[PDF] Chapter 3: Complex Numbers

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Chapter 3: Complex Numbers

Daniel Chan

UNSW

Term 1 2020

Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 1/40

Philosophical discussion about numbers

QIn what sense is1 a number? DISCUSSQIsp1 a number?A from your Kindergarten teacherNot a REAL number.

Why not then a non-real number?After all,

p1 exists as an expression, and as such it pops up all the time when you solve enough equationsEVEN IF you are only interested in REAL numbers (see later).OK. Let's extend our number system by pretending p1 is a number which we'll

denote as usual byi, and see what happens.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 2/40

Thought experiment concerningiWell ifiis a number, then surely so is 3iand 2 + 3i.In fact, for anya;b;c;d2R,a+bi;c+diare numbers too, surely.But then (a+bi) + (c+di) is a number!That's OK, it must be one we've

seen before (a+c) + (b+d)i.But also (a+bi)(c+di) is a number(??).I guess it ought to be (a+bi)(c+di) =ac+bci+adi+bdi2= (acbd) + (bc+ad)isincei2=1.We've seen this number before. QWhen doesa+bi=c+di?AThen (ac)2= (db)2i2=(db)2which occurs precisely whena=c andb=d. (WHY?)Major Question If we keep playing this game blindly, using our usual rules of arithmetic, will we ever end up proving absurd statements like 1 = 0? Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 3/40

Fields

Our new number system should satisfy the \usual rules of arithmetic", and we need to formalise what this means. This uses the followingDenition

Aeldis the data consisting of a non-empty setFtogether withanaddition rule+, which assigns to anyx;y2Fan elementx+y2F.amultiplication rule, which assigns to anyx;y2Fan elementxy2F.such that the axioms on the following page hold.

Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 4/40

Field axioms

1Associative Law of Addition. (x+y) +z=x+ (y+z) for allx;y;z2F.2Commutative Law of Addition.x+y=y+xfor allx;y2F.3Existence of a Zero. There exists an element ofF(usually written as 0 & called

zero) such that 0 +x=x+ 0 =xfor allx2F.4Existence of a Negative. For eachx2F, there exists an elementw2F(usually

written asx& called thenegativeofx) such thatx+w=w+x= 0.5Associative Law of Multiplication.x(yz) = (xy)zfor allx;y;z2F.6Commutative Law of Multiplication.xy=yxfor allx;y2F.7Existence of a One. There exists a non-zero element ofF(usually written as 1 &

called themultiplicative identity) such thatx1 = 1x=xfor allx2F.8Existence of an Inverse for Multiplication. For each non-zerox2F, there exists

an elementwofF(usually written as 1=xorx1& called themultiplicative inverse

ofx) such thatxw=wx= 1.9Distributive Law.x(y+z) =xy+xzfor allx;y;z2F.10Distributive Law. (x+y)z=xz+yz, for allx;y;z2F.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 5/40

Examples

E.g.F=R;Qare elds when endowed with the usual addition and multiplication

of numbers for the addition and multiplication rule.E.g. the eld with 2 elementsLetF=feven;oddg.Dene the addition rule by

even + even = even;even + odd = odd;::::and the multiplication rule by eveneven = even;evenodd = even;::::You can check all eld axioms are satised.

RemarkThis eld is very important in coding theory.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 6/40

What's subtraction and division?

The point of the axioms, is that this is the minimal set of assumptions to ensure

you can do all the usual arithmetic in the usual way.In particular, you can subtract and divide (by non-zero eld elements). To do this

you needFact In a eldF, the zero, negative, one and multiplicative inverse are unique. (What's this mean?)The proof (omitted) is not hard, but many of you might nd it strange. Hence forx;y2Fwe can dene:xy=x+ (y)and ify6= 0,xy =xy1.E.g.Simplify the following expression in a eld x(y+z)yxDaniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 7/40

Complex numbers

Our thought experiment suggests the followingDenition Acomplex numberis a formal expression of the forma+bifor somea;b2R. In particular, two such numbersa+bi;a0+b0iare equal ia=a0;b=b0as real

numbers.Thereal partofa+biisRe(a+bi) =aand theimaginary partisIm(a+bi) =b.Remarks1. Formal means in particular, that the + is just a symbol, it doesn't

mean addition (yet).2. We often writeafora+ 0iandbifor 0 +bi.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 8/40

Arithmetic of complex numbers

Denition

Given complex numbersa+bi;a0+b0ias above, we dene addition and multiplication by(a+bi) + (a0+b0i) = (a+a0) + (b+b0)i

(a+bi)(a0+b0i) = (aa0bb0) + (ab0+a0b)iWarningThere are two clashes of notation. What'sa+bimean?We're OK.

Theorem

The setCof complex numbers with the above addition and multiplication rule is

a eld.Proof.Is long and tedious but elementary. Note zero is 0 + 0i.This means we can perform complex number arithmetic as usual.

N.B.Cextends the real number system since complex numbers of forma+ 0i add and multiply just like real numbers. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 9/40

Examples of complex arithmetic

EgWhat's the negative ofa+bi?Eg(57i)(6 +i)?EgSimplify (2 +i)(13i)1 + 3iDaniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 10/40

Division

To get the inverse we need

Cool Formula

Letz=a+bi2C(witha;b2Rof course). We dene theconjugate ofzto be z=abi.zz=a2+b22R0:This gives the multiplicative inverse ofzas z 1=za

2+b2:This is all we need since we know inverses of real numbers.

Usually though, we divide as follows

E.g. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 11/40

Cartesian form

A complex numberzwritten in the forma+biwitha;b2Ris called the

cartesian form(Later we'll meet the polar form).QExpress1+i1i1i1+iin cartesian form.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 12/40

Properties of conjugation

Proposition

1zis real i (= if and only if)z=z.2z=z.3z+w=z+wandzw=zw.4zw=zwand

zw  =z w .5Re(z) =12

(z+z) andIm(z) =12i(zz).Proof.Easy. Write both sides out e.g.E.g.Show that for anyz2C, (i+ 5)z(i5)zis real.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 13/40

The Argand diagram

Just as real numbers can be represented by points on the real number line, complex numbers can be represented on the complex plane (or Argand diagram) as follows.z=a+biis represented by the point with coords (a;b) = (Rez;Imz).

The axes though are called therealandimaginaryaxes.Adding complex numbers is by adding real and imaginary parts, i.e. coordinatewise

so is represented geometrically by the addition of vectors. Similarly for subtraction. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 14/40

Polar form

Writing a complex number asz=x+yi;x;y2Ris called thecartesian formof

z. It corresponds to rectilinear coordinates.Suppose the polar coordinates forzare given by (r;) as above.z=rcos+ (rsin)i:Denition

Letz=x+iy;x;y2R.1Themodulus ofzis dened to bejzj=r=px

2+y2sozz=jzj2.2Ifz6= 0, anargument forzis any= argzas above i.e. so that tan=yx

and cos;Rezhave the same sign.=:Argzis theprincipalargument if further < .Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 15/40

Examples: modulus and argument

E.g.Find the modulus and principal argument of1p3i.E.g.Find the complex number with modulus 3 and argument=4.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 16/40

Euler's formula

Denition (Euler's formula)

For2R, we deneei= cos+isin.This is reasonable by

Formulas

1e i1ei2=ei(1+2).2(De Moivre's thm) Forn2Z, (ei)n=ein.3d d(ei) =iei.Proof.2) & 3) easy omitted. We only check 1).(cos1+isin1)(cos2+isin2) = cos1cos2sin1sin2+i(cos1sin2+ sin1cos2)

= cos(1+2) +isin(1+2):Challenge QWhat'sii?Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 17/40

Arithmetic of polar forms

Thepolar formofzisz=reiwherer=jzjandis an argument ofz.Our formulas above give r

1ei1r2ei2= (r1r2)ei(1+2);(rei)1=r1ei:Geometrically, this says that when you multiply complex numbers, youmultiply

the moduliandadd the arguments.Inverting inverts the modulus and negates the argument.jz1z2j=jz1jjz2j jz1j=jzj1

Arg(z1z2) =Arg(z1) +Arg(z2) + 2k;Argz1=Argzunlesswherek2Zis chosen so thatE.g.Find the exact value ofArg1+i1+

p3i.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 18/40

Geometry via complex numbers

QLetz2Chavejzj= 1. Show thatw=izi+zispurely imaginaryin the sense

thatRew= 0. Interpret the result geometrically.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 19/40

Square roots of complex numbers

E.g.Find the complex square rootszof 1630iDaniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 20/40

Quadratic formula

E.g.Solvez2+ (1 +i)z+ (4 + 8i) = 0.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 21/40

Cubic formula

In the 16th century Ferro, Tartaglia, Cardano,..., discovered how to solve cubics.

Formula

z

3+pz=qhas solutions

z=3sq 2 +rq 24
+p327 +3sq 2 rq 24
+p327

:QLet's use this to solvez3z= 0 (which we know has solns ???)Bizarre factIf there are 3 real roots, then the formula above ALWAYS involves

non-real numbers.Moral to this storyEven if you only ever cared about real numbers, complex

numbers naturally arise.Even more bizarre factThere's a similar formula for quartics, but can prove no

such exists for higher degree (see Galois theory course MATH5725). Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 22/40

Proof of the cubic formula

Recall theBinomial Thm(a+b)n=Pn

k=0 n k
akbnkwhere n k
=n!k!(nk)!We use Vieta's substitutionx=wp3wq= w

33w2p3w+ 3wp29w2p327w3

+p(wp3w) =w3p327w3This is equivalent to the quadratic inw3

0 =w6qw3p327

which has roots w 3=12 qrq

2+4p327

!

:Substituting back intox=wp3wgives the formula.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 23/40

Powers of complex numbers

Polar form allows us to ndn-th powers andn-th roots of a complex number.E.g.Findw= (1 +i)18.[Dumb way: multiply out (1 +i)(1 +i):::(1 +i)]Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 24/40

Roots of complex numbers

More interestingly, polar forms allow easy computation of roots.E.g.Solvez4=i.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 25/40

Expressing trigonometric polynomials as polynomials in cos;sinAtrigonometric polynomialis a linear combination of functions of the form cosn;sinn.Example:Use De Moivre's thm to show cos(3) = 4 cos33 cos. cos3=Reei3
=Re(cos+isin)3 =Recos3+ 3icos2sin3cossin2isin3
= cos

33cossin2

= cos

33cos(1cos2)

= 4 cos

33 cos:Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 26/40

cos;sinin terms of exponentialsSince e i= cos+isin e i= cos() +isin() = cosisin =e iwe have cos=ei+ei2 ;andisin=eiei2 :Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 27/40

Expressing cos

n;sinnas trig polynomialsE.g.Prove that sin4=18 cos412 cos2+38 .sin

4=eiei2i

4 = ei44ei2+ 64ei2+ei416 = ei4+ei416 4ei2+ei216 +616
= 18 cos412 cos2+38 :ThusZ sin

4d=Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 28/40

Trigonometric sums

QFind  = cos+ cos2+:::+ cosn.AConsider the sum of a geometric progression

S:=ei+ei2+:::+ein=ei(n+1)eie

i1:Then  =ReS=S+S 2 =12  ei(n+1)eie i1+ei(n+1)eie i1 =


OR note

S=eiein1e

i1=eiein=2(ein=2ein=2)e

i=2(ei=2ei=2)=ei(n+1)=2sinn=2sin=2which has real part  = cos((n+ 1)=2)sinn=2sin=2.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 29/40

Triangle inequality

jz+wj2= (z+w)(z+w) = (z+w)(z+w) =zz+zw+wz+ww =jzj2+zw+zw+jwj2 =jzj2+ 2Re(zw) +jwj2  jzj2+ 2jzwj+jwj2 = (jzj+jwj)2Triangle Inequality jz+wj  jzj+jwjThe name comes from the following geometric interpretation. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 30/40

Complex polynomials

Denition

A functionp:C!Cof the form

p(z) =anzn+an1zn1+


+a1z+a0(withcoecientsan;:::;a02C) is called a(complex) polynomial.Thedegreeofp, writtendeg(p), is the highest power with a non-zero coecient.Ifnabove is the degree, thenanis called theleading coecient.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 31/40

The fundamental theorem of algebra

A(complex) rootof a polynomialpis any2Csuch thatp() = 0.Theorem (Gauss)

Every complex polynomial of degree at least one has a root2C.NoteIt does not give any formula for the roots (unlike the quadratic and cubic

formula).About the proofs You will see a proof in your 2nd year complex analysis course.

There is another proof via Galois theory.

Gauss himself gave several proofs, including the following below which requires algebraic topology to make rigorous. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 32/40

Factorising polynomials

Letp;qbe complex polynomials of degree at least 1. Thenqis afactorofpif

there is a polynomialrsuch thatp=qr. We also sayqdividesp.eg.z1 is a factor ofz31 asz31 = (z1)(z2+z+ 1).Theorem (Remainder and Factor)

Letpbe a complex polynomial of degree at least one.The remainder on dividing pbyzisp().In particular,zis a factor ofp(z) if and only ifis a root ofp.Proof.Use the long division algorithm for polynomial division to see that p(z) = (z)q(z) +rfor some polynomialq(z) and remainderrwhich is

constant since its degree must be less that deg(z).Thenp() =rwhich is zero precisely whenis a root or equivalently,zis a

factor. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 33/40

Fundamental theorem of algebra (factor form)

Putting the Factor Theorem and the Fundamental Theorem of Algebra together says that ifpis a polynomial of degreen,then there exists12Csuch that

p(z) = (z1)g1(z), whereg1(z) has degreen1.Ifn11 then there exists22Csuch thatp(z) = (z1)(z2)g2(z).Continuing, you get

Theorem

Any degreencomplex polynomial has a factorisation of the form p(z) = (z1)(z2):::(zn)c

withi;c2C.The terms (zj) are calledlinear factorsofp.This factorisation is unique up to swapping factors around.

E.g.Factorisep(z) =z3+z22 into linear factors.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 34/40

Multiplicity of roots

In an example like

p(z) = (z3)4(zi)2(z+ 1) where the linear factors are not distinct,we say that (z3) is a factor of

multiplicity4, and that 3 is aroot of multiplicity4.Similarly,iis a root of multiplicity 2 and -1 is a root of multiplicity 1.QFind all cubic polynomials which have 2 as a root of multiplicity 3.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 35/40

Proof of uniqueness of factorisation

Consider two factorisations

p(z) = (z1)(z2):::(zn)c= (z1)(z2):::(zn)d:(1)We need to show that we can re-order thei's so that



1=1;:::;n=n;c=d.First notec=dsince they are both the leading

co-ecient ofp.We argue by induction onn. The casen= 0 already has been veried so assume n>0.Substitute inz=1to obtain

0 = (11)(12):::(1n)d:One of the RHS factors, say1i= 0.Swapi;1so1=1.Dividing (1) byz1gives 2 factorisations of

p(z)z1= (z2):::(zn)c= (z2):::(zn)d:By induction, we may assume also2=2;:::;n=n;c=d, so we've won.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 36/40

Example: factorisation

E.gWritep(z) =z4+ 1 as a product of linear factors.N.B.Here, the complex roots occur in complex conjugate pairs. This is general

phenomena forrealpolynomials.Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 37/40

Roots of real polynomials

A polynomial isrealif the co-ecients are real.Theorem Suppose thatis a root of a real polynomialp. Thenis also a root ofp.Proof. Note that in such a case (z) and (z) are both factors.If =2R, then unique factorisation =)p(z) has a quadratic factor (z)(z) =z2(+)z+ =z2(2Re)z+jj2:which is a real quadratic. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 38/40

Factorising real polynomials

We say a real polynomialpisirreducible over the realsif it can't be factored into

a product of two real polynomials of positive degree.E.g.z23z+ 2 is not irreducible butz2+ 1 is.Why?

UpshotA real quadratic polynomial is irreducible overRi it has non-real roots.Using the the fundamental thm of algebra and the previous slide (and our old

inductive argument) we seeTheorem Any real polynomial can be factored into a product of real linear and real irreducible quadratic polynomials. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 39/40

Example: factorisation of a real polynomial

QFactorisep(z) =z4+ 1 into real irreducible factors.Method 1Just try your luck with factorisation facts you know

z

4+ 1 = (z2+ 1)22z2

= (z2+p2z+ 1)(z2p2z+ 1)ORMethod 2Recall complex linear factorisation z

4+ 1 = (zei=4)(zei=4)(ze3i=4)(ze3i=4)RemarkThis is the rst instance of common technique in mathematics, to answer

a question involving real numbers, rst answer it over the complex numbers and deduce your result accordingly. Daniel Chan (UNSW)Chapter 3: Complex NumbersTerm 1 2020 40/40