(1 + 3) = 8. X
z0
=
2,
z1
=
=
This example demonstrates the general principle of taking the N th roots of a number in the complex domain.
The three numbers 1, e2⇡i/3 , e4⇡i/3 are called the cube roots of unity as they each satisfy z 3 = 1. Similarly, the
N th roots of unity, for any positive integer N , are the N numbers e2⇡in/N for n = 0, 1, 2, · · · , N 1. Values
of n outside the range 0 to N 1 repeat the same values. For example the 360th roots of unity are the 360
complex numbers spaced at 1 intervals around the unit circle, starting at z = 1. Notice that it is easy to
verify that z = e2⇡in/N satisfies z N = 1: (e2⇡in/N )N = e2⇡in = 1.
Example 2.3.11. Find the 4th roots of unity, i.e. all solutions of z 4 = 1.
Example 2.3.12. Find all complex numbers z satisfying z 4 =
Exercise.
Answer: 1, i,
1,
i.
1 + i.
p
Let z = r ei✓ . We need r4 e4i✓ = 1 + i. The
modulus
and
argument of the right side
are 2 and tan 1 ( 1) +
p
p
⇡ = 3⇡/4, respectively. i.e. r4 e4i✓ = 2 e3⇡i/4 . Taking the modulus, r4 = 2, and therefore r = 21/8 .
Taking the argument, 4✓ = 3⇡/4 + 2⇡n, where n is any integer. Therefore ✓ = 3⇡/16 + 2⇡n/4, and thus
z = 21/8 e3⇡i/16+2⇡ni/4 . There are 4 distinct values (n = 0, 1, 2, 3), namely: 21/8 e3⇡i/16 , 21/8 e11⇡i/16 , 21/8 e19⇡i/16 ,
21/8 e27⇡i/16 . In the Argand diagram these are equally spaced around the circle of radius 21/8 with spacing ⇡/2,
starting at 21/8 (cos 3⇡/16 + i sin 3⇡/16). Thus they form an inclined square. Using a calculator their values are
approximately: 0.90672 + 0.60585 i,
0.60585 + 0.90672 i,
0.90672 0.60585 i, 0.60585 0.90672 i.
27
Example 2.3.13. Find the square roots of
3 + 4i.
First, put 3 + 4i
3 + 4i = rei✓ , where
p into polarpform:
r = | 3 + 4i| = 9 + 16 = 25 = 5, and ✓ = arg( 3 + 4i) =
tan 1 ( 4/3) + ⇡ = ⇡ tan 1 (4/3) (✓ is in the second quadrant).
The solutions of z 2 = rei✓ are z = r1/2 ei✓/2+2⇡in/2 =
r1/2 ei✓/2 ei⇡n , for any integer n. The factor ei⇡n is equal to ( 1)n ,
which is equal to +1 or 1 for even and odd n, respectively.
1
Thus z = ±r1/2 ei✓/2 = ± 51/2 exp{i( ⇡2
tan 1 43 )} =
2⇤
⇥
1/2
1 4
1 4
⇡
1
⇡
1
±5
cos( 2
tan 3 ) + i sin( 2
tan 3 )
2
2
⇥
⇤
= ±51/2 sin( 12 tan 1 43 ) + i cos( 12 tan 1 43 ) .
This is essentially the answer, as it is now in the form in which
we can evaluate the real and imaginary parts using a calculator.
If you do this you will find that z = ±(1 + 2i). (Can you figure out theoretically how it comes about that the result reduces
to this simple form?). Notice that the N -th order multiplicity of
N th roots reduces for square roots to the usual sign ambiguity.
Check: [±(1 + 2i)]2 = 1 + 4i 4 = 3 + 4i. X
Figure 2.3.7: To determine the square root z1
we take the square root of the modulus and bisect the argument. The other root is z2 = z1 .
↵ = tan 1 (4/3), = (⇡ ↵)/2. z1 = ±51/2 ei .
2
exp z, sin z, cos z, sinh z, cosh z.
2.3.6
In the derivation of the Euler relation in Section 2.3.4 we considered ei✓ for real ✓, showing that ei✓ =
cos ✓ + i sin ✓. However, there was nothing in the derivation requiring ✓ to be real, and exactly the same
relation holds for a general complex number z. Thus we have eiz = cos z + i sin z. The Euler relation is
recovered if z happens to be real. The general relation eiz = cos z + i sin z can be used to express the sine
and cosine functions in terms of exponentials and vice versa. For example, we can write
eiz
e
iz
= cos z + i sin z
= cos( z) + i sin( z) = cos z
i sin z
The fact that sin( z) = sin z and cos( z) = cos z, familiar results for real z, follows from equations (2.3.2)
and (2.3.3). The above equations can be solved for sin z, and cos z, giving the important relations:
cos z
=
sin z
=
1
2
1
2i
eiz + e
iz
e
e
= cosh iz
iz
=
iz
1
i
sinh iz
This explains the, up to now, mysterious fact that sinh and cosh behave so much line sin and cos. Except for
the factors of i, they are the same functions. Replacing z by iz, we also have:
cos iz
=
sin iz
=
1
z
2 (e +
1
z
2 i (e
e
e
z
)
z
= cosh z
) = i sinh z
The most frequently used versions of these relations are those giving sin ✓, cos ✓, for a real angle ✓, in terms
exponentials:
cos ✓
=
sin ✓
=
1
2
1
2i
ei✓ + e
ei✓
e
i✓
i✓
.
Example 2.3.14. Use the complex representations of sin and cos to derive the relation
sin a cos b = 12 [sin(a + b) + sin(a b)].
1
sin a cos b = 2i
eia
⇣
1
= 4i
ei(a+b) + ei(a
=
1
[sin(a
2
e
b)
+ b) + sin(a
ia
e
1
2
eib + e
i(a b)
b)],
ib
1
= 4i
eia eib + eia e ib e ia eib
⌘
1
e i(a+b) = 4i
[2i sin(a + b) + 2i sin(a
as required.
28
e
b)]
ia
e
ib
Example 2.3.15. Obtain an expression for cos4 ✓ in terms of cosines of multiple angles (i.e. in terms of cos n✓ for integer
values of n).
We shall use the identity (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 , another example of the Binomial Theorem,
which comes later, but which also can be obtained by elementary methods.
We have:
⇥
⇤4
⇥ i✓ 4
⇤
1
cos4 ✓ = 12 ei✓ + e i✓
= 16
(e ) + 4(ei✓ )3 e i✓ + 6(ei✓ )2 (e i✓ )2 + 4ei✓ (e i✓ )3 + (e i✓ )4
⇥ 4i✓
⇤
1
= 16
e + 4e3i✓ e i✓ + 6e2i✓ e 2i✓ + 4ei✓ e 3i✓ + e 4i✓
⇥
⇤
1
= 16
e4i✓ + 4e2i✓ + 6 + 4e 2i✓ + e 4i✓
=
1
16
[2 cos 4✓ + 8 cos 2✓ + 6] =
1
8
cos 4✓ +
1
2
cos 2✓ + 38 .
Example 2.3.16. Evaluate sin( 3 + 4 i).
Trigonometric functions satisfy all the same relations in the complex domain as we are used to in the real domain.
Therefore we write
sin( 3 + 4 i)
=
⇡
sin( 3) cos(4 i) + cos( 3) sin(4 i) =
3.85374 27.0168 i.
Alternatively, we could use
⇣
sin( 3 + 4 i) = 21i ei(
1
=
1
2
=
=
⇥2
3+4 i)
ie
4
[cos(3)
e
4
sin(3)
e
i( 3+4 i)
⌘
1
2i
i sin(3)] + 12 i e4
⇤
⇥
e4 sin(3) + 12 i
=
e
sin(3) cosh(4) + i cos(3) sinh(4)
4 3i
[cos(3) + i sin(3)]
e
4
cosh(4) sin(3) + i sinh(4) cos(3)
⇡
3.85374
e4+3 i =
1
2i
⇥
⇤
cos(3) + e4 cos(3)
e
4
e
3i
e4 e3i
⇤
27.0168 i.
[ Note: In the first method we have used the relations cos iz = cosh z and sin iz = i sinh z. The easy way to remember where
the i goes in such relations is to use the fact that for small z sin(something) ⇠ (something), sinh(something) ⇠ (something),
cos(something) ⇠ 1, cosh(something) ⇠ 1. This should be well known to you for real values (think of the graphs of sin x,
cos x, sinh x, cosh x near x = 0). The fact that it remains true for complex values can be seen by looking at the first terms in
the power series for sin z, cos z etc. Obviously this is related to the fact that sin and sinh are odd functions and cos and cosh
are even functions. If you remember this and write, by mistake, sin iz = sinh z or cos iz = i cosh z, you can immediately
see that it is wrong, because it is not valid if you replace cos(something) and cosh(something) by 1 and sin(something) and
sinh(something) by (something). Similarly sinh(iz) = i sin(z) is right, as is sin(z) = i sinh(iz). ]
2.3.7
The logarithm of a complex number:
ln z .
How can we define the logarithm of a complex number? For real numbers we define the logarithm y = ln x
as the solution, y, of the equation ey = x; i.e. ln is the inverse function of exp. It is natural, therefore, to
define the logarithm of a complex number in the same way. For complex numbers w and z we say that
w = ln z means that ew = z. Let u and v be the real and imaginary parts of w, and let r and ✓ be the modulus
and argument of z: w = u + iv, z = r ei✓ . Then
eu+iv = rei✓
i.e.
eu eiv = rei✓ .
From the modulus and argument of this expression, we need
eu = r
and
v = ✓ + 2⇡n,
where n is any integer, and therefore u = ln r = ln |z| and v = arg z + 2⇡n. Provided that we remember that
arg z is multivalued, there is no need to include the 2⇡n term explicitly. Thus we obtain
ln z = ln |z| + i arg z
Notice that the logarithm on the right side is the logarithm of a real positive number, so this is the usual
logarithm. There is a slight notational ambiguity here, as ‘ln’ on the left side is the complex logarithm
that we are defining, and ‘ln’ on the right side is the usual logarithm. We shall adopt the convention that
ln |something| (i.e. ln with vertical bars) represents the usual logarithm of a real positive quantity1 .
1 Humpty Dumpty to Alice “When I use a word, [...] it means just what I choose it to mean – neither more nor less. ‘The question is’, said
Alice, ‘whether you CAN make words mean so many different things’ ” – Lewis Carroll was a mathematician. It’s a very good practice to
scan mathematical expressions asking yourself the meaning of each part of the expression.
29
The fact that arg z is multivalued means that ln z is multivalued. It is possible to define the principal value
of ln z to be that obtained using the principal value of arg, which means that for the principal value of the
logarithm ⇡ < Im(ln z) 6 ⇡. Some treatments introduce the notation Ln z for the multivalued version of
ln and use ln z for the principal value or vice versa. Here we use the single notation ln z assuming that it
represents the multivalued version unless we say otherwise. Let us check that the logarithm of a complex
number possesses the usual property of a logarithm, that the logarithm of a product z1 z2 is equal to the sum
of the logarithms of z1 and z2 :
ln(z1 z2 ) = ln(r1 ei✓1 r2 ei✓2 ) = ln[(r1 r2 )ei(✓1 +✓2 ) ] = ln |r1 r2 | + i(✓1 + ✓2 ) = ln |r1 | + i✓1 + ln |r2 | + i✓2 .
This verifies that
ln(z1 z2 ) = ln z1 + ln z2 .
Notice that this relation is not true in general for the principal value of the logarithm because the fact that
Im(ln z1 ) and Im(ln z2 ) are in the interval ( ⇡, ⇡] does not guarantee that Im(ln z1 ) + Im(ln z2 ) is also in this
interval.
Example 2.3.17. Evaluate ln( 3 + 4i).
As in Example 2.3.13, |
ln( 3 + 4i) = ln |
i.e.
3 + 4i| = 5 and arg( 3 + 4i) = ⇡
3 + 4i| + i arg( 3 + 4i) = ln |5| + i(⇡
ln( 3 + 4i) = ln |5| + i
Example 2.3.18. Evaluate ln(1
Example 2.3.19. Show that tan
Let w = tan
1
tan
i).
1
z=
z, i.e. z = tan w =
Thus, to find tan
1
1 4
3
+ (2n + 1)⇡ .
Exercise
1
ln
2i
✓
sin w
1 eiw e
=
cos w
i eiw + e
z we need to solve the equation
Putting t = e2iw , we have
1
2
iw
iw
tan
1 4
+ 2⇡n.
3
1 4
+
2⇡n)
3
Therefore
As usual, n is any integer.
Answer:
◆
1 + iz
.
1 iz
tan
ln |2| + i(2⇡n
⇡
).
4
1 e2iw 1
.
i e2iw + 1
1
= iz for w.
+1
=
e2iw
e2iw
t 1
= iz.
t+1
1 + iz
It is straightforward to solve this equation for t (Exercise), to obtain: t =
.
1 iz
✓
◆
1 + iz
But t = e2iw and therefore, taking the logarithm, 2iw = ln
.
1 iz
✓
◆
1
1 + iz
Hence w =
ln
, as required.
2i
1 iz
✓
◆
1
1 + ix
Example 2.3.20. Verify the formula tan 1 x =
ln
for real values x.
2i
1 ix
✓
◆
1
1 + ix
1
We have
ln
=
[ln (1 + i x) ln (1 i x)]. Remembering that x is real, (1 + ix) and (1
2i
1 ix
2i
p
be written in polar form (1 + ix) = r ei✓ and (1 ix) = r e i✓ with r = 1 + x2 and ✓ = tan 1 x.
ix) can
✓ is in the first quadrant if x is positive, and in the fourth quadrant if x is negative.
1
1
Thus
[ln (1 + i x) ln (1 i x)] =
ln |r| + i✓ ln |r| + i✓ + 2⇡in = ✓ + ⇡n,
2i
2i
where n is an arbitrary integer.
✓
◆
1
1 + ix
But since ✓ = tan 1 x, we have shown that tan 1 x + ⇡n =
ln
. The additive term ⇡n is the usual
2i
1 ix
1
1
ambiguity in the definition of tan (i.e. tan is a multivalued function, determined only up to an additive
integer ⇥ ⇡), and therefore we have obtained the required result.
30
2.3.8
Complex powers of a complex number: zw
Raising a complex number to a complex power can be defined as
z w = exp(w ln z) = ew ln z
which says “take the logarithm of z, multiply by w and then take the antilogarithm”, which is the usual way
of evaluating powers using logarithms for real numbers.
Of course, the result should make sense for real as well as for complex exponents w, so let’s first test it for
the problem that we have already done in Example 2.3.10:
Example 2.3.21. Find all values of 81/3 .
In the complex domain ln 8 = ln |8| + 2⇡in, for any integer n. (Recall that ‘ln |something|’ means the usual ln).
Thus:
81/3 = e(ln |8|+2⇡in)/3 = e(ln |8|)/3 e2⇡in/3 = |8|1/3 e2⇡in/3 = 2e2⇡in/3
(|8|
means the usual real exponentiation of a real number). This is the same result as we obtained previously,
leading to the three values obtained by setting n = 0, 1, 2, with other values of n giving repeats of the same three
values (if this is unclear, review Example 2.3.10).
1/3
Let’s also redo Example 2.3.12.
Example 2.3.22. Find all values of ( 1 + i)1/4 .
p
ln( 1 + i) = ln | 2| + 3⇡i
+ 2⇡in
4
p
1/4
Therefore ( 1 + i)
= exp{ 14 ln | 2| + 3⇡i
+ 2⇡in } = exp
4
8
1/8
>
|2| cos 3⇡/16 + i |2|1/8 sin 3⇡/16
>
>
>
>
< |2|1/8 sin 3⇡/16 + i |2|1/8 cos 3⇡/16
i.e. ( 1 + i)1/4 =
>
|2|1/8 cos 3⇡/16 i |2|1/8 sin 3⇡/16
>
>
>
>
:
|2|1/8 sin 3⇡/16 i |2|1/8 cos 3⇡/16
1
8
ln |2| exp
3⇡i
16
exp
⇡in
2
= |2|1/8 e3⇡i/16 in .
(n = 0)
(n = 1)
(n = 2)
(n = 3)
where lines 2 to 4 are obtained by multiplying the previous line by i, which represents a ⇡/2 rotation in the
Argand diagram. Clearly, further multiplications by i will repeat the same values. This is the same result as was
obtained in Example 2.3.12.
Now let’s see what happens for a complex exponent.
Example 2.3.23. Find all values of (2
ln(2
3i) =
Therefore
(2
3i)4+5i
1
2
ln |13| + i
tan
3i)4+5i .
1 3
2
1
2
+ 2⇡n
exp (4 + 5i)
=
exp 2 ln |13| + 5 tan
=
exp 2 ln |13| + 5 tan
=
169 exp 5 tan
1 3
2
⇥
⇥
⇤
+ 2⇡n
⇥
⇤
10⇡n + i 4 tan 1 32 + 8⇡n + 52 ln |13|
⇥
⇤
exp i 4 tan 1 32 + 52 ln |13| e 10⇡n e8⇡in
tan
1 3
2
cos( 4 tan
1 3
2
ln |13| + i
=
(Exercise).
1 3
2
1 3
2
+
5
2
ln |13|) + i sin( 4 tan
1 3
2
+
5
2
⇤
ln |13|) e
10⇡n
This is the answer, as it is now in the form that we can evaluate the real and imaginary parts using a calculator,
for specific values of n. Notice that in this case the values do not ‘wrap around’, so there is an infinite number
of possible values as n varies from 1 to 1. So exponentiating with complex exponents can get complicated!
When n = 0 the value turns out to be ⇡ 18175.5 + 14117.6 i and the other values are obtained by multiplying
by e 10⇡n ⇡ (2.27110 ⇥ 10 14 )n for positive and negative integer values of n.
Even raising a real number to a real power can give an unexpected result:
31
p
2
Example 2.3.24. Find all possible values of (1 + 0 i)
in the complex domain.
p
2
ln(1 + 0i) = ln |1| + 2⇡ni = 2⇡ni. Therefore (1 + 0i)
p
2n
= e2⇡i
p
p
= cos(2⇡ 2 n) + i sin(2⇡ 2 n).
For any value of n, the value is a point on the unit circle,
but the values never ‘wrap around’. They could only
p
wrap around if, for some value
p of n, the number 2⇡ 2 n was equal to an integer multiple of 2⇡, 2⇡m, say. But
in this case we would have 2 = m/n for p
two integers m and n, and Pythagoras proved that there were no two
such numbers. Thus the values of (1 + 0i) 2 are infinite in number. They ‘pepper’ the unit circle in such a way
that there is an infinite number of them in any finite segment of the circle! (not proved here).
Note: When we write ez we do not really mean ‘exponentiate the complex number e + 0i to the complex power z’
1
1 2
1 3
in the sense introduced in this section. What we mean is the function exp z = z + 1!
z + 2!
z + + 3!
z · · · . So ez
is actually a shorthand for exp z, and the notation exp z is really more precise. Nevertheless, the notation ez
is universally used, and does not cause any problems, provided that we treat it as only a shorthand notation
for exp z – it means what we want it to mean.
2.3.9
Complex functions of a real variable.
The notion of a function that has both real and imaginary part depending on a real variable x does not
present any particular difficulties. E.g. if f (x) = xe2ix , where x is real, it represents essentially two real
functions g(x) = Re f (x) = x cos 2x and h(x) = Im f (x) = x sin 2x. We can differentiate and integrate f (x)
and obtain valid results for the differentials and the integrals of g(x) and h(x). This is because the definitions
of differentiation and integration carry through directly to the complex domain by virtue of the fact that dx
is real.
Example 2.3.25. Verify that
d
(x e2ix )
dx
Z
x e2ix dx
=
=
d
d
(x cos 2x) + i
(x sin 2x) and
dx
dx
Z
Z
x cos 2x dx + i x sin 2x dx.
We differentiate and integrate complex functions by the usual rules, regarding i as a constant.
Thus
d
(x e2ix )
dx
=
e2ix + x2ie2ix = cos 2x + i sin 2x + 2ix(cos 2x + i sin 2x)
=
(cos 2x
2x sin 2x) + i(sin 2x + 2x cos 2x).
We want to show that the real and imaginary parts of this result are the differentials of the real and imaginary
parts of xe2ix :
We have
d
(x cos 2x)
dx
d
(x sin 2x)
dx
=
cos 2x
2x sin 2x
X
=
sin 2x + 2x cos 2x.
X
Similarly, for integration, in this case using integration by parts:
R
R 1 2ix
1 2ix
x e2ix dx = x 2i
e
e dx = 12 ixe2ix + 14 e2ix + C
2i
=
=
1
ix(cos 2x + i sin 2x) + 14 (cos 2x + i sin 2x) + C
2
( 12 x sin 2x + 14 cos 2x) + i( 12 x cos 2x + 14 sin 2x) +
C.
We want to show that the real and imaginary parts of this result are the integrals of the real and imaginary parts
of xe2ix :
R
R 1
1
We have
x cos 2x dx =
x 12 sin 2x
sin 2x dx =
x sin 2x + 14 cos 2x + C1 X
2
2
R
R
1
1
1
x sin 2x dx =
x 2 cos 2x + 2 cos 2x dx =
x cos 2x + 14 sin 2x + C2 . X
2
Notice that C1 and C2 are real arbitrary constants, which correspond to the real and imaginary parts of the
complex arbitrary constant C.
32
Sometimes forms involving complex numbers give easier ways to evaluate real integrals. See e.g. Problem
Set 12, Q.8. They sometimes give alternative ways. In Example 2.1.31 we did the following integral, which
we can also do using the complex representation of sin and cos:
R
Example 2.3.26. Use the complex representations of the sine and cosine functions to evaluate I = cos x sin 4x dx.
R
R 1 ix
R 1 5ix
1
cos x sin 4x dx =
e + e ix 2i
e4ix e 4ix dx = 4i
e
e 3ix + e3ix e 5ix dx
2⇣
⌘
1
1 5ix
1
1 3ix
1
= 4i
e
e 3ix + 3i
e
e 5ix + C
5i
3i
5i
=
=
1 5ix
1
1 3ix
e
e 5ix 12
e
20
20
1
1
cos
5x
cos
3x
+
C,
10
6
1
e 3ix
12
+C
as in Example 2.1.31.
Another derivation that is basically equivalent but requires somewhat less algebra is:
R
R
R
R 1 5ix
1
cos x sin 4x dx = Re eix sin 4x dx = Re eix 2i
e4ix e 4ix dx = Re 2i
e
⇣
⌘
3ix
1
1 5ix
1
= Re 2i 5i e
e
+C
3i
=
=
Re
1
10
1 5ix
e
10
cos 5x
1
e 3ix
6
1
6
+C
cos 3x + C.
33
e
3ix
dx