MSO202A:IntroductionToComplexAnalysisInstructor:Dr.G.P.KapoorFB565,DepartmentofMathematicsandStatisticsTel.7609,Email:gp@iitk.ac.in

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Lecture1

TextBooks: E.Kreyszig,AdvancedEngineeringMathematics,8thEd.,JohnWiley&Sons.

RuelV.Churchill,etal:ComplexVariablesandApplications,McGrawHill.

JohnB.Conway:FunctionsofOneComplexVariable,IIEd.,SpringerInternationalStudentAddition.

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ReferenceBooks:

JanG.Krzyz:ProblemsinComplexVariableTheory,AmericanElsevierPublishingCompany.

LarsV.Ahlfors:ComplexAnalysis,McGrawHill.

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SupplementaryCourseMaterial

LectureNotes,AssignmentsandCoursePlanwillbeavailableonthiscourseatthewebpagehttp://home.iitk.ac.in/~gpthroughthelinkMSO202A.In the lecturenotes, someproofsaremarked (*).Suchproofswillnotbeaskedintheexams.TutorialClasses

The assignment problems marked (T) on the assignmentsheetswillbediscussedinthetutorialclasses.

The solutions/hints to the assignment problemsmarked (D)willbemadeavailableonthecourseweb‐site.

Theexercisesgiveninthetextbooksareusuallynotdiscussedin the tutorial classesand the studentsareexpected to solvethese problems on their own. However, the students canapproach the tutor if theyhaveanydifficulty in solving suchproblems.

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Evaluationplan

There will be 2 pre‐announced Quizzes of 40‐minutesdurationandaweightageof20%marksforeach.

TheEnd‐CourseExaminationwillbeof2‐hoursdurationwithaweightageof60%marks.

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ReviewofComplexNumberSystemComplex numbers were introduced to have solutions ofequationslike 2 1 0x whichdonotpossessasolutionintherealnumbersystem.Acomplexnumber z isanorderedpair( , )x y ofrealnumbers.If 1 1 1 2 2 2( , ), ( , )z x y z x y , the elementary operations aredefinedas

1 2 1 2 1 2( , )z z x x y y

1 2 1 2 1 2,z z if x x y y

1 1 1( , )z x y

1 2 1 2 1 2 1 2 2 1( , )z z x x y y x y x y

( , )z x y , 2 2z x y

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Notations:

Throughout in the sequel, denote Complexnumber( ,0) , (0,1)a a i . With these notations, RC,where R is set of all real numbers and C is set of allcomplexnumbers.

TheEuclideandistancebetweenanytwopoints 1 2,z z Cisdefinedas 1 2z z andissometimesdenotedby 1 2( , )d z z .

Notethat 2,| | 1, 1.i i i i Thus,thecomplexnumberi isthesolutionoftheequation 2 1 0x .Further,writing ( ,0), ( ,0), (0,1)x x y y i ,itiseasilyseenbyusing the definitions of addition and product of complexnumbers that x + i y = z. This is called the cartesianrepresentation of the complex number z.

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Proposition1. 2

zz z

Proof: 22 2 2 2( , ).( , ) ( ,0)x y x y x y x y z

Proposition2.2 2 2 2

1( , ), 0

x yif z

z x y x y

Proof:2 2 2 2 2

1( , )

z z x y

z z x y x yz

.

Proposition3.Re ,Im2 2

z z z zz z

i

Proof.Wegiveheretheproofofthesecondpartoftheproposition.Thefirstpartfollowssimilarly. (0,2 )(0, 1) (2 ,0)

Im2 2 2

z z i y yy z

.

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PolarrepresentationofComplexNumbers

With cos , sin , (cos sin ) ( cos , sin )x r y r z r i r r iscalledthepolarrepresentationofthecomplexnumberz.

,r z 0

arg

angle between the line segment from to z and

positive real axis

z

Itfollowsimmediatelythat 1 2 1 2 1 2 2z z r r and k .Further,if 1 1 1 1 2 2 2 2(cos sin ), (cos sin )z r i z r i ,itfollowsthat

1 2 1 2 1 2 1 2(cos( ) sin( ))z z r r i .Thus, 1 2 1 2arg( ) arg( ) arg( )z z z z .

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Similarly,usinginduction,if(cos sin ), 1,2,..., ,j j j jz r i j n then

1 2 1 2 1 1... ... (cos( ... ) sin( ... ))n n n nz z z r r r i Thus, 1 2 1 2arg( ... ) arg arg ... argn nz z z z z z .

Inparticular, (cos sin ), 0.n nz r n i n n Toprovethisidentityforn<0,wehave1 1 cos sin 1

(cos sin( )).(cos sin )

ii

z r i r r

Sothat,forn<0,

1 1( ) [ (cos sin( )]

[cos sin ].

n n n

n

z z ir

r n i n

Thus, [cos sin ]n nz r n i n forallintegersn.Takingr=1inthisidentity,(cos sin ) cos sin ,ni n i n for all integersn whichiscalledDe‐Moivre’sTheorem.

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Aspecialwordaboutargumentofacomplexnumberargzisnotafunction,sincefor ,iz re arg z has all the values

, 2 , 4 ,... soitisnotsinglevalued.Theidentity 1 2 1 2arg( ) arg argz z z z hastobeinterpretedinthesensethatforsomevalueofargonLHS, suitable values of 1arg z and 2arg z on RHS so thatequalityholds.Conversely, forgivenvaluesof 1 2arg argz and z on RHS, suitable values of 1 2arg( )z z on RHS so thatequalityholds.For example, if 1 2z z i and the values of their arguments

are given as 1 23 3

arg , arg2 2

z z

, then 1 2 1z z andout of

all the values 3 2 , 0,1,2,...k k . of 1 2arg( )z z , we mustchoose 1 2arg( ) 3z z ,sothat 1 2 1 2arg arg arg( )z z z z holds.

1 2

3arg arg

2z z

i

1

1 2arg( ) 3z z

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Conversely,if 1 2arg( ) 5z z isgiven,thenwecantake

1 23 11 3

arg 2 , arg 42 2 2 2

z z .

Tomakeargzafunctionofzinthestrictsenseofthedefinitionofafunction,werestricttherangeofargzas ( , ] (orwithanother convention, some authors restrict this range as[0,2 ) ). Once the range of arg z is so restricted, arg z isdenotedbyArgz.Thus, Arg z (or,0 2Arg z withtheotherconvention).

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Remark:If z x iy ,theprincipalvalueof 1tany

x ,denoedas

1Tany

x ,satisfies 1

2 2

yTan

x

,while Arg z .The

relationbetweenArg z and 1 yTan

x isthereforegivenbythe

following:

1

1

1

, 0

, 0, 0

, 0, 0

yTan if x

xy

Arg z Tan if x yxy

Tan if x yx

Graphof 1Tan x

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Remark:Notethat,ingeneral, 1 2 1 2( )Arg z z Arg z Arg z (1)Forexample,withtheconvention Arg z

if 1 2 1 21, , ,2

z z i then Arg z Arg z ,sothat

1 23

2Arg z Arg z

.

But, 1 2( )2

Arg z z

.This illustrates (1),when theconvention

is Arg z .Similarly,withtheconvention0 2Arg z ,if

1 2 1 23

1, , ,2

z z i then Arg z Arg z ,sothat

1 25

2Arg z Arg z

.

But, 1 2( )2

Arg z z

.Thisillustrates(1),whentheconventionis

0 2Arg z .

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Solutionoftheequation nz c :Let

0 0 0(cos sin )

(cos sin )

c r i

and

z r i

Theequation nz c gives 0 0 0(cos sin ) (cos( 2 ) sin( 2 ))nr n i n r k i k

0 0

1/ 00

2

2, 0,1,..., 1

n

n

r r and n k

kr r and k n

n

Therefore,thensolutionsoftheequation nz c are

1/ 0 00

2 2[cos( ) sin( )], 0,1,..., 1n

kk k

z r i k nn n

.

Since, 1/

0n

kz r , all the roots of nz c lie on the circle1/ 1/

0 0(0, ) { : }.n nC r z z r Further, since the angles

0 2, 0,1,..., 1

kk n

n

,dividethiscircleintonequalsectors,

alltheserootsareequispacedon 1/0(0, ).nC r

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Example:Alltherootsof 1nz ,callednthrootsofunity,canbewrittenas

2 2 2( 1) 2( 1)cos0 sin0, cos sin ,...,cos sin

n ni i i

n n n n

or

2 1 2 21, , ,..., ; cos sin .n

n n n nw w where w in n

Notethatif 0z isanyrootoftheequation

nz c ,thenalltherootsofthisequationaregivenby 2 1

0 0 0 0, , ,..., nn n nz z z w z w

since, 1/ 0 02 2[cos( ) sin( )], 0,1,..., 1

nk

k kz c i k n

n n

givesthat

1/ 0 02 2 2 2

[cos( ) sin( )],

0 1

nlk n

k l k lz c i

n nl n

,

whosedistinctvaluesareobtainedfor 1.k k l k n

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VectorRepresentationofComplexNumbersAny complex number ( , )z x y can be represented as thevector , ( )z x i y j r say

.

Thisrepresentationhelpsingeometricallyvisualizingadditionandsubtractionofcomplexnumbersasvectors.However,itdoesnothelpinvisualizingtheproductofcomplexnumbers as this is different from the vector product ofcorrespondingvectors.

1 1 1 1 2 2 2 2

1 2 1 2 1 2 1 2 2 1

1 2

1 2 1 1 1 2 1 2 1 2

2 2

( ,

( , )

, , ,

0 ( ) ( )

0

, if z x i y j r and z x i y j r then

z z x x y y x y x y is in xy plane itself

while for the corresponding vectors r r

i j k

r r x y x y y x k z z k

x y

since

.)is perpendicular to xy plane

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RepresentationofPoints,CurvesandRegionsbyComplexNumbersRepresentationofPoints

z

-

r

1/r 1/z

z

-z z

z

1z

1 2z z

2z

ziz

90

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EquationofacircleanddiskintermsofComplexNumbersEquationofacirclewithcenter 0z andradiusr :

0z z r Equationofanopendiskwithcenter 0z andradiusr :

0z z r Equationofacloseddiskwithcenter 0z andradiusr :

0z z r EquationofaLineintermsofComplexNumbersEquationofalineLpassingthrougha

andparalleltovectorb

is r a t b

, t

or,intermsofnotationofacomplexvariablesz,aandb,thisequationis z a t b

z a

tb

Im( ) 0

z a

b

.

Thus,equationofthelineLisgivenby

:Im( ) 0z a

L zb

.

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*AlgebraicStructureofComplexNumbersField:( , , .)X isafieldif

(i) (X,+)isanabeliangroup.(ii) (X–{0},.)isanabeliangroup.(iii) ‘.’isdistributiveover‘+’.

Itiseasilyverifiedthat(C,+,.)isafieldthatcontainsthefield(R,+,.).OrderedSet:(X,<),where,‘<’isarelation,iscalledanorderedsetif

(i) Oneandonlyoneofthestatementsx<y,x=y,y<xholdsforanyxandy.

(ii) ‘<’istransitive.OrderedField:AnorderedsetXiscalledanorderedfieldif

(i) Xisafield(ii) Xisanorderedset(iii) Ify<z,thenx+y<x+zforallx,yandzX(iv) Ifx>0,y>0,thenxy>0.

Itiseasilyverifiedthat(C,+,.)isafieldaswellasanorderedsetwithrespecttodictionaryordering(dictionaryorderon isdefinedby

1 1 2 2 1 2 1 2 1 2( , ) ( , )x y x y if either x x or if x x then y y .

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However, (C, +, .) is not an ordered fieldwith any order,sinceineveryorderedfield1isalwayspositive(for,either1ispositiveor‐1ispositiveand,if‐1ispositive,then(‐1)(‐1)=1ispositive,whichisacontraction),sothat‐1=(‐1,0)isalwaysnegative.Now,either(0,1)>0or‐(0,1)>0

If(0,1) 0 (0,1).(0,1) ( 1,0)then (0,0) , whichimplies(R,+,.)cannotbeanorderedfield.If (0,1) 0 (0,1). (0,1) ( 1,0)then (0,0) ,whichimplies( ,+,.)cannotbeanorderedfield.

Alternatively,ineveryorderedfield,squareofeveryelementispositive. This gives ‐1 is positive being square of (0,1), acontradiction since ‐1 is always negative as in the abovearguments.