THE UNIVERSITY OF NEW SOUTH WALES

SCHOOL OF MATHEMATICS AND STATISTICS

MATH 1131

MATHEMATICS 1A ALGEBRA.

Section 3: - Complex Numbers.

1. The Number Systems.

Let us begin by trying to solve various algebraic equations. Suppose we only know aboutthe set of natural numbers (written as N). Then we can solve the equation x 3 = 2 andobtain the solution x = 5. On the other hand, if we try to solve the equation x+3 = 2 thenthere is no solution! To solve this equation we need a larger set of numbers which includesthe negative whole numbers as well as the positive ones. This set is called the set of integersand is denoted by Z. Continuing this idea:

In Z: x+ 3 = 2 3x = 2x = 1 No solution

In Q: 3x = 2 x2 = 2x = 2

3No solution

In R: x2 = 2 x2 + 1 = 0

x = 2 No solution

N

Z

Q R

C

101

2

23

2

pi

i

2 i

1

At each stage in the above we are able to solve each new type of equation by extending theset of numbers in which we are working. Hence, to solve the equation x2 = 1 we introducea new symbol i (much as we introduced the symbol

2 to solve x2 = 2.) We define i to be

the (complex) number whose square is 1. i.e. i2 = 1. Using this new symbol, we can nowsolve x2 = 1 to obtain solutions x = i. Furthermore, we can define the complex numbersby:

Definition: The set of all numbers of the form a + bi where a, b are real numbers andi2 = 1 is called the set of all complex numbers and denoted by C.

Summary of Basic Rules and Notation:

Let z = a+ ib, w = c+ id be complex numbers. Then

(i) z w = (a c) + i(b d)

(ii) zw = (ac bd) + i(ad + bc).

(iii) zw= a+ib

c+id cid

cid =(ac+bd)+i(bcad)

c2+d2

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

Example: z = 2 4i, w = 3 + i, find z + w, z w, zw, zw.

Ex: Simplify (1 + i)8.

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Equality:

Two complex numbers are equal iff they have the same real and imaginary parts, i.e. ifz = a+ bi = w = c+ di then we can conclude that a = c and b = d.

Proof:

Roots of Unity:

A complex number 6= 1 is called an n-th root of unity if n = 1.

For example, if 3 = 1, and 6= 1 then we can write,

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Polynomial Equations:

We can now solve ALL quadratic equations.

Ex: Solve 5x2 4x+ 1 = 0 and z2 3z + (3 + i) = 0.

Note also that we can find new solutions to old equations such as x3 1 = 0.

Both of these are examples of the following remarkable theorem:

Theorem. (Fundamental theorem of algebra, FTA)

Suppose p(x) = anxn + an1x

n1 + + a1x + a0 is a polynomial, whose co-efficientsan, , a1, a0 are all real (or complex) numbers, then the equation p(x) = 0 has at leastone root in the complex numbers.

Corollary: The equation p(x) = 0 has exactly n (complex) solutions in the complex num-bers (counting multiplicity).

(The last proviso counting multiplicity refers to polynomials which may, for example, havefactors such as (x 2)4 in which case the root x = 2 is counted four times.)

The above result tells us (among other things) that we do not need to find any largerset of numbers if we want to solve polynomial equations. The complex numbers contain allthe roots of every polynomial.

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Square Roots:

Solve z2 = 3 + 4i.

Conjugates: When solving a quadratic equation (with real roots) over the complex num-bers, you will have observed that the solutions occur in pairs, in the form a+ bi and a bi.There are called conjugate pairs. We say that abi is the conjugate of a+bi (and vise-versa).To represent this, we use the notation z = a + bi, z = a bi. This conjugate operation hasthe following properties:

i. z w = z w

ii. zw = z.w

iii.(zw

)= z

w.

iv. z = z if and only if z is real.

v. z + z = 2Re(z).

You will prove these and similar results in the tutorial exercises.

Also note that repeated application of (ii) gives (a+ bi)n = (a + bi)n.

The Argand Plane:

Complex numbers can be represented using the Argand plane, which consists of Cartesianaxes similar to that which you used to represent points in the plane. The horizontal axis isused to represent the real part and the vertical axis, (sometimes called the imaginary axis),is used to represent the imaginary part. For example, the following points have been plotted:3, 2i,3 + 2i, 4 + i.

5

3

2i3 + 2i

4 + i

b

bb

b

Complex numbers then are 2-dimensional, in that we require two axes to represent them.

Observe that a complex number z and its conjugate are simply reflections of each otherin the real axis.

Inequalities.

We lose the notion of comparison in the complex plane. That is, we cannot say whether onecomplex number is greater or lesser than another.

Polar Form:

You have already seen that complex numbers can be expressed in Cartesian Form, a +ib, a, b R. We can also specify a complex number z by specifing the distance of z from theorigin and the angle it makes with the positive real axis.

This distance is called the modulus and written as |z| while the angle is called the argu-ment and written as Arg(z). We insist, to remove ambiguity, that pi < Arg(z) pi.

Pythagoras theorem gives:

If z = a + bi then |z| =a2 + b2.

Care must be taken to find the correct argument. It is easiest to find the related angle suchthat tan =

ba

and then use this to find the argument in the correct quadrant recalling

that we use negative angles in the third and fourth quadrant.

Ex: Find the modulus and argument of z = 1 + i3 and w = 1 2i

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Properties of Modulus:

The modulus function has the following properties:

(i) |zw| = |z||w|

(ii)z

w

=

|z||w| , provided w 6= 0.

(iii) |zn| = |z|n

(iv) |z| = 0 z = 0.

Example: (Sums of squares).

We can write the integer 5 as 22 + 12 = |2 + i|2. We can also write the integer 13 as22 + 32 = |2 + 3i|2. Hence

Try doing the same for 17 and 29.

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Properties of the Argument:

We can distinguish between the principal argument of z, written Arg(z), which is uniquelydefined and takes values between pi and pi (excluding pi), and the more general argument,written arg(z), which is a set of values. We have Arg(z) = arg(z) mod 2pi, which meansthat we can recover Arg(z) from arg(z) by adding or subtracting the appropriate multipleof 2pi.The Argument function has the following properties:

(i) Arg(zw) = Arg(z)+Arg(w) mod 2pi.

(ii) Arg(z/w) = Arg(z)Arg(w) mod 2pi.

Polar Form:

r sin

r cos

z = (r(cos + i sin )

rb

b

b

From the diagram, we can see that the complex number z can be written in the formz = r(cos + i sin ), where r is the modulus of z and is the argument of z. For example,the complex number 1 i can be written as

1 i =2(cos(pi

4) + i sin(pi

4)) =

2(cos

pi

4 i sin pi

4).

This is sometimes called the polar form of z. You will need to be able to convert a com-plex number from cartesian form, (a + bi), into polar form and vice-versa. For example,3(cos pi

3+ i sin pi

3) =

32+ i3

2.

You may have seen the abbreviation cis to represent cos + i sin . You should not usethat here, since your tutor may not know what it is. This form is NOT generally used inbooks beyond High School. Moreover, as we shall see, this polar form, is really a steppingstone to a much better form which involves e. One important fact about the polar form isa remarkable result called:

De Moivres Theorem:

For any real number , and any integer n, we have

(cos + i sin )n = cosn + i sinn.

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The proof of this, looking at the various cases of n is given in the algebra notes. The methodof proof by induction is used. Note that the result also holds for n rational which we willfind useful later than finding roots.

Let us see how useful this result can be:

Ex: Let z = 1 i3. Find z12.

Let us write de Moivres theorem as follows:

Let f() = cos + i sin , then (f())n = f(n). Also f(0) = 1.

Euler did the following:He supposed we can differentiate the function, treating i just like a real number.

If we momentarily ignore the logical difficulties involved then f () = i(cos + i sin ).Comparing this with d

d(ei) = iei, we can see then that this function seems to have prop-

erties that are very similar to the exponential function ei. We will therefore define:

Definition: ei = cos + i sin (and hence ei = cos i sin ).

This formula is sometimes called Eulers formula. We shall take it to be a definitionof the complex exponential.

Thus, any complex number z can be expressed in the polar form

z = rei

where r is the modulus and the argument of z. For example, z = 1 i =

9

and z = 1

This last formula is quite remarkable since it links together the four fundamental constantsof mathematics. In a very important sense, this is the best way to write complex numbers.We have used the term polar form in two different senses. From now on, when I say polarform, I will (generally) mean this new exponential form. You ought to be able to convert acomplex number from cartesian form to polar form and vice-versa.

Note the following important facts:

(i) The conjugate of the complex number z = ei is given by z = ei.

(ii) ei = ei(+2kpi) where k is an integer.

We can write cos and sin in terms of the complex exponential as follows:

cos =ei + ei

2and sin =

ei ei2i

.

From the polar form, we can deduce the properties of modulus and argument which welisted earlier. Let z = r1e

i1 and w = r2ei2 then zw = r1r2e

i(1+2) from which it followsthat

|zw| = |z||w| and Arg(zw) = Arg(z) + Arg(w) mod 2pi.

Ex: Convert z = 2ei5pi

6 , w = 3eipi

3 to Cartesian form:

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Ex: Evaluate the product (1 + i)(1 i3) in two ways to show that cos pi12

= 1+3

22.

The rules for multiplication of complex numbers in polar form tell us that when wemultiply two complex numbers together, rotation and stretching are involved. In particular,since i = ei

pi

2 , multiplying a complex number by i has the effect of rotating z anti-clockwiseabout the origin, through an angle of 90.

iz

zb

b

Ex: Find the complex number obtained by rotating (4 + 2i) anti-clockwise about theorigin through pi

2.

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More generally, to rotate complex number anticlockwise around 0 through an angle , wemultiply it by ei.

Ex: Rotate 3 i anticlockwise about 0 through an angle of pi4.

The Triangle Inequality:

The modulus operation has a number of other useful properties, but two very importantones are:

i. zz = |z|2 and

ii. (The Triangle inequality), |z1 + z2| |z1|+ |z2|.

Powers and Roots of Complex Numbers:

Ex: Find (13i)10.

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To find roots of complex numbers, we will use the polar form. Note that to find the nthroot of a complex number , we are really solving zn = and so we will convert into polarform. Such an equation will have n solutions! (by the Fundamental Theorem of Algebra.)To get all of these solutions we express in polar form, using the general argument, not theprincipal one. An example will make this clear.

Ex: Find the 7th roots of 1.

If we plot these complex numbers we see that they lie on a circle radius 1 and are equallyspaced around that circle.

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z1

z2

z3

z4

z5

z6

z7

b

b

b

b

b

b

b

Ex: Find the 5th roots of 4(1 i).

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Applications to Trigonometry:

Eulers formula gives a dramatic relationship between the exponential and trigonometricfunctions. We can exploit this to deduce useful relationships and identities in trigonometry.

Ex: Find an expression for cos 5 in terms of sines and cosines.

Observe that one can also easily obtain corresponding formula for sin 5 by taking the imag-inary parts of both sides.

Web Activity: Use Google to find some information about the Chebychev Polynomials(there are various spellings of Chebychev) and see how they are related to the above example.

Ex: Taking the problem the other way round, express sin5 in terms of sine and cosinesof multiples of .

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Such a formula is extremely useful in integration, where one might, for example, wish tointegrate sin5 .

Regions in the Complex Plane:

In this section we see how to represent regions in the argand plane algebraically. For exam-ple, the set A = {z C : |z| 2} represents the set of points whose distance from the originis less or equal to 2. (N.B. |z | measures the distance between z and .) Hence this setrepresents a disc radius 2 centre the origin.

2b

Similarly, the set B = {z C : |z+1| < 2} represents the open disc centre (1, 0) radius2.

1 13b bb

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The set C = {x C : 0 Arg(z) pi3} represents a wedge vertex at the origin, and arms

separated by an angle of 60.

o

Note that the origin in NOT included since the argument of 0 is not defined.

Similarly the set D = {z C : 0 Arg(z i) pi6} represents a wedge centre the

point i as shown.

1 o

Here are some further examples:

Ex: Sketch {z C : |z i+ 1| < 2} {z C : Re(z) 0}

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Ex: Sketch {z C : |z 3| < 2} {z C : Im(z 3i) > 0}

More on Polynomials:

The fundamental theorem of algebra, mentioned above, tells us that in the complex plane,all polynomials have all their roots. This is a very powerful theoretical tool, but it does notexplicitly tell us how to find these roots for a given polynomial. Moreover, if we know theroots then we also know how to factor the polynomial. You will need to recall a number ofbasic facts about polynomials from High School, which are:

Remainder Theorem: If p(x) is a polynomial then the remainder r when p(x) is di-vided by x is given by p().

Factor Theorem: If p() = 0 then (x ) is a factor of p(x).

It is important to look at what the underlying set is when we are factoring, for exam-ple, x2 2 does NOT factor over the rational numbers, but it does over the real numbers,(x +

2)(x 2). Similarly, x2 + 1 does not factor over the real numbers but does over

the complex numbers. From the fundamental theorem of algebra, it is clear that over thecomplex numbers, all polynomials completely factor (at least in theory) into linear factors.

Theorem: Every polynomial (with real or complex co-efficients) of degree n 1 has afactorisation into linear factors of the form:

p(z) = a(z 1)(z 2) (z n)

where 1, 2, , n are the (complex) roots of p(z).

This result, still does not tell us how to factor. Nor does it tell us much about factor-ing over the real and rational numbers. For example, does the polynomial x4 +4 factor overthe real numbers or rational numbers?

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The key to factoring over the real numbers, is to firstly factor over the complex numberssince in the complex plane the polynomial falls to pieces into linear factors.

Ex: Factor x4 + 1 over the complex numbers and hence over the real numbers.

Ex: Factor x6 + 8 over the complex and real numbers.

Note that if the co-efficients of the polynomial are real, then the roots occur in conjugatepairs.

Theorem: Suppose p(x) is a polynomial with real co-efficients, then if is a complex(non-real) root, then so is .

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Proof:

From this is follows that:

Theorem: A polynomial with real co-efficients can be factored into a product of real linearand/or real quadratic factors.

Proof: Factor p(x) over the complex numbers in the form

p(x) = a(x b1) (x br)(x 1)(x 1) (x s)(x s)

where the bis are real and the is are complex (non-real). By the above theorem, thesemust occur in conjugate pairs. Now each such pair of factors containing the conjugate pairs,can be expanded, viz:

(x )(x ) = (x2 ( + )x+ ).

Now + = 2Re() and so is REAL, and also = ||2 is also REAL. Hence the quadraticwe obtain has REAL co-efficients.

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Ex: Show that z = i is a root of p(z) = z4 2z3 + 6z2 2z + 5 = 0 and hence factor p overR and C.

The story over the rational numbers is much more complicated. It is possible to have polyno-mials of arbitrary degree which cannot be factored over the rational numbers. For example,if p is a prime, then xp1 + xp2 + ....+ x+ 1 cannot be factored over the rationals.

Moreover, there is no simple test to tell whether a given polynomial can be factored over therationals. (More on this in MATH2400 and Higher Algebra in 3rd year).

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